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The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4, Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 20.06.2019, and it has been decided to implement it from the educational year 2019-20., , Mathematics and Statistics, (Arts and Science), Part - I, STANDARD - XI, , 2019, Maharashtra State Bureau of Textbook Production and Curriculum Research,, Pune - 411 004, , Download DIKSHA App on your smartphone. If you scan the Q.R.Code, on this page of your textbook, you will be able to access full text and the, audio-visual study material relevant to each lesson provided as teaching, and learning aids.

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PREFACE, Dear Students,, Welcome to the eleventh standard!, You have successfully completed your secondary education and have entered the, higher secondary level. You will now need to learn certain mathematical concepts and acquire, some statistical skills to add more applicability to your work. Maharashtra State Bureau of, Textbook Production and Curriculum Research has modified and restructured the curriculum, in Mathematics in accordance with changing needs., The curriculum of Mathematics is divided in two parts. Part 1 covers topics in, Trignometry, Algebra, Co-ordinate Geometry and Statistics. Part 2 covers Complex Numbers,, Sets and Relations, Calculus and Combinatorics. There is a special emphasis on applications., Activities are added in every chapter for creative thinking., Some material will be made available on E-balbharati website (ebalbharati.in). It, contains a list of specimen practical problems on each chapter. Students should complete the, practical exercises under the guidance of their teachers. Maintain a journal and take teacher’s, signature on every completed practical., You are encouraged to use modern technology in your studies. Explore the Internet for, more recent information and problems on topics in the curriculum. You will enjoy learning if, you study the textbook thoroughly and manage to solve problems., On the title page Q.R. code is given. It will help you to get more knowledge and clarity, about the contents., This textbook is prepared by Mathematics Subject Committee and members of study, group. This book has also been reviewed by senior teachers and subject experts. The Bureau, is grateful to all of them and would like to thank for their contribution in the form of creative, writing, constructive criticism and valuable suggestions in making this book useful to you and, helpful to your teachers., The Bureau hopes that the textbook will be received well by all stakeholders in the right, spirit., You are now ready to study. Best wishes for a happy learning experience., , Pune, Date : 20 June 2019, Indian Solar Date : 30 Jyeshtha 1941, , (Dr. Sunil Magar), Director, Maharashtra State Bureau of Textbook, Production and Curriculum Research, Pune.

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XI Mathematics and Statistics (Part I) for, Arts and Science, Sr., No, , 1, , 2, , 3, , Area, , Angle and Its, measurement, , Topic, , Angle, , The student will be able to­ • understand angle of any measure., • understand different systems of, measurement of angle and relations, between them., • convert an angle from one system to, the other., , Trigonometric Trigonometric, Functions, Functions, , • understand definitions of trigonometric, functions of angle of any measure., • find the values of trigonometric, functions of certain angles., • draw graphs of trigonometric functions., , Trigonometric Functions, of compound, angles and, fractorization, formulae, , • find the trigonometric functions of sum, or difference of the angles., • find the trigonometric functions of, multiple and sub-multiple angles., • express the sum or difference of two, trigonometric functions as product, • learn some rules of trigonometric, ratios of angles of a triangle., , Trigonometric, Functions of, compound, angles, , Determinant, , 4, , Competency Statement, , Determinants, and Matrices, Matrices, , • find value of a determinant., • reduce determinant to simple form., • solve linear equations in 2 or 3, variables, • find area of triangle using, determinants., • understand types of matrices., • Perform algebraic operations of the, matrices.

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5, , 6, , 7, , 8, , 9, , Straight Line, , Straight Line, , • understand locus and its equation., • find equation of a straight line in, different forms., • find angle between given two straight, lines and the distance of a point from, given line., , Circle, , • find equation of circle satisfying given, conditions., • learn and use the properties of circle., • find the equation of tangent to the, circle., , Parabola,, Conic Section Ellipse,, Hyperbola, , • find the equations of conic sections, satisfying given conditions., • learn and use the properties of conics., • find the equation of tangent to the, conic., , Circle, , Measures of, dispersion, , Probability, , Measures of, dispersion, , • calculate range, standard deviation, and variance from given data., , Probability, , • calculate probability of an event and, learn conditional probability, • learn and use Baye’s theorem and its, applications, , Note :- Extra examples for competitive section and practice are given on, e-balbharti. The activities which can be conducted as a part of, practicals are also mentioned in pdf form on our website.

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INDEX, , Sr. No., , Chapter, , Page No., , 1, , Angle and its Measurement, , 1, , 2, , Trigonometry - I, , 14, , 3, , Trigonometry - II, , 35, , 4, , Determinants and Matrices, , 59, , 5, , Straight Line, , 103, , 6, , Circle, , 127, , 7, , Conic Sections, , 140, , 8, , Measures of Dispersion, , 179, , 9, , Probability, , 193, , Answers, , 216

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1, , Angle and its measurement, Now we will differentiate between such, angles., , Let's Study, •, •, •, •, •, , Directed angle., Angles of different measurements, Units of measure of an angle, Length of an arc of a circle., Area of a sector of a circle., , Let's Learn, 1.1 Directed Angles:, Consider the, ray OA. Rotate, it about O till it, takes the position, OB as shown in, Fig. 1.3. Then, angle so obtained, due to the rotation, is called directed, angle AOB. We, Fig. 1.3, define the notion, of directed angle as follows:, , Let's Recall, •, •, , We know how to draw the acute angles, of different measures., In a circle we can find arc length and area, of a sector in terms of the central angle, and the radius., , Definition:, The ordered pair of rays (OA,OB ) together, with the rotation of the ray OA to the position, of the ray OB is called the directed angle, ∠AOB., If the rotation of the initial ray is, anticlockwise then the measure of directed, angle is considered as positive and if it is, clockwise then the measure of directed angle, is considered as negative. In the ordered pair, (OA,OB ), the ray OA is called the initial arm, and the ray OB is called the terminal arm., , Activity No. 1, Draw the angle ABC of measure 40°, , Fig. 1.1, , Fig. 1.2, In the Fig.s 1.1 and 1.2 both the angles are, of 40°. But one is measured in anticlockwise, direction and the other is measured in clockwise, direction., , Fig. 1.3(a), 1, , Fig. 1.3(b)

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and, half of one rotation angle is straight, angle., , O is called the vertex as shown in fig 1.3(a), and 1.3(b)., Observe Fig 1.3(b) and note that, , Right angle:, , (OA,OB ) ≠ (OB ,OA), , One fourth of one rotation angle is called, as one right angle, it is also half of a straight, angle. One rotation angle is four right angles., , ∠ AOB ≠ ∠ BOA, ∠ AOB ≠ ∠ BOA even though they have, same amount of rotation., Zero angle:, If the ray OA has zero rotation, that is, it does not rotate, the initial arm itself is a, terminal arm OB, the angle so formed is zero, angle., , Fig. 1.4, One rotation angle:, , Fig. 1.7, , After one complete rotation if the initial, ray OA coincides with the terminal ray OB, then so formed angle is known as one rotation, angle m∠ AOB = 360°., , Angles in Standard position :, In the rectangular co-ordinate system, a, directed angle with its vertex at origin O and, the initial ray along the positive X-axis, is, called angle in standard position., , Fig. 1.5, Straight angle:, After the rotation, if the initial ray OA and, the terminal ray OB are in opposite directions, then directed angle so formed is known as, straight angle (fig. 1.4)., , Fig. 1.8, In adjoint Fig. 1.8, ∠XOP, ∠XOQ and, ∠XOR are in standard positions. But, ∠POQ, is not in standard position., , Fig. 1.6, Note that in this case AOB is a straight, line., 2

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Angle in a Quadrant:, A directed angle in standard position, is said to be in a particular quadrant if its, terminal ray lies in that quadrant., , Fig. 1.10 (b), , In Fig. 1.8, directed angles ∠XOP,, ∠XOQ and ∠XOR lie in first, second and, third quadrants respectively., , In Fig. 1.10(a), the directed angles having, measure 30°, 390°, −330° have the same initial, arm, ray OA and the same terminal arm, ray, OB. Hence, these angles are co-terminal angles., , Quadrantal Angles:, A directed angle in standard position, whose terminal ray lies along X-axis or Y-axis, is called a quadrantal angle., , If the two directed angles are co-terminal, angles then difference between measures of, these two directed angles is an integral multiple, of 360° e.g. in figure 1.10(a), 390° − (−330)°, = 720° = 2 × 360°., 1.1.1, , Measures of angles:, , The amount of rotation from the initial ray, OA to the terminal ray OB gives the measure, of angle AOB. It is measured in two systems., , Fig. 1.9, In Fig. 1.9, ∠XOP, ∠XOQ, ∠XOR and, ∠XOS are all quadrantal angles., , x, , Co-terminal angles:, , Fig. 1.11, , Directed angles of different amount of, rotation having the same positions of, initial, rays and terminal rays are called co-terminal, angles., , 1), 2), , Sexagesimal system (Degree measure), Circular system (Radian measure), , 1.1.2 Sexagesimal System (Degree Measure):, In this system, the unit of measurement, of angle is a degree., One rotation angle is divided into 360, equal parts, the measure of each part is called, as one degreee angle., ∴, , Fig. 1.10 (a), , 1, 360, , th, , part of one complete rotation, , is called one degree and is denoted by 1°., 3

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Theorem :, , 1 th part of one degree is called one, 60, minute and is denoted by 1'., , The radian so defined is independent of the, radius of the circle used and πc = 1800., Proof: Let us consider a circle with, centre at O and radius r. Let AB be an arc of, length r. Join OA and OB. Then ∠AOB = 1c., Produce AO to meet the circle at C., , 1 th part of one minute is called one, 60, second and is denoted by 1''., , , 1° = 60', 1' = 60'', , m ∠ (one rotation angle) = 360°, m ∠ (straight angle) = 180°, m ∠ (right angle) = 90°, 1.1.3 Circular System (Radian Measure):, In this system, the unit of measurement, of an angle is a radian., Fig. 1.13, , Let r be the radius of a circle with centre, O. Let A and B be two points on circle such, that the length of arc AB is r. Then the, measure of the central angle AOB is defined, as 1 radian. It is denoted by 1c., , Clearly, ∠AOC = a straight angle, , , = 2 right angles, , Since measures of the angles at the centre, of a circle are proportional to the lengths of, the corresponding arcs subtending them:, =, , ∴ , , l (arcAB), l (arcABC), , =, , ∴, , m∠AOB =, , ∴ 1c = m∠AOB =, , Fig. 1.12, , =, , . m∠AOC, , (2 right angles), ,, π, , a constant independent of r., , Thus, one radian is the measure of an, angle subtended at the centre of a circle by, an arc whose length is equal to the radius of, the circle., , Hence one radian is well defined., Also, πc = 2 right angles = 1800., Hence, a radian is a constant angle and two, right angles = 180° = πc, 4

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Activity 2 : Verify the above result by taking, the circles having different radii., , v), , Let an angle have its measure r in radian, and θ in degrees. Then its proportion with the, straight angle is the same in either measure., \, , θ, r, =, p, 180, , p, 180, , \ rc = θ° ×, , Relation between angle and time in a, clock. (R is rotation.), Min Hand, , Hr Hand, , 1R = 360°, , 1R = 360°, , 1R = 60 min, , 1R = 12 Hrs, , 60 min = 360°, , 12 Hrs = 360°, , To convert degree measure into radian, p, measure, multiply degree measure by, ., 180, , 1min = 6° rotation 1 Hr = 30°, , 1 Hr = 601, , 60 min = 30°, 1°, , 1 min =, 2, The word ‘minute’ is used for time, measurement as well as 60th part of degree, of angle., , ii) To convert radian measure into degree, 180, measure, multiply radian measure by, ., p, , vi) Please note that “minute” in time and, “minute” as a fraction of degree angle are, different., , We use this relation to convert radian, measure into degree and vice-versa., Notes:, i), , iii) Taking π = 3.14,, , SOLVED EXAMPLES, ,  180 °, c, we have 1 = , ,  π , , Ex. 1) Convert the following degree measures, in the radian measures.,  1 °, i) 70°, ii) -120°, iii)  , 4, , = 57.3248°, Here fractional degree is given in decimal, fraction. It can be converted into minutes, and seconds as follows, , Solution : We know that θ° = θ ×, , 0.3248° = (0.3248 × 60)', , , = 19.488', , , , = 19' + (.488 × 60)'', , , , ≈ 19' 29'', , i) 70°, ∴ 70°, , 7π, , =, , ii) -120°, , Degree, , 15 30 45 60 90 120 180 270 360, , Radian, , π π, 12 6, , π, 4, , π, 3, , π 2π, 2 3, , π, , 3π, 2, , iii), , 2π, , 1, 4, , ∴, 5, , 0, , =, 1, 4, , 180, , c, , = – 120 ×, , ∴ -120°, , 2π, , = -, , 3, , π, 1, ×, 4, 180, 0, , =, , 180, , c, , 18, , Thus, 1c = 57° 19' 29'', iv) In the table given below, certain degree, measures are expressed in terms of radians., , π, , = 70 ×, , π, , π, 720, , c, , c, , c, , π, 180, , c, , c

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Ex. 2) Convert the following radian measures, in the degree measures., i), , c, , 7π, , -π, , ii), , 3, , c, , iii), , 18, , 4, 7, , i) , , =, , 3, , 3, , = 420°, , 3, =, , 18, -π, , , ∴, , -π, , c, , -π, , 180 °, ×, π, , c, , 7π, , , ∴, , ii) , , 7π, , 18, , c, , ×, , = 74°52'12'', , −30.6947° = −[30°+0.6947°], , , , = −[30°+(0.6947×60)'], , , , = −[30°+41.682'], , , , = −[30°+41'(0.682×60)''], , , , = −[30°41'40.92''], , , , = −30°41'41'' approximately, , Ex. 4) The measures of the angles of the, triangle are in A. P. The smallest angle, is 40. Find the angles of the triangle in, degree and in radians., Solution : Let the angles of the triangle be, a − d, a, a + d in degrees., , 180 °, π, , = −10°, , 18, , , ii), , 180 °, Solution : We know that θ = θ ×, π, c, , = 74°52'+(0.2×60)'', , c, , c, , 7π, , , , ∴ a − d + a + a + d = 180°, , Note that,, 180° = πc, , , , ∴ 3a = 180°, , , , ∴ a = 60°, , Also, smallest angle, , Hence,, c, ,  π , 1° = ,  ,,  180 , ,  180 , 1c = , ,  π , , = 40°, , °, , ∴ a - d = 40°, ∴ 60° - d = 40°, , iii), , ∴, , 4, 7, , c, , 4, 7, , =, 4, 7, , c, , =, , ×, , 180 °, π, , 720 °, =, 7π, , ∴ 60° - 40° = d, ∴ d = 20°, , 360 °, 11, , Now, a + d = 60° + 20° = 80°, Hence the angles are 400, 600, 800, , Ex. 3) Express the following angles in degrees,, minutes and seconds., i) 74.87°, , 40, =, 180, q1, 2πc, πc, 60, so, that, q, =, q, =, ×, π, =, 1, 2, π, 9, 3, 180, 4, 80, =, π = 9 πc, 180, , if they are q1c, q2c, q3c, 400 = q1c, then, , ii) −30.6947°, , Solution:, i), , 74.87°, , = 74°+0.87°, , , , = 74°+(0.87×60)', , , , = 74°+(52.2)', , , , = 74°52'+0.2', , Hence the angles are, , 6, , 2πc πc, 4πc, ,, and, ., 9, 3, 9

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The angles of a triangle in degrees are, , ∴ Sum of the remainging three angles is, 360° − 40° = 320°, Since these threee angles are in the ratio, 3:5:8., ∴ Degree measures of these angles are 3k,, 5k, 8k, where k is constant., ∴ 3k + 5k + 8k = 320°, ∴ 16k = 320°, ∴ k = 20°, ∴ The measures of three angles are, , (3k)° = (3 × 20)° = 60°, , (5k)° = (5 × 20)° = 100°, and , (8k)° = (8 × 20)° = 160°, , 2π π, 40°, 60° and 80° and in radians, ,, and, 3, 9, c, 4π, c, , c, , 9, Ex. 5) The difference between two acute angles, c, 7π, of a right angled triangle is, ., 30, Find the angles of the triangle in, degrees., Solution : Let x and y be the acute angles of a, triangle in degrees., 7π, Here, x - y =, 30, , c, , =, , 7π, 180 °, ×, π, 30, , Ex. 7) Find the number of sides of a regular, polygon if each of its interior angle is, 4π c, ., 5, , , = 42°, ∴ x - y = 42° .......... (I), , Solution:, Let the number of sides be ‘n’., 4π c, each interior angle =, 5, c, 180 °, 4π, =, ×, = 144°, π, 5, , The triangle is right angled., ∴ x + y = 90° .......... (II), adding, (I) + (II),, we get x - y + x + y = 42° + 90°, ∴ 2x = 132°, , Exterior angle = 180° − 144° = 36°, °, ∴ 360 = 36°, n, , ∴ x = 66° , Put in (I), 66° - y = 42°, , ∴ 66° - 42° = y, , 360, 36, ∴ n = 10, ∴ number of sides of the regular polygon, is 10., ∴ n =, , ∴ y = 24°, ∴ The angles of a triangle are 66°, 90°, and 24°., 2π, 9, radian and the measures of the other, three angles are in the ratio 3:5:8, find, their measures in degree., Solution : The sun of angles of a quadrilateral, is 360°., c, 2π, One of the angles is given to be, =, 9, 0, 2π, 180, ×, = 40°, π, 9, Ex. 6) One angle of a quadrilateral is, , Ex. 8) Find the angle between hour hand and, minute hand of a clock at, i) Quarter past five, ii) Quarter to twelve, Solution :, 1) When a hour hand moves from one clock, mark to the next one, it turns through an, 360°, = 30º., angle of, 12, 7

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B), , At quarter past 5,, miniute hand is pointing, to 3. Hour hand has, gone past 5. So the, angle between them is, more than 60º. In one, minute hour hand turns, ο, 1, hence in, through, 2, 15 minutes it has turned, , Fig. 1.14, , i) –140°, ix) –500°, , 0c, , πc, , v) 75°30′ , , vi) 40°48′, , 7π c, 12, , -5π c, ii) 3, v), , -1, 4, , iii) 5c, c, , 7π c ,, 36, m∠ B = 120°, find m∠ C in degree and, radian., , Q.5 In ∆ ABC, if m∠ A =, , In, 0° 30° 45° 60° 90° 180° 270° 360, degrees, In, radians, , iv) 65°30′, , Q.4 Express the following angles in degree,, minute and second., 1 c, i) (183.7)°, ii) (245.33)0 iii), 5, , Note:, , πc, 4, , iii) −132° , , 11π c, iv), 18, , ∴ the angle between the hands is equal to, 90º − 7.5º = 82.5º., , π, 2, , ii) 250°, , i), , ii) At quarter to twelve, minute hand is pointing, to 9, hour hand is between 11 and 12 though, it is nearer 12. It will take 15 minutes i.e., 7.5º to reach 12., , π, 3, , i) 85° , , Q.3 Convert the following angles in degree., , Thus the angle between, the hands is equal to, 60º + 7.5º = 67.5º., , πc, 4, , x) –820°, , Q.2 Convert the following angles in to radian., , ο, , πc, 6, , ii) 250° iii) 420° iv) 750°, , v) 945° vi) 1120° vii) –80° viii) –330°, ,  15 , through   = 7.5º.,  2, , Fig. 1.15, , Draw the angles of the following measures, and determine their quadrants., , 5π c, and, Q.6 Two angles of a triangle are, 9, c, 5π, . Find the degree and radian measure, 18, , 2πc, , of third angle., EXERCISE 1.1, , Q.7 In a right angled triangle, the acute angles, are in the ratio 4:5. Find the angles of, the triangle in degree and radian., , Q.1 A) Determine which of the following pairs, of angles are co-terminal., i) 210°, −150°, , ii) 360°, −30°, , iii) −180°, 540°, , iv) −405°, 675°, , v) 860°, 580°, , vi) 900°, −900°, , Q.8 The sum of two angles is 5πc and their, difference is 60°. Find their measures in, degree., Q.9 The measures of the angles of a triangle, are in the ratio 3:7:8. Find their measures, in degree and radian., 8

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Q.10 The measures of the angles of a, triangle are in A.P. and the greatest is, 5 times the smallest (least). Find the, angles in degree and radian., , The area A of a sector is in the proportion, of its central angle θ., If the central angle θ is in radian,, , θ, A, =, 2π Area of the circle, , Q.11 In a cyclic quadrilateral two adjacent, c, angles are 40° and π . Find the angles, 3, of the quadrilateral in degree., , ∴, , Q12 One angle of a quadrilateral has measure, 2π c, and the measures of other three, 5, angles are in the ratio 2:3:4. Find their, measures in degree and radian., , ∴ A=, , ii) Hexagon, , iii) Septagon , , iv) Octagon, , 1, θ r2, =, r 2θ, 2, 2, , The arc length S of a sector is in the, proporation of its central angle. If the central, angle is θ radians., , Q.13 Find the degree and radian measure of, exterior and interior angle of a regular, i) Pentagon , , θ, A, = 2, 2π π r, , θ, S, =, 2π circumference of the circle, ∴, , Q.14 Find the angle between hour-hand and, minute-hand in a clock at, , θ, S, =, 2π 2π r, , ∴ S = rθ., , i) ten past eleven, ii) twenty past seven, , SOLVED EXAMPLES, , iii) thirty five past one, iv) quarter to six, v) 2:20 , , , Ex. 1) The diameter of a circle is 14 cm. Find, the length of the arc, subtending an angle of, 54° at the centre., , vi) 10:10, , Let’s Understand, , Solution : Here diameter = 14 cm, , ∴ Radius = r = 7 cm, , 1.2 ARC LENGTH AND AREA OF A, SECTOR:-, , π, θ = 54 × 180, c, , c, , =, , 3π, , c, , 10, , To find s, we know that s = rθ, = 7 ×, , 3π 7 × 3 22 66, ×, =, =, 10, 10, 7 10, , , ∴ arc length = 6.6 cm, Ex. 2) In a circle of radius 12 cms, an arc PQ, subtends an angle of 30° at the centre. Find the, area between the arc PQ and chord PQ., , Fig. 1.16, 9

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Solution :, r, , To find s and A., 1 2, We know that s = rθ and A = r θ, 2, 2π, ∴ s = 15 ×, = 10π and, 3, 1, 2π, A =, × 15 × 15 ×, = 75π, 2, 3, ∴ s = 10π cm and A = 75π sq.cm., , = 12cms,, , θc = 30°, =  30 × π , 180 , , θ =, , π, , c, , c, , 1 2, rθ, 2, , Area of sector OPQ =, , , =, , 1, π, ×12 ×12 ×, 2, 6, , , , =, , 12π sq.cm. ...... (1), , Draw QR ⊥ OP,, , ∴ sin 30° =, , ∴, , QR = 12 ×, , , Area of ∆ OPQ, , , Ex. 4) The perimeter of a sector is equal to, half of the circumference of a circle. Find the, measure of the angle of the sector at the centre, in radian., , Fig. 1.17, , 6, , 1, 2, , Solution : Let r be the radius of a circle., Perimeter of a sector = half of the, circumference, 1, ∴ l(OA) + l(OB) + l(arc APB) = (2πr), 2, ∴ r + r + r.θ, = πr, , QR, 12, , = 6 cms, = Height of ∆ OPQ, 1, =, × base × height, 2, =, , 1, 2, , 2r + r.θ, , = πr, , ∴ 2 + θ = π, ∴ θ = (π − 2)c, , × 12 × 6, , Ex. 5) A pendulum of, Fig. 1.18, length 21cm oscillates, through an angle of, 36°. Find the length of its path., , , = 36 sq.cm, ........(2), By (1) and (2),, Required Area = A(Sector OPQ)−A (∆OPQ), , = (12π − 36)sq.cm., , = 12(π − 3)sq.cm, , Solution : Here r = 21 cm, π, p c, θ = 36° = 36 × 180 =, 5, Length of its path =, , Ex. 3) The area of a circle is 225π sq. cm., Find the length of its arc subtending an angle, of 120° at the centre. Also find the area of, the corresponding sector., , c, , l(arc AXB), = s = rθ, π, = 21 ×, 5, , Solution : Let ‘r’ be the radius of a circle, whose, area is 225π sq. cm., ∴ πr2 = 225π, ∴ r2 = 225, ∴ r = 15 cm., , , , =, , 21 22 66, ×, =, 5 7, 5, , Length of path =, 13.2 cm, , Fig. 1.19, , Ex. 6) ABCDEFGH is a regular octgon, inscribed in a circle of radius 9cm. Find the, length of minor arc AB., , c, 2π c, π , , c, θ = 120° = 120 ×,  = 3, 180 , , , 10

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Solution : Here r = 9cm, , (7), , OAB is a sector of the circle having, centre at O and radius 12 cm. If, m∠AOB = 45o, find the difference, between the area of sector OAB and, triangle AOB., , (8), , OPQ is the sector of a circle having, centre at O and radius 15 cm. If, m∠POQ = 30o, find the area enclosed, by arc PQ and chord PQ., , (9), , The perimeter of a sector of the circle, of area 25π sq.cm is 20 cm. Find the, area of the sector., , o, πc, 360, = 45° =, , θ =, 4, 8, , l(minor arc AB), = S,   , , = rθ , π , = 9   cm, 4, , Fig. 1.20, , (10) The perimeter of the sector of the circle, of area 64π sq.cm is 56 cm. Find the, area of the sector., , EXERCISE 1.2, (1), , (2), , (3), , Find the length of an arc of a circle, which subtends an angle of 108o at the, centre, if the radius of the circle is 15, cm., , Let's Remember, , The radius of a circle is 9 cm. Find the, length of an arc of this circle which, cuts off a chord of length, equal to, length of radius., Find the angle in degree subtended at, the centre of a circle by an arc whose, length is 15 cm, if the radius of the, circle is 25 cm., , (4), , A pendulum of length 14 cm oscillates, through an angle of 18o. Find the length, of its path., , (5), , Two arcs of the same lengths subtend, angles of 60o and 75o at the centres of, two circles. What is the ratio of radii of, two circles ?, , (6), , The area of a circle is 25π sq.cm. Find, the length of its arc subtending an angle, of 144o at the centre. Also find the area, of the corresponding sector., , •, , If an angle is r radians and also θ degrees, r, θ, then, =, π, 180°, , •, , π  , 1° = (0.01745)c, , θ = θ ×, ,  180 , , •, ,  180  , 1c = 57°17′48″, θ = θ ×, , π , , , •, , Arc length = s = rθ., , •, , c, , o, , o, , c, , θ is in radians., 1 2, Area of a sector A = r θ , where θ is, 2, in radians., , •, , Two angles are co-terminal if and only, if the difference of their measures is an, integral multiple of 360., , •, , Exterior angle of a regular polygon of n, 360 , sides = , ,  n , , •, , 11, , o, , In one hour, hour’s hand covers 30o and, a minutes hand covers 360o.

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•, , I, 1), , In 1 minute, hour hand turns through, o, 1,   and minute hand turns through 6°., 2, , 7), , MISCELLANEOUS EXERCISE - 1, , 8), , c, , 2), , B) 264o, , 9), C) 224o, , D) 426o, , 156 is equal to, , 17π , A) , ,  15 , , c, , D)  7π , , c, , 13π , B) , ,  15 , , c, , 11π , C) , ,  15 , , A) π, , A horse is tied to a post by a rope. If the, horse moves along a circular path, always, keeping the rope tight and describes 88, meters when it traces the angle of 72o at, the centre, then the length of the rope is, , 5), , 13π, 14, , B), , 14π, 13, , C), , 15π, 14, , D), , 4, , 14π, 15, , 20 meters of wire is available for fancing, off a flower-bed in the form of a circular, sector of radius 5 meters, then the, maximum area (in sq. m.) of the flowerbed is, B) 20, , C) 25, , B) 4:5, , C) 5:4, , D) 3:4, , B) 60o, , C) 54o, , D) 65o, , D) 30, 12, , B) 3+π, , C) 6+π, , D) 6, , Find the number of sides of a regular, polygon if each of its interior angle is, ., , 2), , Two circles, each of radius 7 cm,, intersect each other. The distance, between their centres is 7 2 cm. Find, the area of the portion common to both, the circles., , 3), , ∆ PQR is an equilateral triangle, with side 18 cm. A circle is drawn, on the segment QR as diameter. Find, the length of the arc of this circle within, the triangle., , 4), , Find the radius of the circle in which, central angle of 60o intercepts an arc of, length 37.4 cm., , Angle between hands of a clock when it, shows the time 9.45 is, , A) 15, , π, 9, , 1), , A) (7.5)o B) (12.5)o C) (17.5)o D) (22.5)o, 6), , D), , Answer the following., , If a 14cm long pendulum oscillates, through an angle of 12o, then find the, length of its path., A), , π, 2, , II, , 3π c, , A) 70 m. B) 55 m. C) 40 m. D) 35 m., 4), , C), , 10) The central angle of a sector of circle of, area 9π sq.cm is 600, the perimeter of the, sector is, , c, , , ,  15 , , 3), , π, 6, , Find the measure of the angle between, hour-hand and the minute hand of a clock, at twenty minutes past two., A) 50o, , o, , B), , A semicircle is divided into two sectors, whose angles are in the ratio 4:5. Find, the ratio of their areas?, A) 5:1, , is equal to, , A) 246o, , π, 3, , A), , Select the correct option from the given, alternatives.,  22π , , ,  15 , , If the angles of a triangle are in the ratio, 1:2:3, then the smallest angle in radian is

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5), , A wire of length 10 cm is bent so as to, form an arc of a circle of radius 4 cm., , 9), , A train is running on a circular track of, radius 1 km at the rate of 36 km per, hour. Find the angle to the nearest, minute, through which it will turn in 30, seconds., , 10), , In a circle of diameter 40 cm, the length, of a chord is 20 cm. Find the length of, minor arc of the chord., , 11), , The angles of a quadrilateral are in A.P., and the greatest angle is double the, least. Find angles of the quadrilateral in, radian., , What is the angle subtended at the, centre in degrees ?, 6), , If two arcs of the same length in two, circles subtend angles 65o and 110o at, the centre. Find the ratio of their radii., , 7), , The area of a circle is 81π sq.cm., Find the length of the arc subtending an, angle of 300o at the centre and also the, area of the corresponding sector., , 8), , Show that minute hand of a clock gains, 5o30' on the hour hand in one minute., , 13

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2, , Trigonometry - 1, shall now extend the definitions of trigonometric, ratios to angles of any measure in terms of coordinates of points on the standard circle., , Let's Study, •, , Trigonometric functions with the help of, unit circle, , •, , Extensions of trigonometric functions to any, angle, , •, , Range and Signs of trigonometric functions, in different quadrants, , •, , Fundamental Identities and Periodicity of, trigonometric functions, , sinθ =, , adjacent side, opposite side, , cosθ =, ,, hypoteneous, hypoteneous, , •, , Domain, Range and, trigonometric function, , tanθ =, , opposite side, . (see fig 2.1 (a)), adjacent side, , Graph, , of, , Let's Recall, We have studied that, in a right angled, triangle if measure of an acute angle is 'θ', then, , each, , Also, cosecθ =, , • Polar Co-ordinates, , 1, 1, , secθ =, ,, cosθ, sinθ, cotθ =, , 2.1 Introduction, Trigonometry is a branch of Mathematics that, deals with the relation between sides and angles, of triangles. The word ‘trigonometry’ is derived, from the Greek words 'trigonon' and ‘metron’., It means measuring the sides of triangles. Greek, Mathematicians used trigonometric ratios to, determine unknown distances. The Egyptians, used a primitive form of trigonometry for building, pyramids in the second millennium BC. Greek, astronomer Hipparches (190-120 BC) formulated, the general principles of trigonometry and he is, known as the founder of the trigonometry., , 1, ., tanθ, , Let's Learn, 2.1.1 Trigonometric functions with the help of, a circle:, Trigonometric ratios of any angle, We have studied that in right angled ∆ ABC,, 'q' is an acute angle, , We are familiar, with, trigonometric, ratios of acute angles a, in right angled triangle., We have introduced, the, concept, of, directed angle having, any measure, in the, previous chapter. We, , Fig. 2.1(a), , 14, , Fig. 2.1(b)

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cosq =, , adjacent side BC, =, hypoteneous AC, , Since P lies on the circle, OP = r, , sinq =, , opposite side AB, =, hypoteneous AC, , The definitions of sinθ, cosθ and tanθ can, now be extended for θ = 0° and 90° ≤ θ ≤ 360°., We will also define secθ, cosecθ and cotθ., , ∴, , We will now extend this definition to any, angle q, consider q as directed angle,, , Every angle θ, 0° ≤ θ ≤ 360°, determines a, unique point P on the circle so that OP makes, angle θ with X-axis., , Let 'q' be an acute angle. [See fig. 2.1 (b)], consider a circle of radius 'r' with centre at, origin 'O' of the co-ordinate system., , The pair (x,y) of co-ordinates of P is uniquely, determined by θ. Thus x = rcosθ, y = rsinθ are, functions of θ., , OA is the initial ray of angle q,, OB is its terminal ray., , Note :, 1) If P (x, y) lies, on the unit circle, then cosθ = x, and sinθ = y. ∴, P (x, y) ≡ P(cosθ,, sinθ), 2) The trigonometric, functions do not, depend on the, Fig. 2.2, position of the, point on the terminal arm but they depend, on measure of the angle., , P(x,y) is a point on the circle and on ray OB., Draw PM ⊥er to OA., ∴ OM = x, PM = y and OP = r., using ∆ PMO we get,, cos θ =, , PM y, OM x, = ,, = , sin θ =, OP r, OP r, , then we define, x, x − co-ordinate of P, cos θ =, =, r, Distance of P from origin, sin θ =, , x2 + y 2 = r, , y, y − co-ordinate of P, =, r, Distance of P from origin, , Point P(x,y) is, on the circle, of radius r and, Q (x',y') is on the, unit circle., , and r2 = x2 + y2, Hence, cos2θ + sin2θ = 1, For every angle 'q', there is corresponding, unique point P(x,y) on the circle, which is on the, terminal ray of 'q', so trignometric ratio's of q are, also trignometric functions of 'q'., , Considering, results on similar, triangles., y, y', sinθ = r = 1 ,, , Note that : 1) Trignometric ratios / functions are, independent of radius 'r'., 2) Trignometric ratios of coterminal, angles are same., We consider the circle with center at origin, and radius r. Let P (x,y) be the point on the circle, with m∠MOP = θ, , Fig. 2.3, , ∴ y = r sinθ, y1 = sinθ and, x', x, cosθ = r = 1 , x = r cosθ x1 = cosθ, , 15

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2.1.2 Signs of trigonometric functions in, different quadrants :, Trigonometric functions have positive or, negative values depending on the quadrant, in which the point, P(x, y) lies. Let, us find signs of, trigonometric, ratios in different, quadrants. If the, terminal arm of an, angle θ intersects, the unit circle in, Fig. 2.4, the point P(x, y),, then cosθ = x., y, sinθ =y and tanθ = . The values of x and, x, y are positive or negative depending on the, quadrant in which P lies., 1), , 3), , cosθ = y is negative, sinθ = x is negative, tanθ = y is positive, x, Hence only tanθ is, positive sinθ and cosθ are, negative for θ in the third, quadrant., 4), , Hence only cosθ is, positive; sinθ and tanθ, are negative for θ in the, fourth quadrant., , Fig. 2.8, , You can check sinθ & cosecθ, have the, same sign, cosθ & secθ have the same sign and, simillarly tanθ & cotθ have the same sign, when, they exists., , Fig. 2.5, , Remark: Signs of cosecθ, secθ and cotθ are, same as signs of sinθ, cosθ and tanθ respectively., , π, In the second quadrant ( < θ < π), y is, 2, positive and x is negative, hence, , 2.1.3 Range of cosθ and sinθ : P(x, y) is point, on the unit circle. m∠AOB = θ. OP = 1, , cosθ = y is positive, , ∴ x2 + y2 = 1, , sinθ = x is is negative, tanθ = y is negative, x, Hence only sinθ is, positive, cosθ and tanθ, are negative for θ in the, second quadrant., , 3π, < θ < 2π), x is, 2, positive and y is negative, hence, , In the fourth quadrant (, , cosθ = x is positive, tanθ = y is negative, x, , cosθ = x is positive, , 2), , Fig. 2.7, , sinθ = y is negative, , π, In the first quadrant (0 < θ <, ), both x and, 2, y are positive, hence, , sinθ = y is positive, tanθ = y is positive, x, Hence, all, trigonometric functions, of θ are positive in the, first quadrant., , 3π, In the third quadrant (π < θ <, ), both x, 2, and y are negative, hence, , ∴ x2 ≤ 1 and y2 ≤ 1, ∴ -1 ≤ x ≤1 and -1 ≤ y ≤ 1, ∴ -1 ≤ cosθ ≤ 1 and -1 ≤ sinθ ≤ 1, , Fig. 2.6, 16

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SOLVED EXAMPLE, , 2), , Ex.1. Find the signs of the following :, i) sin 300° ii) cos 400° iii) cot (−206°), , Hence x = 0 and y = 1, ∴ sin90° = y = 1, , Solution:, (For given θ, we need to find coterminal angle, which lies between 0° and 360°), i), , cos90° = x = 0, tan90° is not defined, as cos90° = 0, 1, 1, cosec90° = y = 1, =1, , 270° < 300° < 360°, ∴ 300° angle lies in the fourth quadrant., ∴ sin 300° is negative., , ii) 400° = 360° + 40°, ∴ 400° and 40° are co-terminal angles, (hence their trigonometric ratios are, same), Since 40° lies in the first quadrant, 400°, also lies in the first quadrant., ∴ cos 400° is positive., , x, 0, cot90° = y = 1 = 0, (Activity) :, Find trigonometric functions of angles 180°,, 270°., , 154° and −206° are coterminal angles. Since, 154° lies in the second quadrant, therefore, cot (−206°) is negative., , 3), , 2.1.4 Trigonometric Functions of specific angles, , 4), , Angle of measure 0° : Let m∠XOP = 0°., Its terminal arm intersects unit circle in, P(1,0). Hence x = 1 and y = 0., , Angle of measure 360° or (2π)c : Since 360°, and 0° are co-terminal angles, trigonometric, functions of 360° are same as those of 0°., c, Angle of measure 120° or 2π :, 3, , Let m ∠ XOP = 120°. Its terminal arm intersects, unit circle in P(x, y)., , We have defined,, , Draw PQ perpendicular to the X-axis, , sinθ = y, cosθ = x, y, and tanθ =, x, ∴ sin0° = 0, cos0° = 1,, 0, and tan0° = 1 = 0, , Fig. 2.10, , sec90° is not defined as x = 0, , iii) −206° = −360° + 154°, , 1), , π c, Angle of measure 90° or, : Let, 2, m ∠ XOP = 90°. Its terminal arm intersects, unit circle in P(0,1)., , ∴ ∆ OPQ is 30° - 60° - 90° triangle., 1, ∴ OQ =, and, 2, 3, PQ =, and OP = 1, 2, As P lies in the second, 1, quadrant, x = - 2, and y = 3, 2, , Fig. 2.9, , cosec0° is not defined as y = 0, sec0° = 1 and cot0°, is not defined as y = 0, , 17, , Fig. 2.11

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6), , Angle of measure – 60° or –, , π, :, 3, , Trig. Fun., , sinθ , , cosθ , , 360° = 0c, , 0, , 1, , πc, , 1, 2, , 3, 2, , 1, 2, , 1, 2, , 3, , 3, 2, , 1, 2, , πc, , 1, , 0, , 0, , −1, , −1, , 0, , Angles, , Let m ∠ XOP = –60°., Its terminal arm, , 30° =, , intersects unit circle, in P (x, y)., , 45° =, , Draw PQ perpendicular, to the X-axis ., ∴ ∆ OPQ is 30° – 60°, – 90° triangle., OQ =, , 1, 3, and PQ =, and OP = 1, 2, 2, , As P lies in the fourth quadrant, x =, y=–, , 3, 2, , y=–, , cos(–60°), , x=, , =, , =, , cosec(–60°) =, sec(–60°), cot(–60°), , 90° =, 1, and, 2, , =, =, , y, x, 1, y, , 3, 2, , 4, , πc, , 2, , 3π, 2, , (Activity) :, , 1, 2, , Find trigonometric functions of angles 150°,, , - 3, = 2 =–, 1, 2, 2, =–, 3, , 210°, 330°, – 45°, – 120°, – 3π and complete the, 4, table., , 3, , Trig., Fun., , θ , Angle, , 1, x =2, x, y =, , πc, , 180° = π, 270° =, , ∴ sin(–60°) =, , tan(–60°), , 60° =, , Fig. 2.14, , 6, , 1, 2, 3, 2, , 150°, =–, , 210°, , 1, , 330°, , √3, , –45°, –120°, , Note : Angles –60° and 300° are co-terminal, angles therefore values of their trigonometric, functions are same., , 3π, – 4, , The trigonometric functions of 0° – 30° – 45°, 60° – 90° are tabulated in the following table., , 19, , sin θ cos θ tan θ cosecθ sec θ cot θ

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Ex.5 If secx = 13 , x lies in the fourth quadrant,, 5, find the values of other trigonometric, functions., , 3π, ∴ tan2 θ = 1 and, < θ < 2π (the fourth, 2, quadrant), ∴ tanθ = –1. Hence cotθ = – 1, , Solution : Since secx = 13 , we have cosx = 5, 5, 13, 2, 2, Now tan x = sec x – 1, , Now sinθ = tanθ cosθ = (–1), Hence cosec θ = – 2, , 13 2, 144, 169, ∴ tan2 x =, –1=, –1=, 5, 25, 25, ∴ tan2 x = 144, and x lies in the fourth, 25, quadrant., , 1 + tan θ + cos ecθ 1 + (−1) + (− 2 ), =, = −1, 1 + cos θ − cos ecθ 1 + (−1) − (− 2 ), , 3, and 180° < θ < 270° then find, 5, all trigonometric functions of θ., , –12, 5, ∴ tan x =, cot x = –, 5, 12, Further we have, sin x = tan x × cos x, =–, , Ex.8 If sinθ = −, , Solution : Since 180° < θ < 270°, θ lies in the, third quadrant., 5, 3, Since, sinθ = −, ∴ cosecθ = −, 5, 3, 2, 2, Now cos θ = 1 − sin θ, , 12, 5, 12, ×, =–, 5, 13, 13, , And cosec x =, , 1, 13, =–, sin x, 12, , 4, , find the value of, 3, 2 sin A – 3 cos A, 2 sin A + 3 cos A, , Ex.6 If tanA =, , ∴ cos2θ =1 −, ∴ cosθ = −, , Solution : Given expression, sin, 2 sin A – 3 cos A, 2 cos, 2 sin A + 3 cos A =, sin, 2 cos, = 2 tan A – 3, 2 tan A + 3, , A - 3 cos, A, cos, A + 3 cos, A, cos, , 2, Solution : Given secθ = 2 ∴ cosθ =, , ∴ secθ = − 5, 4, , sinθ, , A, A, A, A, , ∴ tanθ =, , 3, 4, ∴ cotθ =, 4, 3, , EXERCISE 2.1, 1) Find the trigonometric functions of, 0°, 30°, 45°, 60°, 150°, 180°, 210°, 300°,, 330°, −30°, −45°, −60°, −90°, −120°, −225°,, −240°, −270°, −315°, , 2, 3π, < θ < 2π then find the, 2 ,, , 2, 1 + tan θ + cos ecθ, value of, ., 1 + cot θ − cos ecθ, , 4, 5, , 9 16, =, 25 25, , Now tanθ = cosθ, , 4, 2�   − 3, 1, 3, =  , = –, 17, 4, 2�   + 3, 3, , Ex.7 If sec θ =, , 1, =– 1, 2, 2, , 2) State the signs of, i) tan380°, ii) cot230°, 1, 2, , iii) sec468°, , 3) State the signs of cos 4c and cos4°. Which of, these two functions is greater ?, , Now tan2 θ = sec2 θ – 1 = 2 – 1 = 1, 21

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4) State the quadrant in which θ lies if, i) sinθ < 0 and tanθ >0, ii) cosθ < 0 and tanθ >0, , 1, values in the domain. For example cosecθ = sinθ, is true for all admissible values of θ. Hence this, is an identity. Identities enable us to simplify, complicated expressions. They are basic tools, of trigonometry which are being used in solving, trigonometric equations., , 5) Evaluate each of the following :, i) sin30° + cos 45° + tan180°, ii) cosec45° + cot 45° + tan0°, iii) sin30° × cos 45° × tan360°, , The fundamental identities of trigonometry,, namely., , 6) Find all trigonometric functions of angle in, standard position whose terminal arm passes, through point (3, −4)., , sin2 θ + cos2 θ = 1,, using this identity we can derive simple, relations in trigonometry functions, , 1), , π, 12, , 0 < θ < , find the value of, 13, 2, 2, 2, sin θ – cos θ, 1, ,, 2 sin θ cos θ, tan2 θ, , 7) If cosθ =, , 8), , e.g. cosθ = ± 1 − sin 2 θ and, 2, sinθ = ± 1 − cos θ, , 2), , Using tables evaluate the following :, , 3), , i) 4cot450 − sec2 600 + sin 300, ii) cos2 0 + cos2, 9), , These relations are called fundamental, identities of trigonometry., , π, π, π, + cos2 + cos2, 6, 3, 2, , Find the other trigonometric functions if, i) If cosθ = −, , quadrant., iii) If cot x =, , 2.2.1 Domain and Range of Trigonometric, functions : Now we will find domain and, range of trigonometric functions expressed, as follows., , 3, and 1800 < θ < 2700 ., 5, , ii) If secA = − , , π, 2, 2, 2, 1 + cot θ = cosec θ, if θ ≠ 0, 1 + tan2 θ = sec2 θ, if θ ≠, , 25, and A lies in the second, 7, , We now study sin θ, cos θ, tan θ as functions, of real variable θ . Here θ is measured in radians., , 3, , x lies in the third quadrant., 4, , We have defined sinθ and cos θ, where θ is, a real number. If a and θ are co-terminal angles, , −5, iv) tan x = , x lies in the fourth quadrant., 12, , and if 0° ≤ α ≤ 360°, then sin θ = sinα, and cos θ, = cosα. Hence the domain of these function is R., , Let's Learn, , Let us find the range sin θ and cos θ, We have, sin2θ + cos2θ = 1, , Fundamental Identities, i), , 2.2 Fundamental Identities :, , Consider y = sinθ where θ ∈ R and, y ∈ [–1. 1], , A trigonometric identity represents a, relationship that is always for all admissible, 22

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The domain of sine function is R and range, , The domain of tanθ is R except, , is [–1, 1]., , θ = (2n + 1), , π, ,, 2, , π–, π+, ,, , tanθ → + ∞ and as θ →, 2, 2, tanθ → – ∞., , As θ →, , when you learn the concept of the limits you, will notice., Since tanθ =, , real number, range of tan function is R., , Fig. 2.15, ii), , y, , value of tanθ can be any, x, , Consider y = cosθ where θ ∈ R and, , iv), , y ∈ [–1, 1], , Consider y = cosecθ, cosecθ does not exist for θ = 0, ±π, ±2π,, , The domain of cosine function is R and, range is [–1, 1]., , ±3 π ..., In general cosecθ does not exist if θ = nπ,, where n ∈ I., The domain of cosecθ is R except θ = nπ,, and range is R., The domain of sine function is R and range, is [–1, 1]., Now as –1 ≤ sinθ ≤ 1, cosecθ ≥ 1, , Fig. 2.16, iii) Consider y = tanθ, tanθ does not exist for, π, 3π 5π, θ=± ,± ,±, ....., 2, 2, 2, In general tanθ does not exist if θ =, π, (2n + 1) , where n ∈ I, 2, , , , or cosecθ ≤ – 1., , ∴ The range of cosecant function is, {y ∈ R : |y| ≥ 1} = R – (–1, 1), , Fig. 2.17, , Fig. 2.18, 23

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π, 5π, π, 2π, 3π, x –π - 6 - 4 - 3 - 2 - 3, , SOLVED EXAMPLES, Ex.1 Find the value of sin 41π ., 4, Solution : We know that sine function is periodic, with period 2π., ∴ sin 41π = sin 10π + π, 4, 4, , y, , 0, , -, , π, π, 4, 6 0, , –0.5 –0.71 –0.87 –1 –0.87 –0.71 –0.5 0, , Take the horizontal axis to be the X– axis and the, vertical axis to be the Y− axis., , π, = sin 4 = 1, 2, , Ex.2 Find the value of cos 765°., Solution : We know that cosine function is, periodic with period 2π., ∴ cos 765° = cos(720° + 45°), = cos(2 × 360° + 45°), = cos 45° =, , 1, 2, , Fig. 2.21, The graph of y = sinx is shown above. Since, the period of sine function is 2π It means that, take the curve and shift it 2π to left or right, then, the curve falls back on itself. Also note that the, graph is with in one unit of the Y− axis. The graph, increases and decreases periodically., , Let's Learn, 2.9 Graphs of trigonometric functions :, Introduction : In this section we shall study the, graphs of trigonometric functions. Consider x, to be a real number or measure of an angle in, radian. We know that all trigonometric functions, are periodic. The periods of sine and cosine, functions is 2π and the period of tangent function, is π. These periods are measured in radian., , (ii) The graph of cosine function: Consider, y = cosx, for –π < x < π. Here x represents, a variable angle. The table of values is as, follows:, x 0, , (i) The graph of sine function:, , π, 6, , π, 4, , π, 3, , π, 2, , 2π, 3, , 3π, 4, , π, 4, , π, 3, , π, 2, , y 1 0.87 0.71 0.5 0, , Consider y = sinx, for – π < x < π. Here x, represents a variable angle. The table of, values is as follows:, x 0, , π, 6, , 2π, 3, , 3π, 4, , 5π, 6, , π, , 0.5 0.71 0.87 1, , Using the result cos(– θ) = cosθ, we have, following table:, , 5π, 6 π, , π, π, π, π, 2π, 5π, 3π, x –π- 6 - 4 - 3 - 2 - 3 - 4 - 6, , y 0 0.5 0.71 0.87 1 0.87 0.71 0.5 0, , Y 1, , Using the result sin(– θ) = –sinθ, we have, following table:, , 0.87, , 0.71, , 0.5, , 0, , 0, , 0.5 0.71 0.87 1, , Take the horizontal axis to be the X– axis and the, vertical axis to be the Y− axis., 25

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The graph of y = cosx is shown below. Since the, period of cosine function is 2π. It means that, take the curve and shift it 2π to left or right, then, the curve falls back on itself. Also note that the, graph is with in one unit of the Y− axis. The graph, increases and decreases periodically., , Fig. 2.23, (Activity) :, 1) Use the tools in Geogebra to draw the, different types of graphs of trigonometric, functions., Geogebra is an open source application, available on internet., , Fig. 2.22, , 2) Plot the graphs of cosecant, secant and, cotangent functions., , (iii) The graph of tangent function:, π, π, Let y = tanx for – < x <, 2, 2, π, Note that does not exist for x = . As x, 2, π, increases from 0 to, :, 2, 1) sinx increases from 0 to 1 and, , SOLVED EXAMPLES, 1, Ex. 1 If tan θ + tan θ = 2 then find the value of, 1, tan2 θ +, tan2θ, 1, Solution : We have tan θ + tan θ = 2, , 2) cosx decreases from 1 to 0., sinx, ∴ tanx = cosx will increase indefinitely as x, π, starting from 0 approaches to . Similarly, 2, π, starting from 0 approaches to – , tanx, 2, decreases indefinitely. The corresponding, , Squaring both sides, we get, 1, +, tan θ, 1, ∴ tan2 θ + 2 +, =4, tan2θ, 1, ∴ tan2 θ +, =2, tan2θ, , tan2 θ + 2 tan θ ×, , values of x ad y are as in the following table:, π, x -3, , -, , π, 4, , y –1.73 –1, , π, 6, , 0, , π, 6, , π, 4, , π, 3, , –0.58, , 0, , 0.58, , 1, , 1.73, , -, , 1, =4, tan2θ, , Ex. 2 Which of the following is true?, i), ii), 26, , 1–tan2 θ, 2 cos2 θ = 1+tan2θ, cot A–tan B, cot B–tan A = cot A tan B

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Solution : LHS =, , secθ − tan θ, secθ + tan θ, , = secθ − tan θ × secθ − tan θ, secθ + tan θ secθ − tan θ, =, , =, , (secθ − tan θ ) 2, sec 2 θ − tan 2 θ, , sec2 θ + tan 2 θ − 2secθ tan θ, 1, , = 1 + tan2 θ + tan2 θ – 2 sec θ tan θ, , Fig. 2.24, , = 1 – 2 sec θ tan θ + 2 tan2 θ = RHS, Ex.12 Prove that (sec A – tan A)2 =, , The Cartesian co-ordinates of the point P(r, θ), will be given by relations :, , 1–sin A, 1+sin A, , x = r cosθ, , Solution : LHS = (sec A – tan A)2, , From these relations we get, , = sec2 A + tan2 A – 2 sec A tan A, =, =, =, , 1, 2, , +, , 2, , sin A, 2, , cos A cos A, , −2, , r=, , sin A, cos A cos A, , cos 2 A, , (1 − sin A), , 1 − sin 2 A, , =, , y, x 2 + y 2 and tanθ =, x, , SOLVED EXAMPLE, , 1 + sin 2 A − 2sin A, 2, , and y = r sinθ, , Ex. Find the polar co-ordinates of the point, whose Cartesian coordinates are (3,3)., , 1 − sin A, = RHS, 1 + sin A, , Solution : Here x = 3 and y = 3, To find r and θ., , Let's Learn, 2.2.4 Polar Co-ordinate system : Consider O as, the origin and OX as X-axis. P (x,y) is any point, in the plane. Let OP = r and m∠XOP = θ. Then, the ordered pair (r, θ) determines the position of, point P. Here (r, θ) are called the polar coordinates, of P. The fixed point O is called the Pole and the, fixed ray OX or X-axis is called as the polar axis., , r, , =, , y, x 2 + y 2 and tanθ =, , r, , =, , x 2 + y 2 = 32 + 32 =, , x, , 18 = 3 2, , ∴r = 3 2, Since point P lies in the first quadrant,θ is an, angle in the first quadrant., y 3, tanθ = = = 1, ∴ θ = 45°, x 3, Polar co-ordinates of P are (r, θ) =(1, 45° ), , 30

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3, , Trigonometry - II, m∠XOP = A, m∠XOQ = B. From figure OP, = OQ = 1, ∴ Co-ordinates of P and Q are (cosA,sinA), and (cosB, sinB) respectively ., ∴ d(PQ), ∴, , Let's Study, ∙, ∙, ∙, ∙, ∙, , Trigonometric functions of sum and, difference of angles., Trigonometric, functions, of, allied, angles., Trigonometric functions of multiple, angles., Factorization formulae., Trigonometric functions of angles of a, triangle., , =, , (cosA − cosB ) 2 + ( sinA − sinB ) 2, , =, , cos 2 A − 2cosAcosB + cos 2 B + sin 2 A − 2 sinAsinB + sin 2 B, , =, , (cos 2 A + sin 2 A) + ( cos 2 B + sin 2 B ) − 2(cosAcosB + sinAsinB ), , =, , 1 + 1 − 2(cosAcosB + sinA sinB), , =, , 2 − 2 ( cosAcosB + sinA sinB ), , Let's Recall, , [d(PQ)]2 = 2 − 2 ( cosAcosB + sinA sinB ) ...(1), , In the previous chapter we have studied, trigonometric functions in different quadrants., , Now consider OQ as new X-axis. Draw new, Y-axis perpendicular to it., , 3.1 Compound angle : Compound angles are, sum or difference of given angles., Following are theorems about trigonometric, functions of sum and difference of two, angles., , ∴ m∠QOP = A-B, ∴Co-ordinates of, sin(A−B)), , P and Q are (cos (A−B),, , and (1,0) respectively., P ≡ (cos (A−B), sin (A−B),Q ≡ (1,0), ∴ d(PQ), , Let’s Derive, Theorem : 1) For any two angles A and B, cos, (A-B) = cos A cosB + sinA sin B, , =, , [cos ( A − B ) − 1]2 + [sin ( A − B ) − 0]2, , Proof :, , =, , cos 2 ( A − B ) − 2 cos ( A − B ) + 1 + sin 2 ( A − B ), , =, , cos 2 ( A − B ) + sin 2 ( A − B ) + 1 − 2 cos ( A − B ), , =, , 1 + 1 − 2cos( A − B), , =, , 2 − 2cos( A − B), , Draw a unit standard, circle. Take points P, and Q on the circle, so that OP makes, an angle A with, positive X-axis and, OQ makes an angle, B with positive, X-axis., , [d ( PQ)]2 =, 2 − 2cos ( A − B).....(2), , Fig. 3.1, 35

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sin A sin B, +, cos A cos B, = 1 − sin A sin B, cos A cos B, =, , Ex. 2) Find the value of tan, Solution : tan, , tan A + tan B, 1 − tan A tan B, tan A + tan B, ∴ tan (A+B) = 1 − tan A tan B, , =, , Theorem : 6) For any two angles A and B,, tan (A-B) = tan A − tan B, 1 + tan A tan B, , =, , (Activity), , Results :, 1) If none of the angles A,B and (A+B) is a, multiple of π, cot A cot B − 1, then, cot (A+B) =, cot B + cot A, , =, , 1 − tanπ tan, 0 + tan, , π, , 12, , π, 12, , 1 + 0 × tan, , π, , 12, , π, 12, , tan, , π, 4, , 1 + tan, , − tan, , π, 4, , π, 6, , π, , tan, , 6, , 1, 3, =, 1, 1 + 1×, 3, = = 2− 3, , Ex. 1) Find the value of cos 15°, , Ex. 3) Show that, , Solution : cos 15° = cos(45°−30°), = cos45°cos30° + sin 45°sin30°, , =, , 12, , 1−, , SOLVED EXAMPLES, , =, , π, , π π , = tan  − , 4 6, , 2) If none of the angles A,B and (A−B) is a, multiple of π, cot A cot B + 1, then, cot (A−B) =, cot B − cot A, , =, ,  c πc , 13π c, tan  π + , 12 , 12, , , tanπ + tan, , = tan, , 13π c, 12, , tan x + tan y, sin( x + y ), =, tan x − tan y, sin( x − y ), , Solution : L.H.S. =, , 1 3, 1 1, +, 2 2, 2 2, , =, , 3, 1, +, 2 2 2 2, , sin( x + y ), sin( x − y ), , sin x cos y + cos x sin y, sin x cos y − cos x sin y, , (dividing numerator and denominator by, cos x cos y), , 3 +1, 2 2, , =, 37, , sin x sin y, +, cos x cos y, sin x sin y, −, cos x cos y

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3π, 3π, + q) = − cos q, cos (, + q) = sinq,, 2, 2, 3π, tan (, + q)= -cot q, 2, , 4) sin (, , 3) If sin A = -5 , π < A < 3π and, 13, 2, cos B = 3 , 3π < B < 2 π, 5 2, find, , i) sin (A+B) , , 5) sin (2π - q) = -sin q, cos (2π - q), , ii) cos (A-B), , =, cosq , tan (2π-q)= -tan q, , iii) tan (A+B), , Above results are tabulated in following table ., , π, 5, 1, 4) If tan A= 6 , tan B = 11 , prove that A+B = 4, , allied, angles/, , Let's Learn, , Allied angles : If the sum or difference of the, measures of two angles is either '0' or an integral, π, multiple of, then these angles are said to be, 2, allied angles., , ±θ, π ±θ,, , 2, angles., , sin, , −sinq cosq, , cosq, , −sinq −sinq, , sinq, , cos, , cosq, , −sinq −cos q −cosq cosq, , cosq, , tan, , −tanq cotq, , −cotq −tan q tanq, , tanq, , sinq, , sinq, , tanq, , SOLVED EXAMPLES, , If q is the measure of an angle the, , π, , π+ q 2π - q 2π + q, , π, π-q, π, +q, 2 -q 2, , Trigonometric, functions, , 3.2 Trigonometric functions of allied angels., , −θ ,, , -q, , Ex. 1) Find the values of, , 3π, ±θ , 2π − θ are its allied, 2, , i) (sin495°), , ii) cos 930°, , iii)tan 840°, , Solution :, , We have already proved the following results :, , i) sin (495°) = sin 495°, , π, π, 1) sin (, − q) = cos q, cos (, − q) = sinq,, 2, 2, π, tan (, − q)= cot q, 2, , = sin (360° + 135°), = sin 135°, , π, π, − q) = cos q, cos (, + q) = -sinq,, 2, 2, π, tan ( +q) = - cot q, 2, , = sin (, , 2) sin (, , Similarly we can also prove the following results :, , = cos (2 × 360° + 210°), = cos 210°, , π, + 450), 2, , = cos 45° =, , ii) cos 930°, , 1, 2, , = cos (π + 30°), = −cos 300 = −, , 3, 2, , iii) tan 840° = tan (2 × 360° + 120°) = tan 120° =, π, tan ( +30°) = - cot30° = - 3, 2, , 1) sin (π - q) = sin q, cos (π - q) = - cosq,, tan (π-q)= -tan q, , Ex. 2) Show that :, , 2) sin (π + q) = -sin q, cos (π + q) = - cosq,, tan (π+q) = tan q, , i) cos 24° + cos 55° + cos 125° + cos 204° +, 1, cos 300° = 2, Solution :, L.H.S. = cos 24° + cos 55° + cos 125° + cos204°, , + cos 300°, , 3π, 3π, - q) = − cos q, cos (, − q) = -sinq,, 2, 2, 3π, tan (, − q) = cot q, 2, , 3) sin (, , 40

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= [4 cos3 q − 3cosq] + i [3sin q - 4sin3q], = cos3q + isin3q, , = R.H.S., , Ex. 7) Show that 4sinq cos q - 4cosq sin q, = sin4q, 3, , L.H.S. =, =, , sec2x = 1 + ( -, , 5, 3, But x lies in II quadrant., ∴ secx is negative., ∴ sec x = − 5, 3, , sin 2 A + cos 2 A + 2 sinAcosA, sin 2 A + cos 2 A − 2 sinAcosA, , ∴ sin x =, , =, , =, , =, , sinA, cosA, sinA, 1−, cosA, 1 + tanA, 1 − tanA, , π, 4, , 1 − tan, , 4, , tanA, , [∴ 1 = tan, , ∴ sin, , π, ], 4, 46, , =, , 5−3, 2×5 =, , tan, , +tanA, , π, , 5+3, =, 2×5, x, cos, 2, , 1+, , tan, , 4, , 5, , x, But sin, =, 2, , ( sinA + cosA), (cosA − sinA) 2, , cosA + sinA, cosA, cosA − sinA, cosA, , =, , 3, 5, , 4, 9, 3, sin x = 1 − cos 2 x = 1 − (− ) 2 = 1 − 25 = ± 5, 5, , =, , =, , ∴ cos x = −, , 1 + sin 2 A, 1 − sin 2 A, , sinA + cosA, cosA − sinA, , =, , 4 2, 16, 25, ) =1+, = 9 + 16 =, 3, 9, 9, 9, , sec x = ±, , 2, , =, , R.H.S., , Solution : we know that 1 + tan2q = sec2q, , 1 + sin 2 A = tan  π + A , , , 4, , 1 − sin 2 A, , Solution :, , =, , Ex. 9) Find sin x , cos x , tan x, 2, 2, 2, 4, if tan x =, , x lies in II quadrant., 3, , 4sinq cos3q - 4cosq sin3q, 4sinq cosq [cos2 - sin2q], 2. (2sinq cosq ) (cos2 q - sin2q), 2. sin2q . cos2q, sin 4q, R.H.S., , Ex. 8) Show that, , π, , tan  + A , 4, , , 3, , Solution:, L.H.S =, =, =, =, =, =, , =, , x, 2, , [∴ x lies in II quadrant], 1 − cosx, =, 2, , 4, 2, =, 5, 5, 1 + cosx, =, 2, ,  3, 1−  − ,  5, 2, , 1, 1, =, 5, 5, , x, 2 =, =, x, cos, 2, , x, =, 2, ,  3, 1−  − ,  5, 2, , sin, , 4, 5, 1 =, 5, , 4 5, × =, 5 1, , 4 =2, , 1, 2, x, x, , cos, =, , tan, =2, 5, 5, 2, 2

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sin (A + B ) + sin (A + B) = 2sin AcosB, sin (A - B) + sin (A - B) = 2 cos A sinB, , SOLVED EXAMPLES, , we get, , Ex. 1) Prove the following :, , C+D , C−D, sin C + sin D = 2 sin ,  cos , ,  2 ,  2 , , i), , sin 40° - cos 70° =, , 3 cos 80°, , ii) cos 40° + cos 50° + cos 70° + cos 80°, , C+D , C−D, sin C - sin D = 2 cos ,  sin , ,  2 ,  2 , Simillarly, , Solution :, , the equations,, , i) LHS = sin 40° - cos 70°, , = cos 20° + cos 10°, , cos ( A + B ) = cos A cos B - sinAsin B ...... (3), , = sin (90° - 50°) - cos70°, , cos ( A - B ) = cos A cos B + sinAsin B ...... (4), , = cos 50° - cos70°, , gives,, , = - 2 sin 60°sin ( - 10°), , C+D , C−D, cosC + cos D = 2 cos ,  cos , ,  2 ,  2 , , = 2 sin 60°sin 10°, , C−D , C+D ,  sin  2 ,  2 , , =, , ∴ cosC - cos D = - 2 sin , , 3 cos 80 =, 2, , 3 cos 80°, , = R.H.S, , ∴ sin (−θ) = sinθ, , L.H.S., ,   C − D , C−D , ∴ −sin , ,  = sin  − ,  2 ,   2 , , , 2×, , = cos 40° + cos 50° + cos 70° + cos 80°, = (cos 80° + cos 40°) + (cos 70° + cos 50°), ,  D −C , = sin , ,  2 , ,  80 + 40 ,  80 − 40 ,  70 + 50 ,  70 − 50 ,  cos ,  + 2cos ,  cos , , 2 ,  2 ,  2 ,  2 , , = 2 cos , , C+D ,  D −C ,  sin , ,  2 ,  2 , , ∴ cosC - cos D = 2 sin , , = 2cos 60° cos 20° + 2cos60° cos10°, = 2 cos60° (cos 20° + cos 10°), , 3.4.2 Formulae for conversion of product in to, sum or difference :, , 1, (cos20° + cos 10°), 2, = cos20° + cos 10° = R. H. S., =2, , For any angles A and B, 1) 2sin A cos B = sin (A + B) + sin (A - B), 2) 2cos A sin B = sin (A + B) - sin (A - B), , Ex. 2) Express the following as sum or difference, of two trigonometric function:, , 3) 2cos A cos B = cos (A + B) + cos (A - B), , i), , 4) 2sin A sin B = cos (A - B) - cos (A + B), , 49, , 2 sin4q cos 2q, , Solution : =, , 2 sin 4q cos 2q, , , , =, , sin (4q + 2q) + sin (4q - 2q), , , , =, , sin 6q + sin 2q

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[sin ( 3 x + 9 x ) − sin ( 3x − 9 x ) ]−[ sin ( x + 5 x ) + sin ( x − 5 x )], , =, , cos ( x + 5 x ) + cos ( x − 5 x )  − cos ( 9 x − 3 x ) − cos ( 3 x + 9 x ) , , =, , sin12 x − sin ( −6 x ) − sin6 x − sin ( −4 x ), cos 6 x + cos ( −4 x ) − cos 6 x + cos12 x, , EXERCISE 3.4, 1) Express the following as a sum or difference, of two trigonometric function., i), , sin12 x + sin 6 x − sin6 x + sin4 x, =, cos 6 x + cos 4 x − cos 6 x + cos12 x, , ii) 2sin, , sin12 x + sin 4 x, = cos12 x + cos 4 x, , π, 2π, cos, 2, 3, , iii) 2cos4q cos2q, iv) 2cos35° cos75°, ,  12 x + 4 x ,  12 − 4 x , 2sin ,  cos , , 2, , ,  2 , =,  12 x + 4 x ,  12 x − 4 x , 2cos ,  cos(, , 2, 2, , , , , =, , 2sin 4x cos 2x, , 2) Prove the following :, i), , sin8 x, = tan8x = R.H.S., cos8 x, , tan ( x + y ), sin 2 x + sin 2 y, =, sin 2 x − sin 2 y, tan ( x − y ), , ii) sin6x + sin 4x - sin2x = 4cosx sin2x cos3x, , 1, iv) cos 20° cos 40° cos 60° cos 80° = 16, Solution : L.H.S., , iii), , sinx − sin3 x + sin5 x − sin7 x, = cot2x, cosx − cos3 x − cos5 x + cos 7 x, , iv) sin18° cos39° + sin6° cos15° = sin24° cos33°, 1, v) cos20° cos40° cos60° cos80° = 16, 3, vi) sin20° sin40° sin60° sin80° = 16, , = cos 20° cos 40° cos 60° cos 80°, 1, = cos 20° . cos 40° ., . cos 80°, 2, 1, =, [cos 20° cos 40° cos 80°], 2, 1, = 4 [cos(20°+40°) + cos (20°- 40°)] cos 80°, 1, = 4 [cos60°) + cos (- 20°)] cos 80°, 1, 1, =, [, cos80° + cos 20° cos 80°], 2, 4, 1, 1, = 4 [, cos80° + cos 20° cos 80°], 2, 1, 1, 1, = 8 cos80° +, ., . 2cos 20° cos 80°, 4, 2, 1, 1, = 8 cos80° + [cos (20 + 80 )+cos (20 - 80)], 8, 1, 1, = 8 cos80° +, [cos 100° + cos (- 60)°], 8, 1, 1, 1, = 8 [cos80° + [cos 180° - 80°)] + 8 × 2, 1, 1, 1, = 8 [cos80° - cos 80°] + 16 = 16, , Let's :Learn, 3.5 Trigonometric functions of angles of a, triangle, Notation: In D ABC; m ∠ BAC = A,, m ∠ ABC = B, m ∠ ACB = C, ∴A+B+C= p, Result 1) In D ABC, A + B + C = p, ∴A+B= p-C, ∴ sin (A + B) = sin (p - C), ∴ sin (A + B) = sin C , Simillarly:, sin (B + C) = sin A, , = R. H.S., 51, , sin (C + B) = sin B, , and

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Result 2) In DABC, A+B+C = p , , ii) In DABC, A+B+C = p ∴ B + C = p - B, π, B, B, A+C , ∴ cos , =, cos, (, ), =, sin, , 2, 2, 2,  2 , B, A+C , ∴ cos , =, sin, , 2,  2 , , ∴ B + C = p - C , ∴ cos (B +C) = cos (p - A), ∴ cos (B + C) = - cos A, Simillarly:, cos (A + B) = - cos C, , Verify., , and, ,  A+ B , 1) cos ,  = sin,  2 ,  B+C , 2) cos ,  = sin,  2 , , cos (C + A) = - cos B, Result 3) for any DABC, C,  A+ B , i) sin ,  = cos 2,  2 , A,  B+C , sin ,  = cos 2 ,  2 , , C, 2, A, 2, , SOLVED EXAMPLES, Ex. 1) In DABC prove that, , C + A, B,  = cos, 2,  2 , , sin , , i) sin2A + sin2B - sin2C = 4cosA cosB sinC, Solution : L.H.S. = sin2A + sin2B - sin2C, , C,  A+ B , ii) cos ,  = sin 2,  2 , ,  2 A − 2B ,  2 A + 2B , = 2sin , cos, ,  - sin2C, , 2, , , 2, , , = 2sin(A + B) cos (A- B) -2sinC cosC, , A,  B+C , cos ,  = sin 2,  2 , , = 2sin (p-C) cos (A-B) - 2sinC cos [p - (A+B )], , C + A, B,  = sin, 2, 2, , , , cos , , = 2sinC cos (A - B) + 2sinC cos(A+ B), , Proof., , = 2sinC [cos (A - B) + cos(A+ B), , i) In DABC, A+B+C = p ∴ A +B = p - C, , A− B + A+ B ,  A− B − A− B , = 2 sinC.2cos , ,  cos , 2, 2, , , π, C, π −C,  A+ B , ∴ , =, =, , 2, 2, 2,  2 , , , , , , = 4 sinC cosA cosB, = 4 cosA cosB sinC, , π, C, C,  A+ B , sin , =, sin, (, ), =, cos, , 2, 2, 2,  2 , , = R.H.S., , C,  A+ B , sin ,  = cos 2,  2 , , A, B, C, ii) cosA+cosB + cosC =1+ 4sin 2 sin 2 sin 2, Solution : L.H.S = cosA + cosB + cosC, , Verify., ,  A+ B ,  A − B  + 1 - 2 sin2 C, = 2 cos ,  cos , , 2,  2 ,  2 , C, π C , = 2 cos  −  cos  A − B  + 1 - 2 sin2 2, 2 2,  2 , , A,  B+C , 1) sin ,  = cos 2 ,  2 , B,  B+C , 2) cos ,  = sin 2,  2 , , , , 52

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C, C, = 1+ 2 sin 2 cos  A − B  - 2 sin2 2,  2 , , = cosC.2sin  A − B + A + B  sin  A + B − A + B , 2, 2, ,  , , = 2cosC sin A sin B, , C, C   A− B , = 1+ 2 sin 2 cos ,  − sin , 2,   2 , , = 2 sin A sin B cos C, = R. H.S., , C,  A− B , π A+ B , = 1+ 2 sin 2 [cos ,  − sin  −, ], 2 ,  2 , 2, , iv) cotA cot B + cot B cot + cot C cotA = 1, Solution : In DABC, A + B + C = p, ∴ A+B =p - C, ∴ tan ( A+ B ) = tan ( p - C), tanA + tanB, ∴, = tan ( p - C), 1 − tanA tanB, , C,  A− B ,  A+ B , = 1+ 2 sin 2 [cos ,  − cos , ],  2 ,  2 , C, = 1+ 2 sin 2 .2sin  A − B + A + B  sin  A + B − A + B , 4, 4, ,  , , A, B, C, = 1+ 4 sin 2 sin 2 sin 2, B, C, A, = 1+ 4 sin 2 sin 2 sin 2, = R.H.S., , ∴ tanA + tan B = - tanC + tanA tanB tanC, ∴ tanA + tan B + tanC = tanA tanB tanC, ∴, , ∴ cot A cot B + cotB cotC + cotC cotA = 1, , iii) sin A + sin B - sin C = 2 sinA sinB cosC, 2, , 2, , 1, 1, 1, 1, 1, 1, +, +, =, ., ., cotA cotB cotC, cotA cot B cotC, , 2, , Solution : L.H.S. = sin2A + sin2B - sin2C, , B, B, C, C, A, A, v) tan 2 tan 2 + tan 2 tan 2 +tan 2 tan 2 =1, , 1 − cos 2 A 1 − cos 2 B, +, − sin 2C, 2, 2, 1, = 2 [2 - cos2A - cos2B] - sin2C, =, , Solution : In DABC, A + B + C = p, ∴ A+B =p - C ∴, , 1, = 1 - 2 [cos2A + cos2B] - sin2c, , A + B π −C π C, =, =2-2, 2, 2, , A B, π, C, tan ( 2 + 2 ) = tan ( 2 - 2 ), ,  2 A − 2B , 1,  2 A + 2B , = 1- 2 . 2 cos ,  - sin2C,  cos , 2, , 2, , , , A, B, + tan, C, 2, 2, ∴, = cot 2, A, B, 1 − tan tan, 2, 2, tan, , = 1 - sin2C - cos( A + B ) + cos (A- B), = cos2C - cos [ p - C ] cos (A- B), , A, B, 1, + tan, 2, 2, ∴, =, C, tan, A, B, 1 − tan tan, 2, 2, 2, A, B, C, ∴ [tan 2 + tan 2 ] tan 2 =1 - tan, A, C, C, B, ∴ tan 2 tan 2 +tan 2 tan 2 =1-tan, tan, , = cos2C + cosC cos (A- B), = cos C [cosC + cos (A- B)], = cos C [cos[ p - (A+B)] + cos (A- B)], = cos C [- cos (A+B ) + cos (A- B)], = cos C [cos (A-B) - cos (A+ B)], 53, , A, B, tan, 2, 2, A, B, tan, 2, 2, , A, C, A, C, B, B, ∴ tan 2 tan 2 +tan 2 tan 2 +tan 2 tan 2 =1

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vi), , A, C, cosA − cosB + cosC + 1, = cot 2 cot 2, cosA + cosB + cosC − 1, , =, , cosA − cosB + cosC + 1, Solution : L.H.S. =, cosA + cosB + cosC − 1, =, , [cosA − cosB ] + [1 + cosC ], [cosA + cos B ] − [1 − cos C ], , =, ,  A+ B   B − A, 2 C, 2sin ,  sin ,  + 2cos, 2,  2   2 , A, B, A, B, +, −, , , ,  , 2 C , 2 cos ,  cos ,  +  −2sin, , 2,  2 ,  2  , , =, , π c   B− A, 2 C, 2sin  −  sin ,  + 2cos, 2,  2 2  2 , c, A, B, π, −, , , ,  , 2 C , 2 cos  −  cos ,  +  −2 sin, , 2,  2 2,  2  , , C, 2, , = cot, , A, C, cot, 2, 2, , In DABC, A + B + C = p show that, , C,  B− A, 2 C, sin ,  + 2cos, 2, 2,  2 , C,  A+ B , 2 C, 2sin cos ,  − 2 sin, 2, 2,  2 , , 1) cos2A + cos2 B + cos 2 C, = -1- 4 cos A cos B cos C, 2) sin A + sin B + sin C, A, B, C, = 4 cos 2 cos 2 cos 2, , C, C,  B− A, [sin ,  + cos ], 2, 2,  2 , C, C,  A− B , sin [cos ,  − sin, 2, 2,  2 , , 3) cos A + cos B - cosC, A, B, C, = 4 cos 2 cos 2 sin 2 - 1, ,   A+ B ,  B − A , sin ,  + sin , , , C,  2 ,   2 , = cot 2, π A+ B,  A− B , [cos , )],  − sin ( −, 2, 2,  2 , , 4) sin2 A + sin2 B + sin2C = 2 + 2cosAcosBcosC, A, B, C, 5) sin2 2 + sin2 2 - sin2 2, A, B, C, = 1- 2cos 2 cos 2 sin 2, ,  A+ B ,  B− A, sin ,  + sin , ,  2 ,  2 , C, = cot 2, A+ B,  A− B , cos , ),  − cos (, 2,  2 , , =, , = cot, , EXERCISE 3.5, , cos, , =, , A, 2, A, sin, 2, , cos, , = R.H.S., , 2 cos, , =, , B, A, cos, C, 2, 2, cot 2 ., A, B, 2sin sin, 2, 2, 2sin, , B, C, A, B, C, A, 6) cot 2 +cot 2 +cot 2 =cot 2 cot 2 cot 2, 7) tan2A + tan2B + tan2C = tan2A tan2B tan2C, ,  A+B B−A ,  A+B B−A , +, −, , , , , 2 , 2 ,  2,  2, 2sin, cos, C, 2, 2, cot 2 ., A−B A+B, A+B A−B, +, −, , , , , 2, 2 , 2, 2 , 2 sin , sin , 2, 2, , 8) cos2 A +cos2 B - cos2C = 1- 2sinA sinB cosC, , 54

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19) 2 sinA cosB = sin (A + B) + sin (A − B), , vii) tan15° = tan, , 2 cosA sinB = sin (A + B) − sin (A − B), , π, 5π, = 3 − 1 = cot75° = cot, 12, 12, 3 +1, , 2 cosA cosB = cos (A + B) + cos (A − B), viii), , 2 sinA sinB = cos (A − B) − cos (A + B), , tan75° = tan, , 3, , π, = 3 + 1 = cot15° = cot, 12, 3 −1, , 20) For ∆ABC,, sin (A + B) = sinC, sin (B + C) = sinA, sin (A + C) = sinB, , ix) tan(22.5°) = tan, =, , cos(A + B) = − cosC, cos (B + C) = − cosA, cos(A + C) = − cosB, , =, , A, B,  A+C , = cos 2 , sin ,  = cos 2,  2 , C,  A+ B ,  B+C , cos ,  = sin 2 , cos , ,  2 ,  2 , , Activity :, , 3π, 8, π, 2 + 1 = cot (22..5°) = cot, 8, , Select correct option from the given, alternatives., , 1), , The value of sin (n+1) Asin (n+2) A +cos, (n+1) A cos (n+2) A is equal to, A) sin A, , 5 −1, 2π, = cos 72° = cos, 4, 5, , 2), , 5 +1, π,  3π , =, = sin 54° = sin  , 5, 4,  10 , , ii) cos36° = cos, iii) sin72° = sin, , 10 + 2 5, 2π, π, =, = cos18° = cos, 4, 5, 10, , iv) sin36° = sin, , 2π, =, 5, , v) sin15° = sin, , 3 −1, π, 5π, =, = cos75° = cos, 12, 12, 2 2, , vi) cos15° = cos, , 3 +1, π, 5π, =, = sin75° = sin, 12, 12, 2 2, , 10 − 2 5, 4, , = cos54° = cos, , 3π, 8, , I), , Verify the following., π, =, 10, , 2 - 1 = cot67.5 = cot, , MISCELLANEOUS EXERCISE - 3, , A, B, A+C , = sin 2 , cos , =, sin, , 2,  2 , , sin18° = sin, , π, 8, , x) tan(67.5°) = tan, , C,  A+ B ,  B+C , sin ,  = cos 2 , sin , ,  2 ,  2 , , i), , π, =2+, 12, , 3π, 10, , 3), , B) cosA, , If tan A − tan B = x and cot B − cotA = y, then cot (A − B ) = …, A), , 1, 1, −, y, x, , B), , 1, 1, −, x, y, , C), , 1, 1, +, x, y, , D), , xy, x−y, , If sinq = nsin (q + 2 a) then tan (q + a), is equal to, A), , 1+ n, tan a, 2−n, , C) tan a , 56, , C) − cos A D) sin2A, , B), , 1− n, tan a, 1+ n, , D), , 1+ n, tan a, 1− n

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4, , , Determinants and Matrices, , Let's Study, , ·, , Definition and Expansion of Determinants, , ·, , Minors and Co-factors of determinants, , ·, , Properties of Determinants, , ·, , Applications of Determinants , , ·, , Introduction and types of Matrices , , ·, , Operations on Matrices, , ·, , Properties of related matrices, , , , a, , b, , 1st row, , c, , d, , 2nd row, , 1st, 2nd, column column, , The value of the determinant, ad – bc., , a b, c d, , is , , SOLVED EXAMPLES, , 4.1 Introduction, We have learnt to solve simultaneous, equations in two variables using determinants., We will now learn more about the determinants, because they are useful in Engineering, applications, and Economics, etc., , Ex. Evaluate i) , , iii), , The concept of a determinant was, discussed by the German Mathematician G.W., Leibnitz (1676-1714) and Cramer (1750), developed the rule for solving linear equations, using determinants., , 7 9, cos θ sin θ, ii), −4 3, − sin θ cos θ, , i, log 42 log 42, where i2 = –1 iv), −2i 7, 2, 4, , 4, , Solution :, i), , 7 9, = 7 × 3 − (−4) × 9 = 21 + 36 = 57, −4 3, , ii), , cos θ sin θ, = cos2θ − (−sin2θ), − sin θ cos θ, , Let's Recall, 4.1.1 Value of a Determinant, In standard X we have studied a method of, solving simultaneous equations in two unknowns, using determinants of order two. In this chapter,, we shall study determinants of order three., , = cos2θ + sin2θ = 1 , iii), , 4, i, −2i 7, , = 4 × 7 − (−2i) × i = 28 + 2i2, = 28 + 2(−1) [Qi2 = −1], = 28 − 2 = 26, , a b, The representation, is defined as the, c d, determinant of order two. Numbers a, b, c, d, are called elements of the determinant. In this, arrangement, there are two rows and two columns., , iv), 59, , log 42 log 42, 2, , 4, , = 4×log4 2 − 2×log42, = log424 − log422

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ii), , x 3 2, x x 1 = 9 , 1 0 1, , = 2(0 + 6) − 5(0−2) – 1(3−2), , ∴ x (x−0) − 3(x−1) + 2(0−x) =9, , = 22 – 1, , ∴ x2 − 3x + 3 − 2x = 9, , = 21, , = 2(6) − 5(−2) – 1(1), = 12 + 10 − 1, , Interpretation: From (a) and (b) it is seen that, the expansion of determinant by both ways gives, the same value., , ∴ x2 − 5x +3 =9, ∴ x2 − 5x − 6 = 0, ∴ (x−6) (x+1) =0, ∴ x−6 = 0 or x+1 = 0, , EXERCISE 4.1, , ∴ x = 6 or x = −1, 1 −1 2, 5, Ex. 3) Find the value of −2 3, −2 0 − 1, , Q.1) Find the value of determinant, , by, , expanding along a) 2nd row b) 3rd column and, Interprete the result., , i), , 2 −4, 2i 3, ii), 7 − 15, 4 −i, , iv), , a h g, h b f, g f c, , a) Expansion along the 2nd row, = a21 c21 + a22 c22 + a23 c23, = –2(−1)2+1, , 1, 2, −1 2, + 3(−1)2+2 −2 − 1, 0 −1, , , , + 5(−1)2+3, , Q.2) Find the value of x if, , 1 −1, −2 0, , x, , i), , = 2(+1 −0) + 3(–1+4) − 5(0−2), , 4i, , = 2 + 9 + 10, , −1 2, 1 − 3 = 29, −4 5, , i 3 2i, , Q.3 Find x and y if 1 3i 2, i2 = −1, 5 −3, , = 21, , 4 = x+iy where, i, , Expansion along 3rd coloumn, = a13 c13 + a23 c23 + a33 c33, , =, , x2 − x + 1 x + 1, = 0 ii) 2 x, x +1, x +1, 3, , = 2(1) + 3(3) − 5(−2), , b), , 3 −4 5, iii) 1 1 − 2, 2 3 1, , 2(−1)1+3, , −2, −2, , − 1(−1)3+3, , Q.4) Find the minor and cofactor of element of, the determinant, , 3, 1 −1, + 5(−1)2+3, +, 0, −2 0, , 2 −1 3, D = 1 2 − 1 , 5 7, 2, , 1, , −1, −2 3, , 63

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R1 → R1 + kR3, , R1 ↔ R2 D = D1, then D1 = −D ....... (property 2) ...... (I), , a1 + ka3 b1 + kb3 c1 + kc3, , But R1 = R2 hence D1 = D .......... (II), , A1 =, , ∴ adding I and II, 2D1 = 0 ⇒ D1 = 0, Property 4 - If each element of a row (or, a column) of determinant is multiplied by a, constant k then the value of the new determinant, is k times the value of given determinant., , 1 + 2(−1), −1, B1 =, 1, -1 6, B1 = -1 2, 1, 2, , For example,, x1 y1 z1, , a3, , b3, , c3, , a3 b3 c3, , c3, , R1 → R1 + 2R2, , Property 5 - If each element of a row, (or column) is expressed as the sum of two, numbers then the determinant can be expressed, as sum of two determinants, , c2 = a2 b1 c1 + a2 b2 c2, , b3, , 1 2 3, Now, B = -1 2 0, 1 2 1, , ii) If corresponding elements of any two rows, (or columns) of determinant are proportional, (in the same ratio) then the value of the, determinant is zero., , b2, , a3, , 3(–2–2) = 2 + 2 – 12, = 4–12 = –8 ----------(i), , Remark i) Using this property we can take, out any common factor from any one row (or any, one column) of the given determinant, , a2, , c2, , 1 2 3, Ex. : Let B = -1 2 0 = 1(2–0) – 2(–1–0) +, 1 2 1, , The operation Ri → kRi gives multiple of the, determinant by k., , a1 b1 c1, , b2, , Simplifying A1, using the previous properties,, we get A1 = A., , i.e. D = 0, , a1 + x1 b1 + y1 c1 + z1, , a2, , 2 + 2(2) 3 + 2(0), 2, 0, 2, 1, 3, 0 = –1(2–0) – 6(–1–0) + 3(–2–2), 1, , = –2 + 6 – 12 = 6 – 14 = –8 ----(ii), , a3 b3 c3, , From (i) and (ii) B = B1, Remark : If more than one operation from, above are done, make sure that these operations, are completed one at a time. Else there can be, mistake in calculation., , Property 6 - If a constant multiple of all, elements of any row (or column) is added to, the corresponding elements of any other row, (or column ) then the value of new determinant, so obtained is the same as that of the original, determinant. The operation Ri ↔ Ri + kRj does, not change the value of the determinant., , Main diagonal of determinant : The main, diagonal (principal diagonal) of determinant A is, collection of entries aij where i = j, , Verification, , OR, , a1 b1 c1, , Main diagonal of determinant : The set of, elements ( a11 , a22, a33 , ---- ann) is called the main, diagonal of the determinant A., , A = a2 b2 c2, a3 b3 c3, 65

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a11 a12 a13, , 1, 2, 3, = 100 505 606 707 by using property, 1, 2, 3, , e.g. D = a21 a22 a23 here a11, a22 , a33 are, a31 a32 a33, element of main diagonal, , = 100 × 0 (R1 and R3 are identical), , Property 7 - (Triangle property) - If each, element of a determinant above or below the main, diagonal is zero then the value of the determinant, is equal to product of its diagonal elements., , = 0, , ii), , that is, a1 b1, , c1, , a1 0, , 0, , b2 c2 = a2 b2, , 0, , 0, , c3, , 0, , 312 313 314, 315 316 317 = 0, 318 319 320, 312 313 314, L.H.S. = 315 316 317, 318 319 320, , 0 = a 1b 2c 3, , a3 b3 c3, , C 2→ C 2 – C 1, , Remark : If all elements in any row or any, column of a determinant are zeros then the value, of the determinant is zero., , 312 1 314, = 315 1 317, 318 1 320, C 3→ C 3 – C 1, , SOLVED EXAMPLES, , 312 1 2, = 315 1 2, 318 1 2, , Ex. 1) Show that, i), , 101 202 303, 505 606 707 = 0, 1, 2, 3, , take 2 common from C3, , 101 202 303, LHS = 505 606 707, 1, 2, 3, , , , = 2 (0) ( C2 and C3 are identical), , R1 → R1 – R3, , =, , =, , 312 1 1, = 2 315 1 1, 318 1 1, , 1 a a2, 1 a bc, 2, Ex. 2) Prove that 1 b ca = 1 b b, 1 c ab, 1 c c2, , 100 200 300, 505 606 707, 1, 2, 3, , 1 a bc, L.H.S. = 1 b ca, 1 c ab, , 100 × 1 100 × 2 100 × 3, 505, 606, 707 , 1, 2, 3, , , 66, , R1 →aR1

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x, y, z, z = k.xyz then find the value , Ex. 3) If - x y, x -y z, of k, , a a 2 abc, 1 1 b ca, =, a, 1 c ab, , Sloution :, , R2 → bR2, , x, y, z, z, L.H.S. = - x y, x -y z, , a a 2 abc, , =, , 1 1 b b 2 abc, ×, a b, 1 c ab, , R2→R2 +R1, , R3 → cR3, , x, y z, = 0 2 y 2 z, x -y z, , 2, , a a abc, 1 1 1, 2, =, × × b b abc, b, a, c, c c 2 abc, , R3 → R3 + R1, x, y, z, = 0 2 y 2 z, 2x 0 2z, , a a2 1, , 1, 2, =, × abc b b 1, abc, c c2 1, , x y z, = 2 × 2 0 y z taking (2 common, x 0 z, , (taking abc common from C3), a a2 1, =, , , , b b2 1, , from R2 and R3) , , 2, , c c 1, , = 4[x(yz) – y(0 – xz)+z(0 – xy)], , C1 ↔C3, , = 4[xyz + xyz – xyz], = 4xyz, , 1 a2 a, , From given condition, , = (–1) 1 b 2 b, , L.H.S. = R.H.S., , 1 c2 c, , 4xyz = k xyz, , C2 ↔C3, , ∴k = 4, 1 a a2, , EXERCISE 4.2, , = (–1)(–1) 1 b b 2, 1 c c2, , Q.1) Without expanding evaluate the following, determinants., , 1 a a2, , 2 7 65, 2 3, 4, 1 a b+c, 8 iii) 3 8 75, i) 1 b c + a ii) 5 6, 5 9 86, 6 x 9 x 12 x, 1 c a+b, , 2, = 1 b b = R.H.S., , 1 c c2, 67

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4.3 APPLICATIONS OF DETERMINANTS, , x+ y y+z z+x, Q.2) Prove that z + x x + y y + z, y+z z+x x+ y, , 4.3.1 Cramer’s Rule, In linear algebra Cramer’s rule is an explicit, formula for the solution of a system of linear, equations in many variables. In previous class, we studied this with two variables. Our goal here, is to expand the application of Cramer’s rule to, three equations in three variables (unknowns) ., Variables are usually denoted by x, y and z., , x y z, = 2 z x y, y z x, Q.3) Using properties of determinant show that, i), , a+b a, a a+c, b, c, 1, , ii), , b, c = 4abc, b+c, , log x y, , log y x, , 1, , log z x, , log z y, , Theorem - Consider the following three linear, equations in variables three x, y, z., a 1x + b 1y + c 1z = d 1, , log x z, , a2x + b2y +c2z = d2, , log y z = 0, , a 3x + b 3y + c 3z = d 3, , 1, , Here ai , bi , ci and di are constants., , Q.5) Solve the following equations., i), , ii), , The solution of this system of equations is, , x + 2 x + 6 x −1, x + 6 x −1 x + 2 = 0, x −1 x + 2 x + 6, , x=, , Dx, Dy, Dz, ,y=, ,z=, D, D, D, , provided D≠0 where, , x -1 x, x-2, 0 x -2 x -3 = 0, 0, 0, x -3, , 4+ x 4− x 4− x, Q.6) If 4 − x 4 + x 4 − x = 0 then find the, 4− x 4− x 4+ x, values of x, , a1 b1 c1, , d1 b1 c1, , D = a2 b2 c2, a3 b3 c3, , Dx = d 2 b2 c2, d3 b3 c3, , a1 d1 c1, , a1 b1 d1, , Dy = a2 d 2 c2, a3 d3 c3, , Dz = a2 b2 d 2, a3 b3 d3, , Remark :, , Q.7) Without expanding determinants show that, , 1) You will find the proof of the Cramer’s Rule, in QR code., , 1 3 6, 2 3 3, 1 2 1, 6 1 4 + 4 2 1 2 = 10 3 1 7, 3 7 12, 1 7 6, 3 2 6, , 2) If D = 0 then there is no unique solution for, the given system of equations., , 68

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Ex. 2) By using Cramer’s rule solve the following, linear equations., x +y – z = 1, 8x +3y – 6z = 1, –4x – y + 3z = 1, Solution : Given equations are, x + y – z = 1 , 8x + 3y – 6z = 1, –4x – y + 3z =1, , SOLVED EXAMPLES, Ex. 1) Solve the following equation by using, Cramer’s rule., x+y+z = 6, x–y+z = 2, x+2y–z = 2, Solution : Given equations are, x+y+z = 6 x–y+z = 2 x+2y–z = 2, D, , 1 1, 1, = 1 -1, 1, 1 2 -1, , = 1(1–2) – 1(–1–1) + 1(2+1), = –1 + 2 + 3, = –1 + 5 , = 4, 6, 1, Dx = 2 - 1, 2 2, , =, , , , , , = 1(9–6) – 1(24–24) –1(–8+12), = 3+0–4, = –1, 1 1 -1, 1 3 -6, 1 -1, 3, , = 1(9–6) – 1(3+6) –1 (–1–3), = 3 – 9 + 4, = –2, , = 6(1–2) – 1(–2–2) + 1(4+2), = –6 + 4 + 6, = 4, , Dy =, , 1 6 1, Dy = 1 2 1, 1 2 -1, , 1 1 -1, 8 3 -6, -4 1, 3, , = 1(3+6) – 1(24–24) – 1(8+4) , = 9 – 0 – 12, = –3, , = 1(–2–2) – 6(–1–1) + 1(2–2), = –4 + 12 + 0, , Dz =, , = 8, 1 1 6, Dz = 1 - 1 2, 1 2 2, , 1, 1 1, 8, 3 1, -4 - 1 1, , = 1(3+1) – 1(8+4) + 1(–8+12), = 4 – 12 + 4, = 8 – 12, = –4, , = 1(–2–4) – 1(2–2) + 6(2+1), = –6 + 0 + 18, = 12, ∴x=, , D, , Dx =, , 1, 1, -1, , 1 1 -1, 8 3 -6, -4 1, 3, , ∴ x =, , Dy, Dx 4, Dz, 8, = = 1, y =, =, =2 and z, =, D, D, D, 4, 4, , Dy, -3, Dx -2, =, = 2, y =, =, = 3 and, D, -1, D, -1, , Dz, -4, =, =4, D, -1, ∴x =2, y = 3, z = 4 are the solutions of the given, equations., ∴z=, , 12, = 3 are solutions of given equation., 4, 69

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Ex. 3) Solve the following equations by using, determinant, , ∴, , 1 2 1, 1 1 1, 2 1 3, + + = –2 , − + =3 , − + = –1, x y z, x y z, x y 2, Solution :, Put, , 1, =p, x, , 1, 1, =q, =r, y, z, , ∴ Equations are , p + q + r = –2, p – 2q + r = 3, 2p – q + 3r = –1, 1 1 1, 1 -2 1, 2 -1 3, , D, , =, , , , , = 1(–6+1) – 1(3–2) + 1(–1+4), = –5 – 1 + 3, = –3, , Dp =, , , , , q =, , -5, Dq, 5, =, =, , , 3, q, -3, , r, , 6, Dr, =, = –2, -3, D, , =, , ∴, , 3, 1, 5, =p=, ∴ x = , , 5, x, 3, , ∴, , -5, 1, -3, =q=, ∴y=, , , 3, 5, y, , ∴, , 1, -1, = r = –2 ∴ z =, z, 2, , 3, -3, -1, ,y=, ,z=, are the solutions of, 5, 5, 2, the equations., Ex. 4) The cost of 2 books, 6 notebooks and, 3 pens is Rs.120. The cost of 3 books,, 4 notebooks and 2 pens is Rs.105. while the, cost of 5 books, 7 notebooks and 4 pens is, Rs.183. Using this information find the cost, of 1 book, 1 notebook and 1 pen., Solution : Let Rs. x, Rs. y and Rs. z be the cost of, one book, one notebook and one pen respectively., Then by given information we have,, , -2 1 1, 3 -2 1, -1 - 1 3, , 2x + 6y + 3z = 120, , 1 -2 1, 1, 3 1, 2 -1 3, , 3x + 4y + 2z = 105, 5x + 7y + 4z = 183, , = 1(9+1) + 2(3-2) + 1(–1–6), = 10 + 2 – 7, = 5, Dr =, , 5, Dp, -5, =, = , , 3, D, -3, , ∴x=, , = –2(–6+1) – 1(9+1) + 1(–3–2), = 10 – 10 – 5, = –5, , Dq =, , p =, , 1 1 -2, 1 -2 3, 2 -1 -1, , = 1(2+3) – 1(–1–6) – 2(–1+4), = 5 + 7 –6 , = 6, , 2 6 3, 3 4 2, 5 7 4, , D, , =, , , , = 2(16–14) – 6(12–10) + 3(21–20), , , , = 2(2) – 6(2) + 3(1), , , , = 4 – 12 + 3 , = 7 – 12, , , 70, , = –5

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Dx =, , 4.3.2 Consistency of three equations in two, variables, , 120 6 3, 40 6 3, 105 4 2 = 3 35 4 2, 183 7 4, 61 7 4, , Consider the system of three linear equations, in two variables x and y, , = 3 [40(16–14) – 6(140–122) + 3(245–244)], , a1x+ b1y +c1 = 0, (I), a2x+ b2y +c2 = 0 (II), a3x+ b3y +c3 = 0 , (III), , = 3[40(2) – 6(18) + 3(1)], = 3[80 –108 +3] , = 3[ 83 – 108], = 3[– 25] = –75, , Dy =, , These three equations are said to be, consistent if they have a common solution., , 2 120 3, 2 40 3, 3 105 2 = 3 3 35 2, 5 183 4, 5 61 4, , Theorem : The necessary condition for the, equation a1x + b1y + c1 = 0 , a2x + b2y + c2 = 0,, a3x + b3y + c3 = 0 to be consistent is, , = 3[2(140–122) – 40(12–10) + 3(183–175)], , a1 b1 c1, a2 b2 c2, = 0, a3 b3 c3, , = 3[ 2(18) – 40(2) + 3(8)], = 3[ 36–80+24], = 3[ 60–80], , Proof : Consider the system of three linear, equations in two variables x and y., , = 3[–20], = –60, Dz =, , a 1x + b 1y + c 1 = 0, a 2x + b 2y + c 2 = 0, a 3x + b 3y + c 3 = 0, , 2 6 120, 2 6 40, 3 4 35 = 3 3 4 105, 5 7 183, 5 7 61, , We shall now obtain the necessary condition, for the system (I) be consistent., , = 3[2(244–245) – 6(183–175) + 40(21–20)], = 3[2(–1) – 6(8) + 40(1)], , Consider the solution of the equations, a2x+b2y = – c2, a3x+b3y = – c3, , = 3[–2 – 48 + 40 ] , = 3[–50+40], = 3[–10], , If , , = –30, ∴x=, , (I), , Dy, Dx, -60, -75, =, = 15, y =, =, = 12,, D, D, -5, -5, , a2, a3, , b2, ≠ 0 then by Cramer’s Rule the system, b3, , of two unknowns, we have, , Dz, -30, z=, =, =6, D, -5, , x =, , ∴ Rs.15, Rs. 12, Rs. 6 are the costs of one book,, one notebook and one pen respectively., , -c2, -c3, a2, a3, , b2, b3, b 2 , y =, b3, , a2, a3, a2, a3, , -c2, -c3, put these, b2, b3, , values in equation a1x + b1y + c1 = 0 then, 71

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=, , 1, [–2(2+8)+3(3+1)+1(–24+2)], 2, , Solution : Given (x1, y1) ≡ (3, 7), (x2, y2) ≡ (4, –3), and (x3, y3) ≡ (5, –13), , =, , 1, [−20+12−22] , 2, , =, , 1, 1, [−42+12] = [–30] = –15, 2, 2, , x1, 1, Area of ∆ =, x2, 2, x3, , Area is positive. , ∴ Area of triangle = 15 square unit, This gives the area of the triangle ABC, in that order of the vertices. If we consider the, same triangle as ACB, then triangle is considered, in opposite orientation. The area then is 15, sq. units. This also agrees with the rule that, interchanging 2nd and 3rd rows changes the sign, of the determinant., , B ≡ (x2, y2) ≡ (5, 7), C ≡ (x3, y3) ≡ (8,9), x1, If x2, x3, , y1 1, y2 1 = 0 then, A, B, C are collinear, y3 1, , 2 5 1, ∴ 5 7 1 = 2(7–9) –5(5−8) +1(45–56), 8 9 1, = –4 +15 − 11 = –15 +15 = 0 , ∴ A, B, C are collinear., , EXERCISE 4.3, , determinant can be of either sign), , 1, [3 × 3k] ∴ ±9 = 3k ∴ k = ±3, 2, , 1, 1, [30+7–37] =, [37–37] = 0 , 2, 2, , Solution : Given A ≡ (x1, y1) = (2, 5) , , , −3 0 1, 1, ∴ ±9 =, 3 0 1 ( Area is positive but the, 2, 0 k 1, , ∴ ±9 =, , =, , A(2, 5), B(5, 7), C(8, 9), , y1 1, y2 1, y3 1, , 1, [–3(0 – k) + 1(3k – 0)], 2, , 1, [3(–3+13) – 7(4–5) + 1(−52+15)], 2, , Ex. 4) Show that the following points are collinear, by determinant method., , Solution : Given (x1, y1) ≡ (–3, 0), (x2, y2), ≡ (3, 0) and (x3, y3) ≡ (0, k) and area of ∆ is 9 sq., unit., , ∴ ±9 =, , =, , A(∆ABC) = 0 ∴ A, B, C are collinear points, , Ex. 2) If the area of triangle with vertices, P(−3, 0), Q(3, 0) and R(0, K) is 9 square, unit then find the value of k., , x1, 1, We know that area of ∆ =, x2, 2, x3, , y1 1, 3 7 1, 1, y2 1 =, 4 −3 1, 2, y3 1, 5 −13 1, , Q.1) Solve the following linear equations by, using Cramer’s Rule., i) x+y+z = 6, x–y+z = 2, x+2y–z = 2, ii) x+y−2z = –10,, 2x+y–3z = –19, 4x+6y+z = 2, , Ex. 3) Find the area of triangle whose vertices are, A(3, 7) B(4, −3) and C(5, –13). Interpret , your answer., , iii) x+z = 1, y+z = 1, x+y = 4, 74

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iv), , −2 1 3, 2 3 1, − − = 3 , − + = –13, x y z, x y z, 2 3, and − = –11, x z, , MISCELLANEOUS EXERCISE - 4 (A), (I) Select the correct option from the given, alternatives., , Q.2) The sum of three numbers is 15. If the, second number is subtracted from the sum, of first and third numbers then we get 5., When the third number is subtracted from, the sum of twice the first number and the, second number, we get 4. Find the three, numbers., , a, b, a+b, c, b+c, Q.1 The determinant D = b, a+b b+c, 0, = 0 lf, A) a,b, c are in A.P. , B) a , b, c are in G.P. , C) a, b, c are in H.P. , D) α is root of ax2 + 2bx + c = 0, , Q.3) Examine the consistency of the following, equations., i) 2x−y+3 = 0, 3x+y−2=0, 11x+2y−3 = 0 , ii) 2x+3y−4=0, x+2y=3, 3x+4y+5 = 0, iii) x+2y−3 =0, 7x+4y−11=0, 2x+4y−6= 0, , Q.2 If, , Q.4) Find k if the following equations are, consistent., i) 2x+3y-2=0, 2x+4y−k=0, x−2y+3k =0, , xk +2, y k +2, z k +2, , x k +3, y k +3, z k +3, , = (x–y) (y–z) (z–x), , 1 1 1, ( + + ) then, x y z, , ii) kx +3y+1=0, x+2y+1=0, x+y=0, , A) k= –3 B) k = –1 C) k = 1 D) k = 3, , Q.5) Find the area of triangle whose vertices are, , sinθ .cosφ, Q.3 Let D = cosθ .cosφ, − sinθ .sinφ, , i) A(5,8), B(5,0) C(1,0), 3, −1, , 1), Q(4, 2), R(4,, ), 2, 2, iii) M(0, 5), N(−2, 3), T(1, −4), ii) P(, , sinθ .sinφ, cosθ .sinφ, sinθ .cosφ, , cosθ, − sinθ then, 0, , A) D is independent of θ, B) D is independent of φ, , Q.6) Find the area of quadrilateral whose, vertices are, A(−3, 1), B(−2, −2), C(3,−1), D(1,4), , C) D is a constant , D), , Q.7) Find the value of k, if the area of triangle, whose vertices are P(k, 0), Q(2, 2), R(4,, 3, 3) is, sq.unit, 2, Q.8) Examine the collinearity of the following, set of points, i) A(3, −1), B(0, −3), C(12, 5), ii) P(3, −5), Q(6, 1), R(4, 2), iii) L(0, 1 ), M (2, −1), N(−4, 7 ), 2, 2, , xk, yk, zk, , dD, at θ =π/2 is equal to 0, d, , Q.4 The value of a for which system of equation, a3x + (a + 1)3y + (a + 2)3z = 0 ax + (a +1) y, + (a + 2) z = 0 and x + y + z = 0 has non zero, Soln. is, A) 0 B) –1 C) 1 D) 2, , 75

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Q.5, , b+c c+a, q+r r+ p, y+z z+x, , c b, A) 2 r q, z y, , a+b, p+q =, x+ y, , a, b, p B) 2 q, x, y, , C) Determinant is number associated to, square matrix , D) None of these , (II) Answer the following questions., , a b c, a c, p r C) 2 p q r, x y z, x z, , 1 −3 12, 2 −5 7, Q.1) Evaluate i) 5 2 1 ii) 0 2 −4, 9 7 2, 9 0 2, , a c b, D) 2 p r q, x z y, , Q.2) Evaluate determinant along second column, 1 −1 2, 3 2 −2, 0 1 −2, , Q.6 The system 3x – y + 4z = 3, x + 2y –3z, = –2 and 6x + 5y + λz = –3 has at least one, Solution when, A) λ = –5, B) λ = 5, C) λ = 3, , 2, 3, 5, Q.3) Evaluate i) 400 600 1000, 48 47, 18, , D) λ = –13, , 101 102 103, ii) 106 107 108 by using properties, 1, 2, 3, , x 3 7, Q.7 If x = –9 is a root of 2 x 2 = 0 has, 7 6 x, other two roots are, , Q.4) Find minor and cofactor of elements of the, determinant., , A) 2, –7 B) –2, 7 C) 2, 7 D) –2, –7, , 1 −1 2, −1 0 4, i) −2 1 3 ii) 3 0 −2, 1 0 3, 0 −4 2, , 6i −3i 1, Q.8 If 4 3i −1 = x + iy then, 20 3, i, , Q.5) Find the value of x if, , A) x = 3 , y = 1 B) x = 1 , y = 3 , C) x = 0 , y = 3 D) x = 0 , y = 0, , 1 4 20, 1 2x 4x, i) 1 −2 −5 = 0 ii) 1 4 16 = 0, 1 2 x 5x2, 1 1 1, , Q.9 If A(0,0), B(1,3) and C(k,0) are vertices of, triangle ABC whose area is 3 sq.units then , value of k is, , Q.6) By using properties of determinant prove, , A) 2 B) –3 C) 3 or −3 D) –2 or +2, , x+ y, that z, 1, , Q.10 Which of the following is correct, A) Determinant is square matrix, B) Determinant is number associated to, matrix, 76, , y+z, x, 1, , z+x, y =0, 1

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Q.7) Without expanding determinant show that, xa, b + c bc b 2 c 2, 2, 2 2, i) c + a ca c a = 0 ii) a, 1, a + b ab a 2b 2, , iii) (k−2)x +(k−1)y =17 , , (k−1)x+ (k−2)y = 18 and x + y = 5, , yb zc, b2 c2, 1 1, , Q.11) Find the area of triangle whose vertices are, i) A(−1,2), B(2,4), C(0,0), ii) P(3,6), Q(−1,3), R (2,−1), , x y z, = a b c, bc ca ab, l m n, iii) e d f =, u v w, , iii) L(1,1), M (−2,2), N (5,4), , n f, l e, m d, , Q.12) Find the value of k, i) If area of triangle is 4 square unit and, vertices are P(k, 0), Q(4, 0), R(0, 2), ii) If area of triangle is 33/2 square unit and, vertices are L (3,−5), M(−2,k), N (1,4), , w, u, v, , Q.13) Find the area of quadrilateral whose, vertices are A (0, −4), B(4, 0) , C(−4, 0),, D (0, 4), , 0 a b, iv) −a 0 c = 0, −b −c 0, , Q.14) An amount of ` 5000 is put into three, investments at the rate of interest of 6% , 7%, and 8% per annum respectively. The total, annual income is ` 350. If the combined, income from the first two investments is, ` 70 more than the income from the third., Find the amount of each investment., , a 1 1, Q.8) If 1 b 1 =0 then show that, 1 1 c, 1, 1, 1, +, +, =1, 1− a 1− b 1− c, , Q.15) Show that the lines x−y=6, 4x−3y=20 and, 6x+5y+8=0 are concurrent .Also find the , point of concurrence, , Q.9) Solve the following linear equations by, Cramer’s Rule., , Q.16) Show that the following points are collinear, by determinant, , i) 2x−y+z = 1, x+2y+3z = 8, 3x+y−4z =1, 1 1 5 1 1 4, ii) 1 + 1 = 3 , + = , + =, y z 6 z x 3, x y 2, , a) L (2,5), M(5,7), N(8,9) , b) P(5,1), Q(1,−1), R(11,4), , iii) 2x+3y+3z=5 , x−2y+z = – 4 , , 3x– y– 2z=3, , Further Use of Determinants, , iv) x–y+2z=7 , 3x+4y–5z=5 , 2x–y+3z=12, Q.10) Find the value of k if the following equation, are consistent., i) (k+1)x+(k−1)y+(k−1) = 0, (k−1)x+(k+1)y+(k−1) = 0, , 1), , To find the volume of parallelepiped and, tetrahedron by vector method, , 2), , To state the condition for the equation, ax2+2hxy+by2 +2gx+2fy+c =0 representing a, pair of straight lines., , 3), , To find the shortest distance between two, skew lines., , 4), , Test for intersection of two line in three, dimensional geometry., , (k−1)x+(k−1)y+(k+1) = 0, ii) 3x+y−2=0 kx+2y−3=0 and 2x−y = 3, 77

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5), , To find cross product of two vectors and, scalar triple product of vectors, , 6), , Formation of differential equation by, eliminating arbitrary constant., , B is a matrix having 3 rows and 2 columns., The order of B is 3×2. There are 6 elements in, matrix B., , 1 + i 8 , iii) C = , , C is a matrix of order 2×2., −3i ,  i, , Let's Learn, ,  −1 9 2 , , iv) D =  3 0 −3 , D is a matrix of order 2×3., In general a matrix of order m x n is represented, by, , 4.4 Introduction to Matrices :, The theory of matrices was developed by, a Mathematician Arthur Cayley. Matrices are, useful in expressing numerical information, in compact form. They are effectively used, in expressing different operators. Hence in, Economics, Statistics and Computer science they, are essential., ,  a11 a12, a,  21 a 22,  a 31 a 32, A = [ aij]mxn = , … …,  a i1 a i 2, , a m1 a m 2, , Definition : A rectangular arrangement of mn, numbers in m rows and n columns, enclosed in, [ ] or ( ) is called a matrix of order m by n., , … a1 j, … a2j, … a3j, … …, … a ij, … a mj, , … a1n , … a 2 n , … a 3n , , … …, … a in , , … a mn , , Here aij = An element in ith row and jth column., , A matrix by itself does not have a value or any, special meaning., ,  2 −3 9 , , , Ex. In matrix A = 1 0 −7  =,  4 −2 1 , , Order of the matrix is denoted by m × n, read as, m by n., ,  a11 a12, a,  21 a 22,  a 31 a 32, , a13 , a 23 , a 33 , , a11 = 2, a12 = –3, a13 = 9, a21 = 1, a22 = 0, a23 =, , Each member of the matrix is called an element, of the matrix., , –7, a31 = 4, a32 = –2, a33 = 1, , Matrices are generally denoted by A, B, C ,… and, their elements are denoted by aij, bij, cij, … etc., e.g. aij is the element in ith row and jth column of, the matrix., , 4.4.1 Types of Matrices :, 1), ,  2 −3 9 , 1 0 −7 , For example, i) A = ,  Here a32 = – 2,  4 −2 1 , , Row Matrix : A matrix having only one, row is called as a row matrix. It is of order, 1 x n, Where n ≥ 1., Ex. i) [–1 2]1×2, , 2), , A is a matrix having 3 rows and 3 columns. The, order of A is 3×3, read as three by three. There, are 9 elements in matrix A., , ii) [0 –3 5]1×3, , Column Matrix : A matrix having only one, column is called as a column matrix. It is of, order m x 1, Where m ≥ 1., 5, 1 ,  , Ex. i)  , ii)  −9 , 0,   2 x1,  −3 3 x1, ,  −1 −5, , , ii) B =  2 0 ,  6 9 , 78

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5), , Note : Single element matrix is row matrix, as well as column matrix. e.g. [5]1x1, 3), , Zero or Null matrix : A matrix in which, every element is zero is called as a zero or, null matrix. It is denoted by O., , 5 0 0 , , , Ex. i) A = 0 0 0  = diag (5,0,9) , 0 0 9  3 x 3, , 0 0 0 , , , Ex. i) O = 0 0 0 , 0 0 0  3 x 3, ,  −1 0 , ii) B = , ,  0 −5 2 x 2, , 0 0 , ii) O = 0 0 , 0 0  3 x 2, 4), , , , Square Matrix : A matrix with equal, number of rows and coloumns is called a, square matrix., Examples, i) A =, , ii) C =, , Diagonal Matrix : A square matrix in which, every non-diagonal element is zero, is called, a diagonal matrix., ,  −1 0 0 , , , iii) C =  0 −2 0 ,  0 0 −3 2 x 2, , Note: If a11, a22, a33 are diagonal elements of a, diagonal matrix A of order 3, then we write the, matrix A as A = Diag., ,  5 −3 i ,  1 0 −7 , , ,  2i −8 9  3 x 3, , 6), ,  −1 0 ,  1 −5, ,  2x 2, , Scalar Matrix : A diagonal matrix in which, all the diagonal elements are same, is called, as a scalar matrix., 5 0 0 , , , For Ex. i) A = 0 5 0 , 0 0 5  3 x 3, , Note : A matrix of order n×n is also called as, square matrix of order n., ,  −2 0 , ii) B = , ,  0 −2  2 x 2, , Let A = [ aij]nxn be a square matrix of order n then, (i) The elements a11, a22, a33,… aii … ann are called the, diagonal elements of matrix A., , 7), , Note that the diagonal elements are defined, only for a square matrix., (ii) Elements aij, where i ≠ j are called non, diagonal elements of matrix A., , Unit or Identity Matrix : A scalar matrix in, which all the diagonal elements are 1(unity) ,, is called a Unit Matrix or an Identity Matrix., An Identity matrix of order n is denoted, by In., , 1 0 0 , 1 0 , Ex. i) I3 = 0 1 0  ii) I2 = , , , , 0 1 , 0 0 1 , Note:, , (iii) Elements aij, where i < j represent elements, above the diagonal., (iv) Elements aij, where i > j represent elements, below the diagonal., , 1., , Statements iii) and iv) can be verified by observing, square matrices of different orders., 79, , Every Identity matrix is a Scalar matrix but, every scalar matrix need not be Identity, matrix. However a scalar matrix is a scalar, multiple of the identity matrix.

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2., , Every scalar matrix is a diagonal matrix but, every diagonal matrix need not be a scalar, matrix., , 8), , Upper Triangular Matrix : A square matrix, in which every element below the diagonal, is zero, is called an upper triangular matrix., Matrix A = [ aij]nxn is upper triangular if, aij = 0 for all i > j., ,  −3 1, ii) B = , ,  1 8 2 x 2, , , , Note:, The scalar matrices are symmetric. A null square, matrix is symmetric., ,  4 −1 2 , , , For Ex. i) A =  0 0 3 ,  0 0 9  3 x 3, , 12) Skew-Symmetric Matrix : A square matrix, A = [ aij]nxn in which aij = − aji, for all i and j,, is called a skew symmetric matrix., ,  −3 1, ii) B = , ,  0 8 2 x 2, 9), ,  2 4 −7 , iii) C =  4 5 −1,  −7 −1 −3 3 x 3, , Here for i = j, aii = − aii, 2aii = 0 aii = 0 for, all i = 1, 2, 3,……n., , Lower Triangular Matrix : A square matrix, in which every element above the diagonal, is zero, is called a lower triangular matrix., , In a skew symmetric matrix each diagonal, element is zero., , Matrix A = [ aij]nxn is lower triangular if, aij = 0 for all i < j., , e.g., ,  2 0 0, For Ex. i) A = , ,  −1 1 0 ,  −5 1 9  3 x 3, ,  0 4 −7 , ii) B =  −4 0 5 ,  7 −5 0  3 x 3, ,  7 0, ii) B = , ,  −1 3 2 x 2, , Note : A null square matrix is also a skew, symmetric., , 10) Triangular Matrix : A square matrix is, called a triangular matrix if it is an upper, triangular or a lower triangular matrix., , 13) Determinant of a Matrix : Determinant of, a matrix is defined only for a square matrix., If A is a square matrix, then the same arrangement, of the elements of A also gives us a determinant,, by replacing square brackets by vertical bars. It is, denoted by |A| or det(A)., , Note : The diagonal, scalar, unit and null matrices, are also triangular matrices., 11) Symmetric Matrix : A square matrix A =, [ aij]nxn in which aij = aji, for all i and j, is called, a symmetric matrix., Ex., ,  0 5, i) A = , ,  −5 0  2 x 2, , If A = [ aij]nxn, then is of order n.,  1 3, Ex. i) If A = , ,  −5 4  2 x 2, , a h g, , , i) A =  h b f ,  g f c  3 x 3, , then |A| =, 80, , 1 3, -5 4

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 2 −1 3, , , ii) If B =  −4 1 5 ,  7 −5 0  3 x 3, , 15) Transpose of a Matrix : The matrix, obtained by interchanging rows and columns, of matrix A is called Transpose of matrix A., It is denoted by A' or AT. If A is matrix of, order m × n, then order of AT is n × m., , 2 -1 3, then |B| = -4 1 5, 7 -5 0, , If AT = A' = B then bij = aji,  −1 5 , , , e.g. i) If A =  3 −2 ,  4 7  3×2, , 14) Singular Matrix : A square matrix A is, said to be a singular matrix if |A| = det(A), = 0, otherwise it is said to be a non-singular, matrix., ,  −1 3 4 , then AT = , ,  5 −2 7  2×3, , 6 3, Ex. i) If A = , , 8 4  2 x 2, 6 3, then |A| =, = 24–24 = 0., 8 4, , 1 0 −2 , , , ii) If B = 8 −1 2 ,  4 3 5  3×3, , , , Therefore A is a singular matrix., ,  1 8 4, then B =  0 −1 3 ,  −2, 2 5  3×3, , 2 3 4, , , ii) If B =  3 4 5  then,  4 5 6  3 x 3, , T, , Remark:, , 2 3 4, |B| = 3 4 5, 4 5 6, , , 1) If A is symmetric then A = AT, 2) If B is skew symmetric, then is B = −BT, , |B| = 2(24–25)–3(18–20) + 4(15–16) , = – 2+6–4 = 0, , Activity :, Construct a matrix of order 2 × 2 where the aij th, (i + j ) 2, element is given by aij =, 2+i, , |B| = 0 Therefore B is a singular matrix.,  2 −1 3, , , iii) A =  −7 4 5 , then,  −2 1 6  3 X 3, , |A| =, , a12 , a, Solution : Let A =  11, be the required, , a 21 a 22  2 X 2, matrix., , 2 -1 3, -7 4 5, -2 1 6, , Given that aij =, , 4, (i + j ) 2, (……..) 2, , a11 =, = ,, 3, 2+i, ……+ 1, , |A| = 2(24-5)-(–1)(-42+10)+3(–7+8) , = 38–32+3 = 9 , , a12 =, , 9, (........)2, = = ......, 3, ..........., , |A| = 9, As |A| ≠ 0 , A is a non-singular, matrix., , a21 =, , (2 + 1) 2, ........, =, , , 4, 2+2, , 81

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a22 =, , ........, (……….) 2, =4, =, ........, 2+2, ,  2 a 3, Ex. 3) Find a, b, c if the matrix A =  −7 4 5 ,  c b 6 , is a symmetric matrix., , ,  4,  3 …, , ∴ A= ,  ..... 4 , ,  ....., ,  2 a 3, Solution : Given that A =  −7 4 5  is a,  c b 6 , symmetric matrix., aij = aji for all i and j, , SOLVED EXAMPLES, x + y, , Ex. 1) Show that the matrix  1,  z, , is a singular matrix., , y+z, 1, x, , As a12 = a21, , z + x, 1 , y , , ∴, , As a32 = a23, ∴, , x + y, , Solution : Let A =  1,  z, , y+z, 1, x, , z + x, 1 , y , , x+ y, ∴ |A| = 1, z, , y+z, 1, x, , z+x, 1, y, , a = –7, b=5, , As a31 = a13, ∴, , c=3, , EXERCISE 4.4, (1) Construct a matrix A = [aij]3×2 whose elements, , Now |A| = (x+y)(y−x)−(y+z)(y−z)+(z+x)(x−z), = y2 – x2 – y2 + z2 + x2 – z2, , aij are given by (i) aij =, , =0, ∴ A is a singular matrix., , (i - j ) 2, 5-i, , (ii) aij = i – 3j (iii) aij =, ,  −1 −5, , , Ex. 2) If A =  2 0 , Find (AT)T.,  3 −4  3X 2, , (i + j )3, 5, , (2) Classify the following matrices as, a row,, a column, a square, a diagonal, a scalar, a, unit, an upper triangular, a lower triangular,, a symmetric or a skew-symmetric matrix., ,  −1 −5, , , Solution : Let A =  2 0 ,  3 −4  3X 2, ,  3 −2 4 , , , (i) 0 0 −5 (ii), , , 0 0 0 , ,  −1 2 3 , ∴ AT = , ,  −5 0 −4  2 x 3, , 5, (iii)  4  ,  ,  −3, ,  −1 −5, , , ∴ (AT)T =  2 0 , =A,  3 −4  3X 2, 82, , 0 4 7,  −4 0 −3, , ,  −7 3 0 , , (iv) 9, , 2, , −3

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6 0 , (v) ,  , 0 6 , , , 3 0 0 , , , (vii) 0 5 0 , , 1, , 0 0, 3, , 1 0 0 , (ix) 0 1 0 , , , 0 0 1 , ,  2 0 0, , , (vi)  3 −1 0 ,  −7 3 1 , ,  10, , (viii)  −15, ,  27, , , −15, 0, 34, , 3 1, 7, , , (6) If A =  −2 −4 1 , Find (AT)T.,  5 9 1, , 27 , , 34 , 5 , , 3 , , 3, , , 1 5 a, , , (7) Find a, b, c if  b −5 −7  is a symmetric,  −4 c, 0, , , matrix., , , , 0, , (8) Find x, y, z if  y, 3, , 2, symmetric matrix., , 0 0 1 , (x) 0 1 0 , , , 1 0 0 , , (3) Which of the following matrices are singular, or non singular ?, ,  2 5 1,  1 2 −5, , , , , (i), (ii)  −5 4 6 ,  2 −3 4 ,  −1 −6 3,  −5 4 9 , ,  3 5 7,  −2 1 4 , , ,  3 2 5 , , 1 + 2i i − 2 ,  0, , (iii) −1 − 2i, 0, −7 , ,  2 − i, 7, 0 , ,  7 5, (iv) , ,  −4 7 , , (10) Construct the matrix A = [ aij]3×3 where, aij = i−j. State whether A is symmetric or, skew symmetric., , (4) Find k if the following matrices are singular, ,  7 3, (i) , ,  −2 k , , − 2, , , x, , z  is a skew, , 0, , , (9) For each of the following matrices, using its, transpose state whether it is a symmetric, a, skew-symmetric or neither., , b, c ,  a, , q, r , (i),  p,  2a − p 2b − q 2c − r , 5 , 5 0, 1 99 100 , (ii) ,  (iii), 6 99 105, , −5i, 0, ,  4 3 1, , , (ii),  7 k 1, 10 9 1, , 4.5 Algebra of Matrices :, (1) Equality of matrices, (2) Multiplication of a matrix by a scalar, (3) Addition of matrices , (4) Multiplication of two matrices. , , k − 1 2 3, , (iii), 1 2 ,  3,  1, −2 4 , , (1) Equality of matrices : Two matrices A and, B are said to be equal if (i) order of A = order, of B and (ii) corresponding elements of, A and B are same, that is if aij=bij for; all i,j, and symbolically written as A=B., , 5 1 −1, (5) If A = , , Find (AT)T., , 3 2 0 , 83

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(3) Addition of Two matrices : A and B are, two matrices of same order. Their addition, denoted by A + B is a matrix obtained, by adding the corresponding elements of A, and B., , 15 14 , Ex. (i) If A , , 12 10  2 x 2, 15 14 , B = ,  and C =, 10 12  2 x 2, , 15 14 , 10 12 , , 2x 2, , Note: A+B is possible only when Aand B are, of same order. , , Here A ≠ B, A ≠ C but B = C by definition of, equality., , A+B is of the same order as that of A and B., , 4 ,  2a − b 4   1, Ex. (ii) If , =, 2   −7 a + 3b ,  −7, , Thus if A = [ aij]mxn and B = [ bij]mxn then A+B, = [ aij + bij]mxn , , then using definition of equality of matrices, we, have 2a – b = 1 …...(1) and a + 3b = 2 …… (2) , 5, Solving equations (1) and (2), we get a = and , 7, 3, b=, 7, ,  2 3 1, Ex. A = ,  and ,  −1 −2 0  2 x 3,  −4 3 1 , B= ,  Find A+B.,  5 7 −8 2 x 3, Solution : By definition of addition,, , (2) Multiplication of a Matrix by a scalar:, If A is any matrix and k is a scalar, then, the matrix obtained by multiplying each, element of A by the scalar k is called the, scalar multiple of the given matrix A and is, denoted by kA. , , 3+3, 1+1 ,  2+( − 4), A+B = , ,  −1+5 −2+7 0+( − 8)  2 x 3,  −2 6 2 , = , ,  4 5 −8 2 x 3, Note : If A and B are two matrices of the same, order then subtraction of the two matrices is, defined as, A–B = A+(–B), where –B is the, negative of matrix B., , Thus if A = [aij]mxn and k is any scalar then, kA = [kaij]mxn., Here the order of matrix A and kA are same., , 1, Ex. A = 3, 4, then kA =, , 3, =, 2, , 1, 3, 4, , 5, 2, 7, , 3, and k = , , 2, , Ex. If A =, , 3 x2, , Find A–B., , 3, A, 2, 5, 2, 7, , 1, 3, 0, , 4, 2, 5, , and B =, 3 x2, , 1, 2, 4, , Solution : By definition of subtraction,, , =, , 3, 2, 9, 2, , 3 x2, , 6, ,  −1 4   1 −5, , , A–B = A+(–B) =  3 −2  +  −2 6 , , ,  0 5   −4 −9 , , 15, 2, 3, 21, 2, ,  −1 + 1 4 + (−5)   0 −1,  , , 4 , =  3 + (−2) −2 + 6  =  1, 0 + (−4) 5 + (−9)   −4 −4 , , 3 x2, , 84, , 5, 6, 9, , ,, 3 x2

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Some Results on addition and scalar, multiplication : If A, B, C are three matrices, conformable for addition and α, β are scalars,, then, , 10 −6   −6 −21,  , , , =  2 0  +  9 −3 ,  −8 −4   −6 6 , 10 − 6 −6 − 21  4 −27 , , 0 − 3  =  11, −3 , =  2+9,  −8 − 6 −4 + 6   −14 2 , , (i) A+B = B+A, That is, the matrix addition is, commutative., (ii) (A+B)+C = A+(B+C), That is, the matrix, addition is associative., , Ex. 2) If A = diag(2, –5, 9), B = diag(–3, 7, –14), and C = diag(1, 0, 3), find B−A−C., , (iii) For matrix A, we have A+O = O+A = A, That, is, a zero matrix is conformable for addition, and it is the identity for matrix addition., , Solution : B−A−C = B−(A+C), Now, A+C = diag(2, –5, 9) + diag(1, 0, 3), = diag(3, −5, 12), , (iv) For a matrix A, we have A+(−A) = (−A) +A, = O, where O is a zero matrix conformable, with matrix A for addition. Where (−A) is, additive inverse of A., , B−A−C= B–(A+C), =diag(–3, 7, –14)–diag(3, –5, 12), 0 ,  −6 0,  0 12 0 , = diag(−6, 12, −26) = , ,  0 0 −26 , , (v) α(A±B) = αΑ ± αB , (vi) (α ± β)A = αΑ ± βA, (vii) α(β·A) = (α·β)·A , ,  2 3 −1, Ex. 3) If A = ,  ,B=, 4 7 5 , , (viii) OA = O, , 1 3 2 ,  4 6 −1 and, , , , 1 −1 6 , C= ,  , find the matrix X such that, 0 2 −5, , SOLVED EXAMPLES, , 3A−2B+4X = 5C., Solution : Since 3A–2B+4X = 5C, ,  5 −3, , , Ex. 1) If A =  1 0  and,  −4 −2 , , ∴ 4X = 5C–3A+2B, 1 −1 6 ,  2 3 −1, ∴ 4X = 5 , –3 , , , 0 2 −5, 4 7 5 , , 2 7, , , B =  −3 1  , find 2A – 3B.,  2 −2 , , 1 3 2 , +2 , ,  4 6 −1, , Solution : Let 2A – 3B, , , , 3 , 5 −5 30   −6 −9, = , +,  , , 0 10 −25  −12 −21 −15, ,  5 −3, 2 7, , 1 0, , = 2,  –3  −3 1 ,  −4 −2 ,  2 −2 , , 2 6 4 , + , , 8 12 −2 , 85

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1, (A – B), 3, , 30 + 3 + 4 ,  5 − 6 + 2 −5 − 9 + 6, = , , 0 − 12 + 8 10 − 21 + 12 −25 − 15 − 2 , , By (1) – (2) , 3Y = A – B , ∴ Y =, ,  1 −8 37 , = , ,  −4 1 −42 , ,   2 −1  −2 1  ,  4 −2 , 1 ,  1, , , , ∴ Y =   1 3  −  3 −1  =  −2 4 , 3,  3,   −3 −2   4 −2  ,  −7 0 , , 37 , 1, −2, , 1, −, 8, 37, , 1, 4, 4 , , , =, ∴X= , 1, 21 , 4  −4 1 −42  , −1, −, , 4, 2 , ,  4,  3, , 2, = −,  3, , − 7,  3, ,  2 x + 1 −1  −1 6   4 5 , +, =, ,, Ex. 4) If , 4 y   3 0   6 12 ,  3, find x and y., , From (1) X + Y = A, ∴ X = A – Y, , ,  2 x + 1 −1  −1 6 , Solution : Given , +, 4 y   3 0 ,  3, , , 2, − , 3, , 4 , 3 , , 0 , , ,  4, ,  2 −1  3, , , 2, ∴ X =  1 3  – −, ,  −3 −2   3, − 7,  3, , 4 5 , = , ,  6 12 , , 2 x 5  4 5 , ∴ ,  = , ,  6 4 y   6 12 , , 4, ,  2− 3, , 2, X =  1+, , 3, ,  −3 + 7, , 3, , ∴ Using definition of equality of matrices,, we have, 2x = 4, 4y = 12 ∴ x = 2, y = 3,  2 −1, , , Ex. 5) If X + Y =  1 3  and,  −3 −2 , X – 2Y, , 2  2, −1 +  , 3, 3,  , 4,  5, 3−  = , 3, 3,   2, −2 + 0   −,   3, , 1, − , 3, , 5 , 3 , , −2 , , , EXERCISE 4.5, ,  −2 1 , , , =  3 −1 then find X ,Y.,  4 −2 ,  2 −1, , , Solution : Let A =  1 3  and B =,  −3 −2 , , 2, − , 3, , 4 , 3 , , 0 , , ,  2 −3, , , (1) If A =  5 −4  , B =,  −6 1 , ,  −2 1 ,  3 −1, , ,  4 −2 , ,  −1 2 ,  2 2,  and , ,  0 3 , ,  4 3, , , C =  −1 4  Show that (i) A +B = B+A,  −2 1 , , Let, X + Y = A …. (1), X – 2Y = B ……(2), , Solving (1) and (2) for X and Y, , (ii) (A+B)+C = A+(B+C), 86

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1 −3,  4 −7  , then find the, , , ,  i 2i ,  2i i , (8) If A = , and B = , ,  , where,  −3 2 ,  2 −3, , matrix A –2B+6I, where I is the unit matrix, of order 2., , -1 = i, find A+B and A–B. Show that A+B is, a singular. Is A–B a singular ? Justify your, answer., , 1 −2 , (2) If A = ,  , B =, 5 3 , ,  1 2 −3, , , (3) If A =  −3 7 −8 , B =,  0 −6 1 , ,  9 −1 2 ,  −4 2 5 , , ,  4 0 −3, ,  2 x + y −1 1 , (9) Find x and y, if , 4 y 4 ,  3,  −1 6 4   3 5 5 , + ,  = , ,  3 0 3  6 18 7 , , then find the matrix C such that A+B+C is a, zero matrix.,  1 −2 , , , (4) If A =  3 −5 , B =,  −6 0 , ,  2a + b 3a − b , 2 3 , (10) If = , = , ,  , find a, b,,  c + 2 d 2c − d ,  4 −1, ,  −1 −2 , 4 2,  and , ,  1 5 , , c and d., (11) There are two book shops owned by Suresh, and Ganesh. Their sales (in Rupees) for, books in three subject – Physics, Chemistry, and Mathematics for two months, July and, August 2017 are given by two matrices, A and B., , 2 4, , , C = =  −1 −4  , find the matrix X such that,  −3 6 , 3A –4B+5X = C., (5) Solve the following equations for X and Y, if,  1 −1, 3X–Y= ,  and X–3Y=,  −1 1 , , July sales (in Rupees), Physics Chemistry , Mathematics., , 0 −1, 0 −1, , , , 5600 6750 8500 , A= ,  First Row Suresh/, 6650 7055 8905, Second Row Ganesh, , (6) Find matrices A and B, if,  6 −6 0 , 2A–B = ,  and,  −4 2 1 , , August sales (in Rupees), Physics Chemistry , Mathematics , , 3 2 8, A–2B = , ,  −2 1 −7 , , 6650 7055 8905 , B =,  First Row, 7000 7500 10200 , Suresh/ Second Row Ganesh then,, , (7) Simplify,, , (i) Find the increase in sales in Rupees from, July to August 2017., ,  cosθ sinθ , cosθ ,  +,  −sinθ cosθ , , , , (ii) If both book shops got 10 % profit in the, month of August 2017, find the profit for, each book seller in each subject in that, month., ,  sinθ −cosθ , sinθ , , cosθ sinθ , , 87

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(4) Algebra of Matrices (continued), Two Matrices A and B are said to be, conformable for the product AB if the number of, columns in A is equal to the number of rows in B., i.e. A is of order mxn and B is of order nxp., In This case the product AB is amatrix, defined as follows., Am×n × Bn×p = Cm×p , where Cij =, , n, , ∑a, k =1, , Solution : Product AB is defined and order of, AB is 1., 3,  , ∴ AB = [1 3 2]  2  = [1 × 3 + 3 × 2 + 2 × 1] , 1 , = [11]1×1, , b, , ik kj, , a11, a,  21, If A = [ aik]m×n = a 31, , a i1, a,  m1, , a12 ... a1k ... a1n , a 22 ... a 2k ... a 2n , a 32 ... a 3k... a 3n , , th, a i2 ... a ik ... a in → i row, a m2 ... a mk ... a mn , ,  b11, ,  b 21, B = [ bkj]n×p = , b,  31,  b p1, , b12 ... b1j ... b1p , , b 22 ... b 2j ... b 2p , b32 ... b3j ... b3p , , b p2 ... b nj ... b np , , then, , 3,  , Ex.2 : Let A= [1 3 2]1×3 and B =  2  , find AB., 1  3×1, Does BA exist? If yes, find it., , Again since number of column of B = number of, rows of A=1, ∴ The product BA also is defined and order of, BA is 3., 3,   [1 3 2], BA =  2  [1, 3 2], 1×31×3=, 1  3×1, ,  3 ×1 3 × 3 3 × 2 ,  2 ×1 2 × 3 2 × 2, , , 1×1 1× 3 1× 2  3×3, , 3 9 6, , , = 2 6 4, 1 3 2  3×3, , ↓, jth column, Cij = ai1 b1j + ai2 b2j + ........ + ain bnj, , Remark : Here AB and BA both are defined but, AB ≠ BA.,  −1 −2 , 1 2, , , Ex.3 : A =  −3 2  , B = , −1 −2  2×2, ,  1 0  3×2, , SOLVED EXAMPLES,  b11 ,  , Ex.1 : If A = [a11 a12 a13] and B = b21 , b31 , Find AB., , Find AB and BA which ever exist., , Solution : Since number of columns of, A = number of rows of B = 3, ,  −1 −2 ,  1 2 , , ∴ AB =  −3 2  , −1 −2 , ,  1 0 , , Solution : Here A is order of 3 × 2 and B is of, order 2 × 2. By conformability of product, AB is, defined but BA is not defined., , Therefore product AB is defined and its order, is 1. (A)1×3 (B)3×1 = (AB)1×1, , 1,  −1 + 2 −2 + 4 ,  −3 − 2 −6 − 4 , 5, = ,  =, 1,  1 + 0, 2 + 0 , , AB = [a11 × b11 + a12 × b21 + a13 × b31], 88, , 2, 10, 2

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Note :, ,  3 2 −1, Ex.4 : Let A = ,  ,,  −2 5 4  2×3, , From the above solved numerical Examples, for, the given matrices A and B we note that, , i) If AB exists , BA may or may not exist., ii) If BA exists , AB may or may not exist., iii) If AB and BA both exist they may not, be equal. , ,  3 −3, B= , ,  −4 2  2×2, Find AB and BA which ever exist., Solution : Since number of columns of A ≠, number of rows of B. Product of AB is not, defined. But number of columns of B = number, of rows of A = 2, the product BA exists,, , 4.6 Properties of Matrix Multiplication :, 1), ,  3 −3  3 2 −1, ∴ BA = ,  , ,  −4 2   −2 5 4 , , 2), , 6 − 15 −3 − 12 ,  9+6, = , ,  −12 − 4 −8 + 10 4 + 8 , , 1 2 , 1 −1 2 , e.g. Let A= , ,B= , ,  ,, 4 3, 0 −1 3 , ,  15 −9 −15, = , ,  −16 2 12 , ,  −2 1 , , , C=  3 −1,  0 2 , ,  4 −3,  −1 3 , Ex.5 : Let A = , and B = , ,  Find, 5 2 ,  4 −2 , AB and BA which ever exist., , 1 2  1 −1 2 , Then AB = , , ,  4 3  0 −1 3 , , Solution : Since A and B are two matrix of same, order 2 × 2., , 1 + 0 −1 − 2 2 + 6  1 −3 8 , = ,  = , ,  4 + 0 −4 − 3 8 + 9   4 −7 17 , , ∴ Both the product AB and BA exist and are of, same order 2 × 2, ,  −2 1 , 1 −3 8  , 3 −1, (AB)C = , , ,  4 −7 17   0 2 , , , ,  4 −3  −1 3 , AB = ,  , ,  5 2   4 −2 ,  −4 − 12 12 + 6   −16 18, = , ,  = ,  −5 + 8 15 − 4   3 11,  −1 3   4 −3, BA = , =, ,  4 −2   5 2 , , For matrices A and B, matrix, multiplication is not commutative that, is AB ≠ BA., For three matrices A,B,C. Matrix, multiplication is associative. That is, (AB)C = A(BC) if orders of matrices, are suitable for multiplication., ,  −2 − 9 + 0 1 + 3 + 16   −11 20 , = ,  = , ,  −8 − 21 + 0 4 + 7 + 34   −29 45, …..(1),  −2 1 , 1 −1 2  , 3 −1, ∴ BC = , , , 0 −1 3   0 2 , , , ,  −4 + 15 3 + 6 , 16 − 10 −12 − 4 , , , , 11 9 , = , ,  6 −16 , Here AB ≠ BA, , , 89, ,  −2 − 3 + 0 1 + 1 + 4   −5 6 , = ,  = , ,  0 − 3 + 0 0 + 1 + 6   −3 7 

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5) For any matrix A there exists a null, matrix O such that a) AO = O and, b) OA = O., , 1 2   −5 6 , A(BC) = , , ,  4 3   −3 7 ,  −5 − 6 6 + 14   −11 20 , = ,  = ,  ..…(2),  −20 − 9 24 + 21  −29 45, , 6) The product of two non zero matrices, can be a zero matrix. That is AB = O, but A ≠ O, B ≠ O, , From (1) and (2), (AB)C = A(BC), , 1 0, e.g. Let A = ,  , B =, 2 0, , 3) For three matrices A,B,C, multiplication, is distributive over addition., i) A(B+C)= AB + AC , (left distributive law), , 1 0  0 0 , Here A ≠ 0, B ≠ 0 but AB= , , ,  2 0   −1 1 , , ii) (B+C)A= BA + CA , (right distributive law), , 0 0 , That is AB = ,  =O, 0 0 , , These laws can be verified by examples., 4), ,  0 0,  −1 1  ,, , , , 7) Positive integral powers of a square, matrix A are obtained by repeated, multiplication of A by itself. That is, A2 = AA, A3 = AAA, ……,, , For a given square matrix A there exists, a unit matrix I of the same order as that, of A, such that AI=IA=A., I is called Identity matrix for matrix, multiplication., , An = AA … n times, (Activity), ,  3 −2 −1, , , e.g. Let A =  2 0 4  ,, 1 3 2 , , 1 −5,  2 −3, If A = , , B = , , ,, 6 7 , 4 8 , Find AB-2I, where I is unit matrix of, order 2., , 1 0 0 , , , and I = 0 1 0 , 0 0 1 , , 1 −5,  2 −3, Solution : Given A = , , B = , , , 6 7 , 4 8 , ,  3 −2 −1 1 0 0 , ,  , , Then AI =  2 0 4  0 1 0 , 1 3 2  0 0 1 , , 1 −5  2 −3, Consider AB−2I = ,  , –2, 6 7   4 8 , , ... ..., ... ..., , , , −3 − 40  ... 0 ,  ......, ∴AB−2I = , –, ......   0 ..., 12 + 28, , 3 + 0 + 0 0 − 2 + 0 0 + 0 − 1 , , , = 2 + 0 + 0 0 + 0 + 0 0 + 0 + 4, 1 + 0 + 0 0 + 3 + 0 0 + 0 + 2 , , , ,  3 −2 −1, , , =  2 0 4  = IA = A, 1 3 2 , ,  ... −43 ... 0 , = ,  – , ,  40 ...   0 ..., ,  … −43, ∴AB−2I = , ,  40 … , 90

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SOLVED EXAMPLES, , , ,  −2 1 , 1 −1 2 , , , Ex. 1: If A= ,  , B =  3 −1 ,, 0, −, 1, 3, , ,  0 2 , show that matrix AB is non singular., , 14 0 0 , , , ∴ A2 –5A =  0 14 0  = 14I,  0 0 14 , ∴ A2–5A is a scalar matrix., ,  −2 1 , 1 −1 2  , 3 −1, Solution : let AB = , , , 0 −1 3   0 2 , , , , ,  −2 − 3 + 0 1 + 1 + 4 , = , ,  0 − 3 + 0 0 + 1 + 6, , , ,  −5 6 , = , ,,  −3 7 , , , , ∴|AB| =, ,  3 −2 , 2, Ex. 3 : If A= ,  , Find k, so that A –kA+2I,  4 −2 , = O, where I is an identity matrix and O is null, matrix of order 2., Solution : Given A2 –kA+2I = O,  3 −2   3 −2 , ∴Here, A2 = AA = ,  , ,  4 −2   4 −2 , , -5 6, -3 7, ,  9 − 8 −6 + 4  1 −2 , = ,  = , , 12 − 8 −8 + 4   4 −4 , , = –35 + 18 = –17 ≠ 0, ∴ matrix AB is nonsingular., , ∴ A2–kA+2I = O, 1 −2 , ∴ ,  –k,  4 −4 , , 1 3 3, , , Ex. 2 : If A = 3 1 3 prove that A2–5A is a, scalar matrix. 3 3 1, ,  3 −2 ,  4 −2  + 2, , , , 1 −2   3k, ∴ ,  –,  4 −4   4k, , Solution : Let A2 = A.A, , 1 0 , 0 1  = O, , , , −2k   2 0  0 0 , +, =, −2k   0 2  0 0 , , −2 + 2k  0 0 , 1 − 3k + 2, ∴ , =, −4 + 2k + 2  0 0 ,  4 − 4k, , 1 3 3 1 3 3,  , , , =  3 1 3  3 1 3, 3 3 1 3 3 1, , ∴ Using definition of equality of matrices, we, have, 1 – 3k + 2 = 0, ∴ 3k=3, –2 + k = 0, ∴ 2k=2 k=1, 4 – 4k = 0, ∴ 4k=4 , –4 + 2k + 2=0, ∴ 2k=2, , 1 + 9 + 9 3 + 3 + 9 3 + 9 + 3, = 3 + 3 + 9 9 + 1 + 9 9 + 3 + 3, 3 + 9 + 3 9 + 3 + 3 9 + 9 + 1, 19 15 15, = 15 19 15, 15 15 19 , 19 15 15, , , ∴ A2 –5A = 15 19 15 – 5, 15 15 19 , , 19 15 15  5 15 15, , , = 15 19 15 – 15 5 15, , , 15 15 19  15 15 5 , , Ex. 4 : Find x and y, if,   6 3,  −4 −1 , , , , [2 0 3] 3 −1 2 + 2  1 0   = [x y], , , , ,   5 4,  , , −, −, 3, 4, , ,  , , 1 3 3,  3 1 3, , , 3 3 1, 91

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Solution :, , an, a 0, n, Ex. 6 : If A = ,  , prove that A =  0, , 0 b, for all n∈N., a 0, Solution : Given A= , , 0 b, ,   6 3,  −4 −1 , , , , Given [2 0 3] 3 −1 2 + 2  1 0   = [x y], , , , ,   5 4,  , , −, −, 3, 4, , ,  ,  18 9   −8 −2  , , ,  , ∴ [2 0 3]   −3 6  +  2 0   = [x y],  15 12   −6 −8 , ,  , , , an, We prove An = , 0, , 10 7 , , , ∴ [2 0 3]  −1 6  = [x y] ,  9 4 , , an, Let P(n) be = , 0, , ∴ [20 + 27 14 + 12] = [x y], , a 0, P(1) is A1 = A= , , 0 b, , ∴ [47 26] = [x y] ∴, , 0,  for all n∈N using, bn , , mathematical induction, 0,  for n∈N., bn , , To prove that P(n) is true for n=1, , x = 47 , y = 26 by, , ∴ P(1) is true., , definition of equality of matrices., , Assume that P(K) is true for some K∈N, ,  sinθ , Ex. 5 : Find if ,  [sinθ cosθ, cosθ ,  θ , , a K, That is P(K) is A = , 0, ,  sinθ , , , Solution : Let cosθ  [sinθ cosθ,  θ , , 0, , bn , , k, , θ] = [17], , 0, , bK , , To prove that P(K) → P(K+1) is true consider, L.H.S. of P(K+1), That is Ak+1, θ] = [17],, , = Ak.A, a K, =, 0, , ∴ [sin2θ + cos2θ + θ2 ] = [17], ∴ By definition of equality of matries, , 0   a 0   a K +1 + 0, 0+0 , =, , , , b K   0 b   0 + 0, b K +1 + 0 , ,  a K +1, 0 , = ,  = R.H.S of P(K+1), b K +1 ,  0, , ∴ 1 + θ2 = 17 ∴ θ2 = 17–1 ∴ θ2 = 16,, ∴ θ = ± 4, , Hence P(K+1) is true., , Remark, Using the distributive laws discussed earlier, we can derive the following results,, , ∴ P(K) ⇒ P(K+1) for all K∈N, Hence by principle of mathematical induction,, the statement P(n) is true for all n∈N., , If A and B are square matrices of the same, order, then, , That is P(n) is true → P(2) is true → P(3) is true , and so on → P(n)is true, n∈N., , i) (A + B)2 = A2 + AB + BA + B2, ii) (A – B)2 = A2 – AB – BA + B2, , an, ∴=A , 0, n, , iii) (A + B) (A– B)= A2 + AB – BA – B2, 92, , 0,  for all n∈N., bn 

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Ex. 7 : A school purchased 8 dozen Mathematics, books, 7 dozen Physics books and 10 dozen, Chemistry books, the prices are Rs.50,Rs.40, and Rs.60 per book respectively. Find the total, amount that the book seller will receive from, school authority using matrix multiplication., , Ex. 8 : Some schools send their students for extra, training in Kabaddi, Cricket and Tennis to a sports, standidium. There center charge fee is changed, pen student for Coaching as well as 4 equipment, and maintenances of the court. The information of, students from each school is given below-, , Solution : Let A be the column matrix of books, of different subjects and let B be the row matrix, of prices of one book of each subject., , Modern School, Progressive School, Sharada Sadan, Vidya Niketan, , Kabaddi Cricket Tennis, , 35, 15 ,  20,  18, 36, 12 , ,  24, 12, 8 , , , 20, 6 ,  25, The charges per student for each game are, given below-, ,  8 × 12   96 ,   , , A =  7 × 12  =  84  B = [50 40 60], 10 ×12  120 , ∴ The total amount received by the, bookseller is obtained by matrix BA., , , E and M is for, E & M equipment and, Kabaddi Coach, , 10  maintain, Cricket  40 ,  50, 50 , Tennis ,  60, 40 , ,  96 , ∴ BA = [50 40 60]  ,  84 , 120 , = [50 × 96 + 40 × 84 + 60 × 120], = [4800 + 3360 + 7200] = [15360], Thus the amount received by the bookseller, from the school is Rs.15360., , To find the expense of each school on Coaching and E and M can be fourd by multiplication of the above, matrics., Kabaddi, Cricket Tennis, Coach E & M, Modern, 35, 18 , Kab  40, 10 ,  20, , , , Progreesive, 36, 12  X cri  50, 50 ,  18,  24, , Sharda School, 12, 8 , Ten  60, 40 ,  25, Vidya Niketan, 20, 6 , , Modern, Progressive, Sharada, Vidya, , =, ,  20 × 40 + 36 × 50 + 18 × 60,  18 × 40 + 36 × 50 + 12 × 60,  24 × 40 + 12 × 50 + 8 × 60,  25 × 40 + 20 × 50 + 6 × 60, ,  800 + 1750 + 108,  720 + 1800 + 720,  960 + 600 + 480, 1000 + 1000 + 360, , 200 + 1750 + 720 , , 180 + 1800 + 480 =, , 240 + 600 + 320 , , 250 + 1000 + 240 , , 20 × 10 + 35 × 50 + 18 × 40 , 18 × 10 + 36 × 50 + 12 × 40 , 24 × 10 + 12 × 50 + 8 × 40 , 25 × 10 + 20 × 50 + 6 × 40 , ,  2650,  3240,  2040,  2360, 93, , 2670 , 2960 , 1160 , 1490 

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 2 −2 ,  2 4 3, ii) A = , ,, B, =,  3 3  and, ,  −1 3 2 , , ,  −1 1 , 3 1 , C= , ., 1 3, , EXERCISE 4.6, , 1), , 4, 3,  2  2 −4 3, Evaluate i)   [, ] ii) [ 2 −1 3]  3, 1 , 1 , , 2), , 1 −3, If A = ,  ,B=, 4 2 , , 7), , 4 1 ,  3 −2  show that, , , , i), , AB ≠ BA., , 3), ,  −1 1 1 , , , If A =  2 3 0  ,,  1 −3 1 , , , , 2 1 4, , , B =  3 0 2  . State whether AB=BA?, 1 2 1 , , 4), , Show that AB=BA where,, , i), ,  −2 3 −1, , , A =  −1 2 −1 , B =,  −6 9 −4 , , 5), 6), , i), ,  −1 1 ,  3 −2  and, , , , 1 −1 3 , ii) A = , , B=, 2 3 2, ,  1 0,  −2 3,  and, ,  4 3, , 1 2, , , C =  −2 0 ,  4 −3, , 1 3 −1,  2 2 −1, , ,  3 0 −1, , sinθ , cos , , B=, , cosθ , sin , , 8), , 1 −2 , 3 −1, If A = , , B= , , ,, 5 6 , 3 7 , Find AB-2I,where I is unit matrix of order 2., , sin, cos , , 9), , 4 8, 2, If A = ,  , prove that A = 0., −, 2, −, 4, , , Verify A(BC) = (AB)C in each of the, following cases., 1 0 1 , , , A = 2 3 0 , B =,  0 4 5 , ,  4 −2 , A= , , B =, 2 3 , 4 1 , C= , ,  2 −1, , Justify your answer., , cosθ, ii) A = ,  sinθ, , Verify that A(B+C)=AB+BC in each of the, following matrices, , 1 2,  4 3 2, , , If A = , , B =  −1 0  show, ,  −1 2 0 ,  1 −2 , that matrix AB is non singular., , 1 2 0 , , , 10) If A = 5 4 2  , find the product, 0 7 −3, (A+I)(A−I)., ,  2 −2 ,  −1 1 ,  and, ,  0 3 , , ∝ 0 , 11) A = , , B=,  1 1, ,  3 2 −1, C= , ,  2 0 −2 , 94, , 1 0, 2,  2 1  find α, if A = B., , 

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21) Find x and y, if, , 1 2 2 , , , 12) If A=  2 1 2  , Show that A2–4A is a,  2 2 1 , scalar matrix., , 2,   2 −1 3   3 −3 4      x , 4 , −,    −1 =  ,  1 0 2  2 1 1    1   y ,  , 22) Find x, y, z if, , 1 0, 13) If A = ,  , find k so that,  −1 7 , A2 –8A –kI = O, where I is a unit matrix and, O is a null matrix of order 2., , , ,  cosα, 23) If A = ,  − sinα, ,  8 4,  5 −4 , 14) If A = , , B = , ,  show that, 10 5 , 10 −8, , , , (A + B) = A + AB + B ., 2, , 2, , 2, , 25) Jay and Ram are two friends in a class. Jay, wanted to buy 4 pens and 8 notebooks, Ram, wanted to buy 5 pens and 12 notebooks., Both of them went to a shop. The price, of a pen and a notebook which they have, selected was Rs.6 and Rs.10. Using Matrix, multiplication, find the amount required, from each one of them., ,  2 1,  −1 2  , show that, , , , (A+B)(A-B)= A2– B2., 1 2,  2 a, 17) If A = , , B = , ,  and if ,  −1 −2 ,  −1 b , (A + B)2 = A2+B2. find values of a and b., , 4.7 Properties of transpose of a matrix :, , 18) Find matrix X such that AX=B, where,  1 −2 , A= ,  and B =,  −2 1 , , sin 2α , ., cos 2α , , AB ≠ BA, but |AB| = |A|.|B|, , where I is unit matrix of order 2., , , ,  cos 2α, A2 = ,  − sin 2α, , sinα , , show that, cosα , , 1 2 , 0 4 , 24) If A = , ,B= , ,  , show that, 3 5 ,  2 −1, ,  3 1, 2, 15) If A = ,  , prove that A – 5A + 7I=0,,  −1 2 , ,  3 4, 16) If A = ,  and B =,  −4 3 , ,  2 0,  1 1 ,  x − 3,  1  ,  , , , , , 3  0 2  − 4  −1 2     =  y − 1 ., 2,  2 2,  3 1      2 z , ,  , , Note :, ,  −3,  −1 .,  , , (1) For any matrix A, (AT)T = A., (2) If A is a matrix and k is constant,, then (kA)T = kAT, ,  3 −2 , 2, 19) Find k, if A= ,  and if A = kA-2I., 4, 2, −, , , , (3) If A and B are two matrices of same order,, then (A + B)T = AT + BT, (4) If A and B are conformable for the product, AB, then (AB)T = BT AT, 2 3, 1 2 1, , , Example, Let A = , , B = 1 2  ,, ,  3 1 3, 1 2 , , 1 2 3  1 ,   , , 20) Find x, if [1 x 1]  4 5 6   −2  = 0.,  3 2 5   3 , , 95

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∴ AB is defined and, , (7) If A is a square matrix, then (a) A + AT is, symmetric. (b) A – AT is skew symmetric., , 2 + 2 + 1 3 + 4 + 2  5 9 , AB = ,  = , ,,  6 + 1 + 3 9 + 2 + 6  10 17 , , 3 5 7 , , , For example, (a) Let A =  2 4 −6  ,,  3 8 −5, , 5 10 , ∴ (AB)T = ,  ……… (1), 9 17 , , 3 2 3 , , , ∴ AT =  5 4 8 , 7 −6 −5, ,  1 3, 2 1 1, , , Now AT =  2 1 , BT = , , , 3 2 2 , , 1 3,  1 3, 2 1 1 , , , ∴ BT AT = ,   2 1, 3, 2, 2, ,   1 3, , , , 3 5 7 , , , Now A + AT =  2 4 −6  +,  3 8 −5, ,  2 + 2 +1 6 +1+ 3 , ∴ BT AT = , , 3 + 4 + 2 9 + 2 + 6 , , , , 5 10 , = ,  …….. (2), 9 17 , T, ,  6 7 10 , , , = 7 8 2 , 10 2 −10 , , ∴ A + AT is a symmetric matrix, by definition., , ∴ From (1) and (2) we have, (AB) = B A, T, , 3 2 3 , 5 4 8 , , , 7 −6 −5, , T, , 3 5 7 , , , (b) Let A – AT =  2 4 −6  –,  3 8 −5, , In general (A1 A2 A3, ....... An)T = AnT .... A3T A2T, A 1T, (5) If A is a symmetric matrix, then AT = A., , 3 2 3 , 5 4 8 , , , 7 −6 −5, , 4 , 0 3,  −3 0 −14 , = , ,  −4 14 0 , ,  2 −3 4 , , , For example, let A =  −3 5 −2 ,  4 −2 1 ,  2 −3 4 , , , AT =  −3 5 −2  = A,  4 −2 1 , , A − AT is a skew symmetric matrix, by definition., Note:, A square matrix A can be expressed as the, sum of a symmetric and a skew symmetric matrix, 1, 1, as A = (A + AT) + (A –AT ), 2, 2, , (6) If A is a skew symmetric matrix, then, AT = – A., 0 5 4, , , For example, let A =  −5 0 −2 ,  −4 2 0 , ,  4 −5 3 , , , e.g. Let A =  −6 2 1  ,,  7 8 −9 , , 0 5 4,  0 −5 −4 , , , , , ∴ AT =  5 0 2  = −  −5 0 −2 ,  −4 2 0 ,  4 −2 0 , ,  4 −6 7 , , , ∴ AT =  −5 2 8 ,  3 1 −9 , , = –A, ∴ AT = –A, 96

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 4 −5 3   4 −6 7 ,  , , , A + AT =  −6 2 1  +  −5 2 8 ,  7 8 −9   3 1 −9 , , EXERCISE 4.7, (1) Find AT, if (i) A =  1 3,  −4 5, , , ,  8 −11 10 , , 9 , =  −11 4,  10, 9 −18, ,  2 −6 1, (ii) A = , ,  −4 0 5, ,  8 −11 10 , 1, 1,  −11 4, 9 , Let P =, (A + AT) =, , 2, 2,  10, 9 −18, , (2) If [ aij]3x3 where aij = 2(i–j). Find A and AT., State whether A and AT are symmetric or, skew symmetric matrices ?, , 11, , , −, 5,  4, 2, , =  11, 9, −, 2,  2, 2, , , 9,  5, −9 , , , 2, ,  5 −3, , , (3) If A =  4 −3 , Prove that (2A)T = 2AT.,  −2 1 ,  1 2 −5, , , (4) If A =  2 −3 4  , Prove that (3A)T,  −5 4 9 , T, = 3A ., , The matrix P is a symmetric matrix.,  4 −5 3   4 −6 7 ,  , , , Also A – AT =  −6 2 1  –  −5 2 8 ,  7 8 −9   3 1 −9 , , , 1 + 2i i − 2 ,  0,  −1 − 2i, 0, −7 , (5) If A = ,  2 − i, 7, 0 , ,  0 1 −4 , , , =  −1 0 −7 ,  4 7 0 , , where i =, ,  0 1 −4 , 1, 1 , T, −1 0 −7 , Let Q =, (A – A ) =, , 2, 2,  4 7 0 , ,  0, , 1, = −,  2, ,  2, , , 1, 2, 0, 7, 2, , -1 ,, Prove that AT = –A., ,  2 −3, , , (6) If A =  5 −4  , B =,  −6 1 , , , −2 , , 7, −, 2, , 0 , , , 2 1,  4 −1,  and, ,  −3 3 , ,  1 2, , , C =  −1 4  then show that,  −2 3 , (i) (A + B) = AT + BT (ii) (A – C)T = AT – CT,  5 4, (7) If A = ,  and B =,  −2 3 , , The matrix Q is a skew symmetric matrix., Since P+Q = symmetric matrix + skew symmetric, matrix., ,  −1 3 ,  4 −1 , then find, , , , CT, such that 3A – 2B + C = I, where I is the, unit matrix of order 2., , Thus A = P + Q., 97

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7 3 0 ,  0 −2 3 , (8) If A , ,B= , ,  then,  0 4 −2 ,  2 1 −4 , , Let's Remember, • The value of a determinant of order 3 × 3, , find (i) AT + 4BT (ii) 5 AT – 5BT., 1 0 1 , (9) If A = , , B =, 3 1 2 , , a1, a2, a3, ,  2 1 −4 ,  3 5 −2  and, , , , b1, b2, b3, , c1, c2 = a1(b2c3–b3c2)–b1(a2c3–a3c2)+, c3, c1(a2b3– a3b2), ,  0 2 3, C= ,  , verify that,  −1 −1 0 , , • The minors and cofactors of elements of a, determinant, , (A + 2B + 2C)T = AT + 2BT + 3CT., , ∆ =, ,  −1 2 1 , (10) If A = ,  and B =,  −3 2 −3, ,  2 1,  −3 2 ,  , prove, ,  −1 3 , , (11) Prove that A + AT is a symmetric and A – AT, is a skew symmetric matrix, where, , •, ,  5 2 −4 ,  3 −7 2 , , ,  4 −5 −3, , a13, a23, a33, , Properties of determinant, , Property (i) - The value of determinant remains, unchanged if its rows are turned into columns, and columns are turned into rows., Property (ii) - If any two rows (or columns) of, the determinant are interchanged then the value, of determinant changes its sign., , (12) Express the following matrices as the sum of, a symmetric and a skew symmetric matrix.,  4 −2 , (i) ,  (ii),  3 −5, , a12, a22, a32, , Minor Mij of the element aij is determinant, obtained by deleting the ith row and jjth column of, determinant D. The cofactor Cij of element aij is, Cij = (–1)i+j Mij, , that (A + BT )T = AT + B., ,  1 2 4, , , (i) A =  3 2 1  (ii) A =,  −2 −3 2 , , a11, a21, a31, ,  3 3 −1,  −2 −2 1 , , ,  −4 −5 2 , , Property (iii) - If any two rows (or columns), of a determinant are identical then the value of, determinant is zero, ,  2 −1,  0 3 −4 , (13) If A = , and B = , , ,,  2 −1 1 ,  3 −2 ,  4 1 , verify that (i) (AB)T = BT AT, , Property (iv) - If any element of a row (or column), of determinant is multiplied by a constant k then, the value of the new determinant is k times the, value of old determinant, , (ii) (BA)T = AT BT, , Property (v) - If each element of a row (or, column) is expressed as the sum of two numbers, then the determinant can be expressed as sum of, two determinants, ,  cosα sinα , T, (14) If A = ,  , show that A A = I,, −, sin, α, cos, α, , , where I is the unit matrix of order 2., 98

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- If a constant multiple of all, row (or column) is added to the, elements of any other row (or, the value of new determinant, the same as that of the original, , • Addition of matrices:, Matrices A = [aij] and B = [bij] are said to, conformable for addition if orders of A and B, are same., , Property (vii) - (Triangle property) - If each, element of a determinant above or below the main, diagonal is zero then the value of the determinant, is equal to the product of its diagonal elements., , • Multiplication of two matrices:, A and B are said to be conformable for the, multiplication if number of columns of A is, equal to the number of rows of B., , Property (vi), elements of a, corresponding, column) then, so obtained is, determinant., , A+B = [aij + bij] . The order of A+B is the same, as that of A and B., , • A system of linear equations, using Cramer’s, Rule has solution =, x, , That is If A = [aik]m×p and B = [bkj]p×n, then AB, is defined and AB = [cij]m×n where, , Dy, Dx, D, =, ,y, , z = z ; provided D≠0, D, D, D, , n, , cij =, , • Consistency of three equations., a1x +b1y+c1 = 0, a2x +b2y+c2 = 0, b1, b2, b3, , .bkj, , i =1,2,….,m, j =1,2,…..,n., , i) A + AT is a symmetric matrix., , c1, c2 = 0, c3, , ii) A − AT is a skew-symmetric matrix., • Every square matrix A can be expressed as, the sum of a symmetric and skew-symmetric, matrix as, 1, 1, [A + AT] +, [A – AT]., A=, 2, 2, , y1 1, y2 1, y3 1, , MISCELLANEOUS EXERCISE - 4 (B), , • Test for collinear of points (x1,y1), (x2,y2),, x1, (x3,y3) if x2, x3, , ik, , • If A is a square matrix, then, , • Area of triangle whose vertices are (x1, y1),, (x2,y2) (x3,y3) is, x1, 1, x, A (∆) =, 2 2, x3, , k =1, , • If A = [aij]m×n is any matrix, then the transpose, of A is denoted by AT = B = [bij]m×n and bji = aij, , a3x +b3y+c3 = 0 are consistent , a1, Then a2, a3, , ∑a, , (I) Select the correct option from the given, alternatives., , y1 1, y2 1 = 0, y3 1, , 1), , • Multiplication of a matrix by a scalar:, , 1 3 , 1 0 , Given A = , , I =, ,  if A–λI is a, 2 2, 0 1 , singular matrix then ………., , If A [aij] is a matrix and k is a scalar, then kA =, [kaij]., 99, , A) λ = 0, , B) λ2 – 3λ – 4 = 0, , C) λ2 + 3 – 4 = 0, , D) λ2 – 3λ – 6 = 0

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2), ,  4 6 −1, , , Consider the matrices A =  3 0 2  , , 1 −2 5 , , 7), ,  2 4, 3,  0 1,  , B =,  , C =1  out of the given,  −1 2 ,  2 , ,  3 1, A) ,  ,  4 3, , then the value of A is ……., , 6 2, C) ,  , 8 6 , , matrix product ………..., i), , (AB)TC, , ii) CTC(AB)T, , iii) CTAB, A), B), C), D), 3), , iv) ATABBTC, , 4), , 8), , Exactly one is defined, Exactly two are defined, Exactly three are defined, all four are defined , , 9), , B) A + B = A – B, D) AB = BA, , D) (A + B)T = AT + BT, , 6), ,  −2 1, 2, 10) If A = ,  and f(x) = 2x – 3x, then, 0, 3, , , , B) (–2, 1), D) (–2, –1), , B) ±2, , C) ±5, , f(A) = ………, , D) 0, , 5 7  1 2  4 5 ,  , , , If  x 1  –  −3 5  =  4 −4  then ……,  2 6   2 y   0 4 , A) x = 1, y = –2, C) x = 1, y = 2 , , For suitable matrices A, B, the false statement, is ……., , C) (A − B)T = AT − BT, , α 2 , 3, If A = ,  and |A | = 125,, 2, α, , , then α = ……., A) ±3, , x = 3, y = 7, z = 1, w = 14, x = 3, y = −5, z = −1, w = −4 , x = 3, y = 6, z = 2, w = 7, x = −3, y = –7, z = –1, w = –14, , B) (AT)T = A, , equation AAT = 9I, where I is the identity, matrix of order 3, then the ordered pair (a, b), is equal to ……., , 5), , 7 6 , D) , , 8 12 , , A) (AB)T = ATBT, , 1 2 2 , , , If A =  2 1 −2  is a matrix satisfying the,  a 2 b , , A) (2, –1), C) (2, 1), ,  4 3, B) , , 4 6, , 3x − y   3 2,  x, If ,  = ,  then ……….,  zx + z 3 y − w   4 7 , a), a), a), a), , If A and B are square matrices of equal, order, then which one is correct among the, following?, A) A + B = B + A, C) A – B = B – A, , 7 4 , 1 2 , If A + B = , and A – B = , , , 8 9 , 0 3, , 14 1 , A) , ,  0 −9 , ,  −14 1 , B) , ,  0 9, , 14 −1, C) , , 0 9, ,  −14 −1, D) , ,  0 −9 , , (II) Answer the following question., 1), , B) x = –1, y = 2, D) x = –1, y = –2, 100, , If A = diag [2 –3 –5], B = diag [4 –6 –3], and C= diag [–3 4 1] then find i) B + C – A, ii) 2A + B – 5C.

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2), , cosα, , If f (α) = A =  sinα,  0, i) f(–α), , 3), , − sinα, cosα, 0, , 0, 0  , Find, 1 , , 8), , show that BA=6I., , ii) f(–α) + f(α)., , Find matrices A and B, where, , 9), , 1 −1, 1 −1, i) 2A – B= , and A +3 B= , , , 0 1 , 0 1 , , ω 2 1 ,  1 ω, 11) If A =  2, ,B= ,  , where is, 1 ,  1 ω, ω, a complex cube root of unity, then show that, AB+BA+A−2B is a null matrix., ,  2 −3,  −3 4 1 , , , If A =  3 −2  , B = , ,  2 −1 −3,  −1 4 , ,  2 −2 −4 , , , 12) If A =  −1 3 4  show that A2 =A.,  1 −2 −3, , Verify i) (A + BT)T = AT + 2B., ii) (3A – 5BT)T = 3AT – 5B., If A = cosα,  sinα, , , − sinα  and A + AT = I, where, cosα , , 4, 13) If A = 3, 3, , I is unit matrix 2 × 2, then find the value of α., , 6), ,  1 2, , , If A=  3 2  and B =,  −1 0 , , 7), , 1, 0, 1, , 4, 4 , show that A2 = I., 3, ,  3 −5, 2, 14) If A = ,  , show that A – 5A – 14I = 0., −, 4, 2, , , , 1 3 2 ,  4 −1 −3 , show, , , ,  2 −1, 15) If A = , , show that A2– 4A + 3I = 0., ,  −1 2 , , that AB is singular., 1 2 3, , , If A =  2 4 6  , B=, 1 2 3, , sinα , ,, cosα , , show that Aα. Aβ = Aα+β., ,  0 0 1, A + 5B= , ,  −1 0 0 , , 5), ,  2 1, 1 2 , If A = , , B= , ,  , verify that,  0 3, 3 −2 , |AB|= |A||B|., ,  cosα, 10) If Aα = ,  − sinα, ,  −1 2 1, ii) 3A – B= ,  and,  1 0 5, , 4), , 1 −1 0 ,  2 2 −4 , , 2 3 4, , If A= ,  , B=  −4 2 −4  ,,  0 1 2 ,  2 −1 5 , ,  1 −1 1 ,  −3 2 −1,  , show, ,  −2 1 0 , ,  −3 2 , 16) If A = , ,B=,  2 −4 , , that AB and BA are both singular matrices., , 1 x,  y 0  , and, , , , (A + B)(A – B) = A2 – B2, find x and y., , 101

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 2 −4 ,  1 −1 2 , , , 22) If A =  3 −2  , B = , ,, −2 1 0 , ,  0 1 , , 0 1 , 0 −1, 17) If A= , and B = , ,  show that, 1 0 , 1 0 , (A + B)(A – B) ≠ A2 – B2., , show that (AB)T = BT AT., ,  2 −1, 3, 18) If A = ,  , find A ., 3, −, 2, , , , 3 −4 , 23) If A = ,  , prove that, 1 −1, 1 + 2n −4n , An = , , for all n∈N., 1 − 2n ,  n, , 19) Find x, y if,,  4 5  4 3  ,  , ,  , i) [0 –1 4] 2  3 6  + 3 1 4  ,   2 −1  0 −1 , ,  ,  , = [x, ,  cosθ, 24) If A = ,  − sinθ, , y]., ,  cosnθ, An = ,  − sinnθ, , ii) , 5,  1 2 1  2 −3 7      x , −1 ,  + 3 1 −1 3    0  =  , 2, 0, 3, y, ,  , , ,  −1  , , sinnθ , , for all n∈N., conθ , , 25) Two farmers Shantaram and Kantaram, cultivate three crops rice,wheat and, groundnut. The sale (In Rupees) of these, crops by both the farmers for the month of, April and may 2008 is given below, , , 20) Find x, y, z if,  0 1   2 1  ,  , ,  , i) 5 1 0  − 3  3 −2  ,  1 1   1 3  , ,  ,  , , sinθ , , prove that, cosθ , ,  x − 1, 2 , , 1  =  y + 1,    2z , , , , April sale (In Rs.), Rice Wheat, Shantaram 15000 13000, Kantaram 18000 15000, ,  1 3 2 ,  3 0 2   1   x ,     , , ii)  2 0 1  + 2 1 4 5    2  =  y , , , , , 3 1 2,    3   z , , 2, 1, 0, , , , , , Groundnut, 12000, 8000, , May sale (In Rs.), Rice , Shantaram 18000, Kantaram 21000, ,  2 1 −3, 1 0 −2 , 21) If A = , , B = , ,  ,, 0 2 6 , 3 −1 4 , , Wheat, 15000, 16500, , Groundnut, 12000, 16000, , Find, i) The total sale in rupees for two months of, each farmer for each crop., , find ABT and ATB., , ii) the increase in sale from April to May for, every crop of each farmer., , 102

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5, , Straight Line, , Let's Study, •, •, •, •, •, , Let's Learn, , Locus of a points in a co-ordinate plane, Equations of line in different forms, Angle between two lines, perpendicular, and parallel lines, Distance of a point from a line, Family of lines, , 5.1 Locus : A set of points in a plane, which satisfy certain geometrical condition, (or conditions) is called a locus., L = {P | P is a point in the plane and P, satisfies given geometrical condition}, Here P is the representative of all points in L., L is called the locus of point P. Locus is a, set of points., , Let's Recall, , The locus can also be described as the route of, a point which moves while satisfying required, conditions. eg. planets in solar system., , We are familiar with the properties of, straight lines, the bisector of an angle, circle, and triangles etc., We will now introduce co-ordinate, geometry in the study of a plane. Every point, has got pair of co-ordinates and every pair of, co-ordinate gives us a point in the plane., We will use this and study the curves with, the help of co-ordinates of the points., What is the perpendicular bisector of a, segment  ? ............... A line., , Illustration :, • The perpendicular bisector of segment AB, is the set, M = { P | P is a point such that PA=PB}., • The bisector of angle AOB is the set :, D = { P | P is a point such that, P is equidistant from OA and OB }, = { P | ∠ POA = ∠ POB }, , What is the bisector of an angle ? .......A ray., , Verify that the sets defined above are the same., , These geometrical figures are sets of points, in plane which satisfy certain conditions., • The perpendicular bisector of a segment is, the set of points in the plane which are, equidistant from the end points of the, segment. This set is a line., • The bisector of an angle is the set of points, in the plane which are equidistant from the, arms of the angle. This set is a ray., , The plural of locus is loci., , Fig. 5.1, , Activity : Draw segment AB of length 6 cm., Plot a few points which are equidistant from, A and B. Verify that they are collinear., , • The circle with center O and radius 4 is, the set L = { P | OP = 4}, 103

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5.1.1 Equation of Locus : If the set of points,, whose co-ordinates satisfy a certain equation in, x and y, is the same as the set of points on, a locus, then the equation is said to be the, equation of the locus., , ∴, ∴, ∴, (y, ∴, ∴, ∴, , SOLVED EXAMPLES, , PA = PB, PA2 = PB2, (x + 3)2 + (y − 0)2 = (x − 3)2 +, −0)2, x2 + 6x + 9 + y2 = x2 − 6x + 9 + y2, 12x = 0, x = 0. The locus is the Y-axis., , 5.1.2 Shift of Origin : Let O′(h, k) be a point, in the XY plane and the origin be shifted to, O′. Let O′X′, O′Y′ be the new co-ordinate, axes through O′ and parallel to the axes OX, and OY respectively., , Ex.1 We know that the y co-ordinate of every, point on the X-axis is zero and this is true, for points on the X-axis only. Therefore, the equation of the X-axis is y = 0., Ex.2 Let L = {P | OP = 4}. Find the equation, of L., , Fig. 5.2, Solution : L is the locus of points in the, plane which are at 4 unit distance from, the origin., , Fig. 5.3, Let (x, y) be the co-ordinates of P referred to, the co-ordinates axes OX, OY and (x', y') be, the co-ordinates of P referred to the co-ordinate, axes O′X′, O′Y′. To find relations between, (x, y) and (x', y')., , Let P(x, y) be any point on the locus L., As OP = 4, OP2 = 16, ∴ (x − 0)2 + (y − 0)2 = 16, ∴ x2 + y2 = 16, , Draw PL ⊥ OX and suppose it intersects O′X′, in L′., , This is the equation of locus L., , Draw PM ⊥ OY and suppose it intersects O′Y′, in M′., , The locus is seen to be a circle, , Let O′Y′ meet line OX in N and O′X′ meet, OY in T., , Ex.3 Find the eqation of the locus of points, which are equidistant from A(−3, 0) and, B(3, 0). Identify the locus., , ∴ ON = h, OT = k, OL = x, OM = y,, , Solution : Let P(x, y) be any point on the, required locus., , O′L′ = x', O′M′ = y', Now x = OL = ON + NL = ON + O′ L′, , P is equidistant from A and B., 104

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Ex. 3) Obtain the new equation of the locus, x2 − xy − 2y2 − x + 4y + 2 = 0 when the, origin is shified to (2, 3), the directions of the, axes remaining the same., , = h + x', and y  =  OM = OT + TM = OT + O′ M′ =, k + y', ∴ x = x' + h, y = y' + k, These equations are known as the formulae for, shift of origin., , Solution : Here (h, k) = (2, 3) and if new, co-ordinates are (X,Y)., , Note that the new co-ordinates can also be, given by (X, Y) or (u,v) in place of (x',y')., , ∴ x = X + h,, , y = Y + k gives, , ∴ x = X + 2,, , y = Y + 3, , The given equation, x2 − xy − 2y2 − x + 4y + 2 = 0 becomes, (X+2)2 − (X + 2)(Y + 3) − 2(Y + 3)2 −, (X + 2) + 4(Y + 3) + 2 = 0, ∴ X2 − XY − 2Y2 − 10Y − 8 = 0, This is the new equation of the given locus., , SOLVED EXAMPLES, Ex. 1) If the origin is shifted to the point, O'(3, 2) the directions of the axes remaining, the same, find the new co-ordinates of the, points, (a)A(4, 6) (b) B(2,−5)., , EXERCISE 5.1, , Solution : We have (h, k) = (3,2), x = x' + h, y = y' + k, , 1., , ∴ x = x' + 3. and y = y' + 2 .......... (1), (a) (x, y) = (4, 6), ∴ From (1), we get, 1, y + 2, , PA = PB., 4 = x1 + 3, 6 =, , 2., , A(−5, 2) and B(4, 1). Find the equation of, the locus of point P, which is equidistant, from A and B., , 3., , If A(2, 0) and B(0, 3) are two points, find, the equation of the locus of point P such, that AP = 2BP., , 4., , If A(4, 1) and B(5, 4), find the equation, of the locus of point P if PA2 = 3PB2., , 5., , A(2, 4) and B(5, 8), find the equation of, the locus of point P such that, , ∴ x' = 1 and y' = 4., New co-ordinates of A are (1, 4), (ii) (x, y) = ( 2,-5), from (1), we get 2 = x' + 3, −5 = y' + 2, ∴ x' = −1 and y' = −7. New co-ordinates of, B are (−1,−7), Ex. 2), (−2, 1),, axes. If, (7, −4),, , If A(1,3) and B(2,1) are points, find the, equation of the locus of point P such that, , The origin is shifted to the point, the axes being parallel to the original, the new co-ordinates of point A are, find the old co-ordinates of point A., , PA2 − PB2 = 13., 6., , A(1, 6) and B(3, 5), find the equation of, the locus of point P such that segment, AB subtends right angle at P. (∠APB, = 90°), , 7., , If the origin is shifted to the point, O′(2, 3), the axes remaining parallel to the, original axes, find the new co-ordinates of, the points, , Solution : We have (h, k) = (−2, 1) and if, new co-ordinates are (X,Y)., x = X + h, y = Y + k, ∴ x = X − 2, y = Y +1, (X, Y) = (7, −4), we get x = 7 − 2 = 5, y = −4 + 1 = − 3., ∴ Old co-ordinates A are (5, −3), , (a) A(1, 3) (b) B(2, 5), 105

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8., , Let's Think :, , If the origin is shifted to the point, O′(1, 3) the axes remaining parallel to the, original axes, find the old co-ordinates of, the points, , • Can intercept of a line be zero ?, • Can intercept of a line be negative ?, 5.1.2 Inclination of a line : The smallest angle, made by a line with the positive direction of, the X-axis measured in anticlockwise sense is, called the inclination of the line. We denote, inclination by θ. Clearly 0° ≤ θ < 180°., , (a) C(5, 4) (b) D(3, 3), 9., , If the co-ordinates A(5, 14) change, to B(8, 3) by shift of origin, find the, co-ordinates of the point where the origin, is shifted., , Remark : Two lines are parallel if and only, if they have the same inclination., , 10. Obtain the new equations of the following, loci if the origin is shifted to the point, O'(2, 2), the direction of axes remaining, the same :, , The inclination of the X-axis and a line, parallel to the X-axis is Zero. The inclination, of the Y-axis and a line parallel to the Y-axis, is 900., , (a) 3x − y + 2 = 0, (b) x2 + y2 − 3x = 7, , 5.2.2 Slope of a line : If the inclination of, a line is θ then tanθ (if it exist) is called, the slope of the line. We denote it by m., ∴ m = tanθ., , (c) xy − 2x − 2y + 4 = 0, (d) y2 − 4x − 4y + 12 = 0, 5.2 Straight Line : The simplest locus in a, plane is a line. The characteristic property of, this locus is that if we find the slope of a, segment joining any two points on this locus,, then the slope is constant., If a line meets the X-axis in the point, A (a, 0), then 'a' is called the X-intercept of, the line. If it meets the Y-axis in the point, B(0, b) then 'b' is called the Y-intercept of, the line., , Activity : If A(x1, y1), B(x2, y2) are any two, points on a non-vertical line whose inclination, is θ then verify that, y −y, tan θ = 2 1 , where x1 ≠ x2., x2 − x1, The slope of the Y-axis is not defined., Similarly the slope of a line parallel to the, Y-axis is not defined. The slope of the X-axis, is 0. The slope of a line parallel to the X-axis, is also 0., Remark : Two lines are parallel if and only, if they have the same slope., SOLVED EXAMPLES, Ex. 1) Find the slope of the line whose, inclination is 60°., , Fig. 5.4, , Solution : The tangent ratio of the inclination, of a line is called the slope of the line., Inclination θ = 60°., , Remarks :, (1) A line parallel to X-axis has no X-intercept., (2) A line parallel to Y-axis has no Y-intercept., , ∴ slope = tanθ = tan60° =, 106, , 3.

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Ex. 2) Find, through the, Solution :, through the, by, , the slope of the line which passes, points A(2, 4) and B(5, 7)., The slope of the line passing, points (x1, y1) and (x2, y2) is given, , y2 − y1, x2 − x1, , Slope of the line AB =, Note that x1 ≠ x2., , α − β = 900 or α − β = −900, α − β = ± 900, ∴ cos(α − β ) = 0, ∴ cosα cosβ + sinα sinβ = 0, ∴ sinα sinβ = −cosα cosβ, ∴ tanα tanβ = −1, ∴ m1m2 = −1, , y2 − y1, 7−4, =, = 1, 5−2, x2 − x1, , Ex. 3) Find the slope of the line which passes, through the origin and the point A(- 4, 4)., Solution : The slope of the line passing, through the points (x1, y1) and (x2, y2) is given, by, m=, , y2 − y1, ., x2 − x1, , Fig. 5.5, , Fig. 5.6, , , , Here A(−4, 4) and O(0, 0)., Slope of the line OA =, Note that x1 ≠ x2., , y2 − y1, 0−4, =, = −1., 0, +4, x2 − x1, , SOLVED EXAMPLES, Ex. 1) Show that line AB is perpendicular to, line BC, where A(1, 2), B(2, 4) and C(0, 5)., Solution : Let slopes of lines AB and BC be, m1 and m2 respectively., , 5.2.3 Perpendicular Lines : We know that, the co-ordinate axes are perpendicular to each, other. Similarly a horizontal line and a vertical, line are perpendicular to each other. Slope of, one of them is zero whereas the slope of the, other one is not defined., , ∴ m1 =, m2 =, , Let us obtain a relation between slopes of nonvertical lines., , 4−2, = 2 and, 2 −1, , y2 − y1 5 − 4, 1, =, =−, x2 − x1 0 − 2, 2,  1, , Now m1 × m2 = 2 ×  −  = −1,  2, ∴ Line AB is perpendicular to line BC., , Theorem : Non-vertical lines having slopes m1, and m2 are perpendicular to each other if and, only if m1×m2= −1., , Ex. 2) A(1,2), B(2,3) and C(−2,5) are vertices, of ∆ ABC. Find the slope of the altitude drawn, from A., , Proof : Let α and β be inclinations of lines, having slopes m1 and m2. As lines are non, vertical α ≠ π and β ≠ π, 2, 2, , Solution : The slope of line BC is, m1 =, , tan α = m1 and tanβ = m2, ∴, From Fig. 5.5 and 5.6 we have,, , y2 − y1, 5−3, 2, 1, =, =− =−, x2 − x1 −2 − 2, 4, 2, , Altitude drawn from A is perpendicular to, BC., 107

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The y co-ordinate of every point on the X-axis, is 0 and this is true only for points on the, X-axis. Therefore, the equation of the X-axis, is y = 0, , EXERCISE 5.2, 1., , 2., , Find the slope of each of the following, lines which passes through the points :, (a) A(2,−1), B(4,3) (b) C(−2,3), D(5,7), (c) E(2,3), F(2,−1) (d) G(7,1), H(−3,1), , The x co-ordiante of every point on the, Y-axis is 0 and this is true only for points, on the Y-axis. Therefore, the equation of the, Y-axis is x = 0 ., , If the X and Y-intercepts of line L are 2, and 3 respectively then find the slope of, line L., , 3., , Find the slope of the line whose inclination, is 30°., , 4., , Find the slope of the line whose inclination, is π ., 4, , 5., , A line makes intercepts 3 and 3 on the, co-ordiante axes. Find the inclination of, the line., , 6., , Without using Pythagoras theorem show, that points A(4,4), B(3, 5) and C(−1, −1), are the vertices of a right angled triangle., , 7., , Find the slope of the line which makes, angle of 45° with the positive direction, of the Y-axis measured anticlockwise., , 8., , Find the value of k for which points, P(k,−1),Q(2,1) and R(4,5) are collinear., , 9., , Find the acute angle between the X-axis, and the line joining points A(3,−1) and, B(4,−2)., , The equation of any line parallel to the, Y-axis is of the type x = k (where k is a, constant) and the equation of any line parallel, to the X-axis is of the type y = k. This is all, about vertical and horizontal lines., Let us obtain equations of non-vertical, and non -horizontal lines in different forms:, 5.3.1 Point-slope Form : To find the equation, of the line having slope m and which passes, through the point A(x1, y1)., , Fig. 5.9, Proof : Let L be the line passing through the, point A(x1, y1) and which has slope m., , 10. A line passes through points A(x1, y1) and, B(h, k). If the slope of the line is m then, show that k − y1 = m(h − x1)., , Let P(x, y) be any point on the line L other, than A., , 11. If points A(h, 0), B(0, k) and C(a, b) lie, on a line then show that, , Then the slope of line L =, , a b, + = 1., h k, , y − y1, ., x − x1, , But the slope of line L is m. (given), , 5.3 Equation of line in standard forms : An, equation in x and y which is satisfied by the, co-ordinates of all points on a line and no, other points is called the equation of the line., , ∴, , y − y1, = m, x − x1, , ∴ The equation of the line L is, (y − y1) = m, 109, , (x − x1)

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The equation of the line having slope m, and passing through A(x1, y1) is, (y − y1) = m (x − x1)., , This is the equation of line L., The equation of line having slope m and which, makes intercept c on the Y-axis is y = mx + c., , Remark : In particular if the line passes, through the origin O(0,0) and has slope m,, then its equation is y − 0 = m (x − 0), ∴ y = mx, , Ex. Obtain the equation of line having slope, 3 and which makes intercept 4 on the Y-axis., Solution : The equation of line having slope, m and which makes intercept c on the Y-axis, is, y = mx + c., ∴ the equation of the line giving slope 3, and making Y-intercept 4 is y = 3x + 4., , Ex. Find the equation of the line passing, through the point A(2, 1) and having slope −3., Soln. : Given line passes through the point, A(2, 1) and slope of the line is −3., The equation of the line having slope m, and passing through A(x1, y1) is, (y − y1) = m (x − x1)., , 5.3.3 Two-points Form : To find the equation, of line which passes through points A(x1, y1) and, B(x2, y2)., , The equation of the required line is, y − 1 = −3(x − 2), ∴ y − 1 = −3x + 6, ∴ 3x + y − 7 = 0, 5.3.2 Slope-Intercept form : To find the, equation of line having slope m and which, makes intercept c on the Y-axis., , Fig. 5.11, Proof : A(x1, y1) and B(x2, y2) are the two, given points on the line L. Let P(x, y) be, any point on the line L, other than A and B., Now points A and P lie on the line L., The slope of line L =, , Fig. 5.10, , y − y1, .......... (1), x − x1, , Also points A and B lie on the line L., , Proof : Let L be the line with slope m and, which makes Y-intercept c. Line L meets the, Y-axis in the point C(0, c)., , ∴ Slope of line L =, From (1) and (2), , Let P(x, y) be any point on the line other than, C. Then the slope of the line L is, , ∴, , y−c, = m (given), x−0, , y2 − y1, .......... (2), x2 − x1, , we get, , x − x1, y − y1, =, x2 − x1 y2 − y1, , y2 − y1, y − y1, =, x2 − x1, x − x1, , This is the equation of line L., As line is non-vertical and non-horizontal,, x1 ≠ x2 and y1 ≠ y2., , y−c, = m, ∴, x, , ∴ y = mx + c, 110

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∴ The slope of the line L = Slope of AB, , b−0, −b, =, =, 0−a, a, −b, y, x y, =, ∴, ∴, + =1, x −a, a, a b, , The equation of the line which passes through, x − x1, y − y1, =, x2 − x1 y2 − y1, , points A(x1, y1) and B(x2, y2) is, , Ex. Obtain the equation of the line passing, through points A(2, 1) and B(1, 2)., Solution : The equation of the line which passes, through points A(x1, y1) and B(x2, y2) is, , This is the equation of line L., The equation of the line which makes, intercepts a and b on the co-ordiante axes, x y, is, + = 1 (a, b ≠ 0), a b, , x − x1, y − y1, =, ., x2 − x1 y2 − y1, , ∴ The equation of the line passing through, points A(2, 1) and B(1, 2) is, ∴, , x − 2 y −1, =, 1− 2 2 −1, , Ex. Obtain the, makes intercepts, axes., Solution : The, makes intercepts, x y, axes, + =1, a b, , x − 2 y −1, =, −1, 1, , ∴ x − 2 = −y + 1, ∴ x + y − 3 − 0, 5.3.4 Double-Intercept form : To find the, equation of the line which makes non-zero, intercepts a and b on the co-ordinate axes., , equation of the line which, 3 and 4 on the co-ordiante, equation of the line which, a and b on the co-ordiante, , The equation of the line which makes intercepts, x y, 3 and 4 on the co-ordiante axes is + = 1, 3 4, ∴ 4 x + 3 y − 12 = 0 ., 5.3.5 Normal Form : Let L be a line and, segment ON be the perpendicular (normal), drawn from the origin to line L., If ON = p and ray ON makes angle α with, the positive X −axis then to find the equation, of line L., , Fig. 5.12, Let a be the X-intercept and b be the, Y-intercept of line L., ∴ Line L meets the X-axis in A(a, 0) and the, Y-axis in B(0, b)., Let P(x, y) be any point on line L other than, A and B., ∴ The slope of line L = slope of line AP, , y−0, y, =, , =, x−a, , x−a, , Fig. 5.13, 111

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From Fig. 5.13 we observe that N is, , ∴, , (p cosa, p sina), Therefore slope of ON is p sin α − 0 = tan α, pco s α − 0, , ∴ 3x + y − 10 = 0, Ex. 2) Reduce the equation, 3x − y − 2 = 0, into normal form. Find the values of p and α ., , As ON ierpendicular to line L,, Slope of line L =, , 3, 1, x+ y =5, 2, 2, , tanq = - cot a, , Solution : Comparing, , 3x − y − 2 = 0 with, ax + by + c = 0 we get a = 3 , b = −1 and, , And it passes through (p cosa, p sina), ∴ By the point − slope form, the equation, of line L is,, − cos α, y − p sina =, ( x - p cosα ), sin α, , c = −2 ., a 2 + b2 = 3 + 1 = 2, Divide the given equation by 2., , y sin a - p sin2 a = x cos a + p cos2 a, x cos a + y sin a = (sin2 a cos2 a), , 3, 1, x − y =1, 2, 2, , x cos a + y sin a = p (p > o), , ∴ cos330° x=sin330° y = 1 is the required, normal form of the given equation., , The equation of the line, the normal to which, from the origin has length p and the normal, makes angle α with the positive directions of, the X-axis, is x cosα + y sinα = p., , p = 1 and θ = 330ο ., Ex. 3) Find the equation of the line :, , SOLVED EXAMPLES, , (i) parallel to the X -axis and 3 unit below it,, (ii) passing through the origin and having, inclination 30°, , Ex. 1) The perpendicular drawn from the origin, to a line has length 5 and the perpendicular, makes angle with the positive direction of the, X-axis. Find the equation of the line., , (iii) passing through the point A(5,2) and having, slope 6, , Solution : The perpendicular (normal) drawn, from the origin to a line has length 5., , (iv) passing through the points A(2-1) and, B(5,1), 3, (v) having slope − and y −intercept 5,, 4, (vi) making intercepts 3 and 6 on the co-ordinate, axes., , ∴ p=5, The perpendicular (normal) makes angle 30°, with the positive direction of the X-axis., ∴ θ = 30°, , (vii) passing through the point N ( −2,3) and, the segment of the line intercepted between, the co−ordinate axes is bisected at N., , The equation of the required line is x cosa +, y sina = p, ∴ x cos30° + y sin30° = p, 112

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Solution : (i) Equation of line parallel to the, X-axis is of the form : y = k ,, , (vi) By using the double intercept form, x y, + =1, a b, , ∴ the equation of the required line is, , y −intercept = b = 6 ., x y, the equation of the required line is, + =1, 3 6, 2x + y − 6 = 0, x −intercept = a = 3 ;, , y = 3, (ii) Equation of line through the origin and, having slope m is of the form : y = mx ., slope = m = tan = tan 300 =, , (vii) Let the given line meet the X −axis, in A(a,0) and the Y −axis in B(0,b)., , 3, , ∴ the equation of the required line is, y = 3x, , ∴, , The mid−point of AB is, , 3x − y = 0, , (iii) By using the point−slope form, ,  a +0 0+b,  2 , 2  = (−2, 3), , , , d, , y-y1 = m (x-x1), , ∴, , slope = m = -6, Equation of the required line is, , a, = - 2 and, 2, , ∴ a = - 4, , (y-2) = -6 (x+5), , b, 2, , = 3, , b = 6, , ∴ By using the double intercept form :, , 6 x + y + 28 = 0, , x y,  + =1, a b, , (iv) By using the two points form, , x y,  = 0, + = 1 ∴ 3x − 2 y +12, −4 6, , An interesting property of a straight line., , x − x1, y − y1, =, x2 − x1 y2 − y1, , Consider any straight line in a plane., It makes two parts of the points which are, not on the line., , Here (x1, y1) = (2-1); (x2, y2) = (5,1), ∴ the equation of the required line is, x − 2 y +1, =, 5 − 2 1+1, ∴ 2 ( x − 2 ) = 3 ( y + 1), ∴ 2 x −3 y − 7 = 0, (v) By using the slope intercept form y = m x + c, 3, Given m = − , c = 5, 4, , Fig. 5.14, Thus the plane is divided into 3 parts, the, points on the line, points on one side of the, line and points on the other side of the line., , ∴ the equation of the required line is, , 3, y = − x+5, 4, , ∴ 3x + 4 y − 20 = 0, , 113

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If the line is given by ax+by+c = 0, then, for all points (x1,y1) on one side of the line, ax1+by1+c > 0 and for all points (x2,y2) on the, other side of the line, ax2+by2+c < 0., , 4., , 5., , origin, , and, , having, , b) passing through the origin and parallel, to AB, where A is (2,4) and B is (1,7)., 1, c) having slope, and containing the, 2, point (3,−2)., , Draw the straight lines given by 2y + x, = 5, x = 1, 6y − x + 1 = 0 give 4 points on, each side of the lines and check the property, stated above., , d) containing the point A(3,5) and having, slope 2 ., 3, e) containing the point A(4,3) and having, inclination 120°., , EXERCISE 5.3, , f) passing through the origin and which, bisects the portion of the line 3x + y = 6, intercepted between the co−ordinate, axes., , Write the equation of the line :, a) parallel to the X−axis and at a distance, of 5 unit form it and above it., b) parallel to the Y− axis and at a distance, of 5 unit form it and to the left of it., c) parallel to the X− axis and at a distance, of 4 unit form the point (−2, 3)., Obtain the equation of the line :, , 6., , Line y = mx + c passes through points, A(2,1) and B(3,2). Determine m and c., , 7., , Find the equation of the line having, inclination 135° and making X-intercept 7., , 8., , The vertices of a triangle are A(3,4), B(2,0), and C(−1,6). Find the equations of the lines, containing, (a) side BC, (b) the median AD, , a) parallel to the X−axis and making an, intercept of 3 unit on the Y−axis., b) parallel to the Y−axis and making an, intercept of 4 unit on the X−axis., 3., , Find the equation of the line, a) containing the, inclination 60°., , Activity :, , 2., , a) passing through the points A ( 2, 0 ) and, B(3,4)., b) passing through the points P(2,1) and, Q(2,-1), , For example, consider the line given by, y − 2x − 3 = 0. Points P(−2,0), Q(−2,4),, 1, R( ,5) lie on one side and at each of those, 2, points, y − 2x − 3 > 0. The points A(0,0),, 1, B( ,3), C(8,4) lie on the other side of the line, 2, and at each of those points y − 2x − 3 < 0., , 1., , Find the equation of the line, , (c) the mid points of sides AB and BC., , a) A ( 2, −3) and parallel to the Y−axis., , Find the x and y intercepts of the, following lines :, 3x 2 y, x y, (a), (b), +, =1, + =1, 2, 3, 3 2, , b) B(4,-3) and parallel to the X−axis., , (c) 2 x − 3 y + 12 = 0, , 9., , Obtain the equation of the line containing, the point :, , 114

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10. Find equations of lines which contains, the point A(1,3) and the sum of whose, intercepts on the co−ordinate axes is zero., , Remark : If a=0 then the line is parallel to, the X-axis. It does not make intercept on the, X-axis., , 11. Find equations of lines containing the, point A(3,4) and making equal intercepts on, the co−ordinates axes., , If b=0 then the line is parallel to the Y-axis., It does not make intercept on the Y-axis., , 12. Find equations of altitudes of the, triangle whose vertices are A(2,5), B(6,-1), and C(-4,-3)., , SOLVED EXAMPLES, Ex. 1) Find the slope and intercepts made by, the following lines :, , 13. Find the equations of perpendicular, bisectors of sides of the triangle whose, vertices are P(−1,8), Q(4,−2) and R(−5,−3)., , (a) x + y +10 = 0, (c) x + 3 y − 15 = 0, , Solution : (a) Comparing equation x+y+10=0, , 14. Find the co-ordinates of the orthocenter of, , with ax + by + c = 0 ,, , the triangle whose vertices are A(2,−2),, B(1,1) and C(−1,0)., , we get =, a  1 ,, =, b  1,c = 10, ∴ Slope of this line = − a = −1, b, c, 10, = −10, The X–intercept is − = −, a, 1, c, The Y-intercept is = − = − 10 = −10, b, 1, , 15. N(3,−4) is the foot of the perpendicular, drawn from the origin to line L. Find the, equation of line L., 5.4 General form of equation of a line: We, can write equation of every line in the form, ax+by+c=0, , (b) Comparing equation 2 x + y + 30 = 0 , with ax + by + c = 0 ., , This form of equation of a line is called the, general form., , we get =, a  2=, ,b  1 ,c = 30, a, ∴ Slope of this line = − = −2, b, 30, c, The X–intercept is − = −, = −15, 2, a, c, 30, The Y-intercept is − = −, = −30, b, 1, , The general form of y = 3 x + 2 is 3 x − y + 2 = 0, x y, + = 1 is 3 x + 2 y − 6 = 0, 2 3, The slope of the line ax + by + c = 0, is, , The general form of, , −, , a, if b ≠ 0, ., b, , The X–intercept is −, , c, if a ≠ 0 ., a, , The Y-intercept is −, , c, if b ≠ 0 ., b, , (b) 2 x + y + 30 = 0, , (c) Comparing equation x + 3 y − 15 = 0, with ax + by + c = 0 ., we get a =1,  ,b = 3,c = −15, ∴ Slope of this line = −, 115, , 1, a, = −, 3, b

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Solution : Slopes of the given lines are, m1 = 3 and m2 = 1 ., 3, , −15, c, = −, = 15, 1, a, c, −15, The y – intercept is − = −, = 5, b, 3, The x – intercept is −, , The acute angle θ between lines having, slopes m1 and m2 is given by, , Ex. 2) Find the acute angle between the, following pairs of lines :, a) 12 x −4 y = 5 and 4 x + 2 y = 7, , m1 − m2, 1 + m1m2, , tan θ =, , b) y = 2 x + 3 and y = 3x + 7, Solution : (a) Slopes of lines 12 x −4 y = 5 and, , tanθ =, , 4 x + 2 y = 7 are m1 = 3 and m2 = 2 ., If θ is the acute angle between lines having, slope m1 and m 2 then, tan θ =, , 3, , 1 + ( 3) ( −2 ), , =, , ∴ θ =45°, , 1− 3, , =, , 2 3, , −1, 3, , 2, =, , 1, 3, , 1, 3, , ∴ θ = 300, , b) y = 2 x + 23 and 2 x + 4 y = 27, Solution : (i) Slopes of lines 2 x −4 y = 5 and, 2 x + y =17 are m = 1 and m2 = −2, 1, 2, 1, × ( −2 ) = −1, given lines, Since m1.m 2 =, 2, are perpendicular to each other., , m1 − m2, 1 + m1m2, , (ii) Slopes of lines y = 2x + 23 and, , 2−3, −1 1, =, =, 1 + ( 2 )( 3), 7, 7, , 2x+4y = 27 are m1 = −, , 1, ∴ θ = tan −1   ., 7, , 1, m, 2 and 2 = 2., , 1, Since m1.m2 = − × ( 2 ) = −1, given lines, 2, are perpendicular to each other., , Ex. 3) Find the acute angle between the lines, y − 3x + 1 = 0 and, , 3−, , a) 2 x −4 y = 5 and 2 x + y = 17 ., , The acute angle θ between lines having, slopes m1 and m 2 is given by, , ∴ tan θ =, , 1, 3 =, 1+ 1, , 3−, , Ex. 4) Show that following pairs of lines are, perpendicular to each other., , 5, = 1, −5, , (b) Slopes of lines y = 2 x + 3 and y = 3x + 7, are m1 = 2 and m2 = 3, , tan θ =, , 1, 3 =, 1, 1+ 3 ×, 3, 3−, , ∴ tanθ = 1, , 3 − ( −2 ), , ∴ tan θ =1, , 3, , =, , =, , m1 − m 2, 1 + m1m 2, , ∴ tan θ =, , 1, , 3y − x + 7 = 0 ., , 116

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Ex. 5) Find equations of lines which pass, through the origin and make an angle of 45°, with the line 3x - y = 6., , ∴ It’s equation is given by the formula, , ( y − y1 )= m( x − x1 ), ∴, , Solution : Slope of the line 3x - y = 6. is 3., Let m be the slope of one of the required, lines. The angle between these lines is 45° ., 0, ∴ tan45 =, , ∴, , ∴ m = −2 or, , 1, 2, , x + 3 y = 23 ., , A (1,1) ., , 1, m2 = 2, , ∴ It’s equation is given by the formula, , Required lines pass through the origin., , ( y − y1 )= m( x − x1 ), , ∴ Their equations are y = −2 x and, , ∴, , ∴, , 3, Solution : Slope of the line 3x + 2 y − 1 is − ., 2, Required line is perpendicular it., 2, The slope of the required line is ., 3, Required line passes through the point, , 1 + 3m= m–3 , , Slopes of required lines are m1 = -2 and, , y =, , 3 y − 21= x + 2, , A (1,1) . Find its equation., , or 1 + 3m = − ( m − 3), , ∴ 1 + 3m = m − 3, , ∴, , Ex. 8) A line is perpendicular to the line, 3x + 2 y − 1 = 0 and passes through the point, , m−3, 1 + ( m )( 3), , m−3, ∴ 1 =, 1 + 3m, , 1, ( y − 7 )= − ( x −2 ), 3, , 2, ∴ ( y − 1)=  ( x −1), 3, ∴ 3 y −3= 2 x − 2, , 1, x, 2, , ∴, , 2 x + y = 0 and x − 2 y = 0, , Note:, , Ex. 6) A line is parallel to the line 2 x + y = 7, and passes through the origin. Find its equation., , Point of intersection of lines :, The co-ordinates of the point of intersection, of two intersecting lines can be obtained by, solving their equations simultaneously., , Solution : Slope of the line 2 x + y = 7 is −2., Required line passes through the origin., ∴ It’s equation is y = −2 x, , Ex. 9) Find the co-ordinates of the point of, intersection of lines x + 2 y = 3 and 2 x − y =1 ., , ∴ 2 x + y = 0., , Ex. 7) A line is parallel to the line x + 3 y = 9, and passes through the point A(2,7). Find its, equation., 1, 3, Required line passes through the point, A(2,7)., , 2x − 3y +1 = 0 ., , Solution : Solving equations x + 2 y = 3 and, 2 x − y = 1 simultaneously, we get x = 1 and, , y = 1., , Solution : Slope of the line x + 3 y = 9 is −, , ∴ Given lines intersect in point (1,1) ., , 117

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Ex. 10) Find the equation of line which is, parallel to the X-axis and which passes through, the point of intersection of lines x + 2 y = 6, and 2 x − y = 2, , 2, 2, 2, 2,  c   c  = c2  a + b , AB =   +  ,  2 2 ,  a b , a   b, 2, , ∴ AB =, , Solution : Solving equations x + 2 y = 6 and, 2 x − y = 2 simultaneously, we get x = 2, and y =2 ., ∴, point, , c, ab, , a 2 + b2, , Now,, , The required line passes through the, , ( 2, 2 ) ., , Area of ∆OAB, , =, , , , =, , As it is parallel to the X-axis, its equation, is y = 2., , But, Area of ∆OAB =, , 5.4.1 The distance of the Origin from a, , , , Line :, The distance of the origin from the line, ax + by + c = 0 is given by p =, , =, , 1 AB × ON, 2, 1 c, 2 ab, , a 2 + b2 × p ... (I), , 1, OA × OB, 2, , 1 c c, c2, ... (II),  =, 2 a b 2ab, , From (I) and (II), we get, , c, , p=, , a 2 + b2, , c, a + b2, 2, , 5.4.2 The distance of the point (x1,y1) from a, line: The distance of the point P (x1,y1) from, line ax + by + c = 0 is given by, p=, , ax1 + by1 + c, a 2 + b2, , Fig. 5.15, Proof : Let A and B be the points where line, ax + by + c = 0 cuts the co-ordinate axes., c, ,  c , ∴ A  − , 0  and B  0, − , b, ,  a , OA =, , c, a, , and OB =, , c, b, , By Pythagoras theorem AB = OA +OB, 2, , 2, , Fig. 5.16, , 2, , 118

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Proof : If line ax + by + c = 0 cuts co-ordiante,  c , axes in B and C respectively then B is  − , 0 ,  a , and C is  0, − c  ., b, , Let PM be perpendicular to ax + by + c = 0 ., Let PM = p, A ( ∆PBC ) =, , 1, 1, c2 c2, pc a 2 + b2, BC × PM = p  2 + 2 =, 2, a, b, 2, 2ab, , Fig. 5.17, Proof : To find the distance between parallel, lines, we take any one point on any one of, these two lines and find its distance from the, other line., , .(I), , c, c, Now P ( x1 , y1 ) , B  − , 0  and C  0, − , b,  a , , are vertices of PBC., ∴, , 1, A ( ∆PBC ) =, , 1, x1, 2, , 1, c, −, a, , y1, , 0, , . .(II), , 1, 0 =, −, ,  c , A  − 1 , 0  is a point on the first line.,  a , , c c2 , 1 c, x1 + y1 +, a ab, 2 b, , Its distance from the line ax + by + c2 = 0, is given by, , c, b, ,  c , a  − 1  + b ( 0 ) + c2, −c1 + c2, a, p= , =, =, 2, 2, a +b, a 2 + b2, , From (I) and (II) we get, , c1 − c2, a 2 + b2, , 1 c, c c2 , pc a 2 + b2, x1 + y1 +, =, 2 b, a ab, 2ab, 2, , 2, , SOLVED EXAMPLES, , 2, , ∴ pc a + b =| acx1 + bcy1 + c |, , Ex. 1) Find the distance of the origin from, the line 3x + 4 y + 15 = 0, , ∴ p a 2 + b 2 =| ax1 + by1 + c |, ∴ p=, , ax1 + by1 + c, , Solution : The distance of the origin from the, , a 2 + b2, , line ax + by + c = 0 is given by, , 5.4.3 The distance between two parallel, lines :, , p=, , Theorem : The distance between the parallel, , a + b2, , ∴ The distance of the origin from the line, , lines ax + by + c1 = 0 and ax + by + c2 = 0 is give, by p =, , c, 2, , 3x + 4 y + 15 = 0 is given by, , c1 − c2, a 2 + b2, , p=, 119, , 15, 32 + 4 2, , =, , 15, = 3, 5

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lines is of the form ( y + 3)= m( x −2 ) . This set, is also form a family of lines., , Ex. 2) Find the distance of the point P ( 2, 5), from the line 3x + 4 y + 14 = 0, , Consider the set of all lines which are, parallel to the line y = x . They all have the, same slope., , Solution : The distance of the point P ( x1 ,y1 ), from the line ax + by + c = 0 is given by, p=, , The set of all lines which pass through, , ax1 + by1 + c, , a fixed point or which are parallel to each, other is a family of lines., , a 2 + b2, , ∴ The distance of the point P ( 2, 5) from, the line 3x + 4 y + 14 = 0 is given by, p=, , 3 ( 2 ) + 4 ( 5 ) + 14, 2, , 3 +4, , =, , 2, , Interpretation, , of, , u, , +, , kv, , =, , 0, , :, , Let, , u ≡ a1 x + b1 y + c1 and v ≡ a2 x + b2 y + c2, , 40, =8, 5, , Equations u=0 and v=0 represent two lines., Equation u + kv = 0, k ∈ R represents a family, of lines., , Ex. 3) Find the distance between the parallel, lines 6 x + 8 y + 21= 0 and 3x + 4 y +7 = 0 ., Solution : We write equation 3x + 4 y + 7 = 0 as, 6 x + 8 y +14 = 0 in order to make the, coefficients of x and coefficients of y in, both equations to be same., Now by using formula, , p=, , c1 − c2, a 2 + b2, , We get the distance between the given, parallel lines as, p=, , 21 − 14, 2, , 6 +8, , 2, , =, , Fig. 5.18, , 7, 10, , In equation u + kv = 0 let, , u ≡ a1 x + b1 y + c1 , v ≡ a2 x + b2 y + c2, , 5.4.4 Family of Lines : A set of lines which, have a common property is called a family, of lines. Consider the set of all lines passing, through the origin. Equation of each of these, lines is of the form y = mx . This set of lines, is a family of lines. Different values of m give, different lines., , We get (a1 x + b1 y + c1 ) + k (a 2 x + b2 y + c2 ) = 0, , Consider the set of all lines which pass through, , other in P ( x1 ,y1 ) then, , (a1 + ka 2 ) x + (b1 + k b2 ) y + (c1 + k c2 ) = 0, , Which is a first degree equation in x and y., Hence it represents a straight line., i) If lines, , the point A ( 2, −3) . Equation of each of these, , u= 0 and v= 0 intersect each, , a 2 x1 + b2 y1 + c2 = 0, , 120, , a1 x1 + b1 y1 + c1 = 0 and

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Therefore, , ∴ 7 k = 16, , (a1 x1 + b1 y1 + c1 ) + k (a 2 x1 + b2 y1 + c2 ) = 0 + k 0 = 0, , ∴ the, , Thus line u + kv = 0  passes through the point, of intersection of lines u = 0 and v = 0 for, every real value of k., ii) If lines u = 0 and v= 0  are parallel to, each other then their slope is same., ∴ −, , a1 + ka 2, b1 + kb2, , ∴ Slopes of lines =, u, u + kv = 0 are the same., , 16, ( 2 x − y − 2 ) = 0, 7, , ∴, , ( x + 2 y + 6) +, , ∴, , ( 7 x + 14 y + 42 ) + ( 32 x − 16 y − 32 ) = 0, , 3x + 2 y − 6 = 0 , x + y + 1 = 0 and the point, A ( 2,1) ., , 0=, ,v  0  and, , Solution : Since the required line passes, through the point of intersection of lines, 3x + 2 y − 6 = 0 and x + y + 1 = 0 , its, equation is of the form u + k v = 0 ., ∴ (3 x + 2 y − 6) + k ( x + y + 1) = 0, , SOLVED EXAMPLES, , ∴ (3 + k ) x + ( 2 + k ) y + ( −6 + k ) = 0, , Ex. 1) Find the equation of the line which, passes through the point of intersection of, lines x + 2 y + 6 = 0 , 2 x − y = 2, and which, , This line passes through the point A(2,1) ., ∴, , makes intercept 5 on the Y-axis., , ( 2, 1), , satisfy this equation., , ∴ (3 + k ) ( 2 ) + ( 2 + k )(1) + ( −6 + k ) = 0, , Solution : As the required line passes through, the point of intersection of lines x + 2 y + 6 = 0, and 2 x − y = 2 , its equation is of the form, , ∴ 4k + 2 = 0, , ∴ k=−, , 1, 2, , ∴ The equation of the required line is, , u + k v = 0 ., ,  1, (3 x + 2 y − 6) +  −  ( x + y + 1) = 0,  2, , ( x + 2 y + 6 ) + k (2x − y − 2) = 0, , ∴ (1 + 2k ) x + ( 2 − 1k ) y + ( 6 − 2k ) = 0, , 5 x + 3 y − 13 = 0, , The Y-intercept of this line is given 5., 6 − 2k, =5, 2−k, , equation of the required line is, , through the point of intersection of lines, , ∴ Line u + kv = 0  is parallel to lines, =, u  0=, ,v  0 ., , ∴ −, , 16, 7, , Ex. 2) Find the equation of line which passes, , = slope of the line u+kv=0, , ∴, , k =, , ∴ 39 x − 2 y + 10 = 0, , a1, a , =− 2, b1, b2, , ∴ each ratio = −, , ∴, , ∴ −6 + 2 k = 10 5 k, 121

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10) Find the equation of the line whose, X-intercept is 3 and which is perpendicular, 0., to the line 3x − y + 23 =, , EXERCISE 5.4, 1), , Find the slope, X-intercept, Y-intercept, of each of the following lines., , 11) Find the distance of the origin from the line, 7x +24y-50 = 0 ., , 0 b) 3x−y−9=0, a) 2 x + 3 y − 6 =, 0, c) x + 2 y =, 2), , 12) Find the distance of the point A ( −2,3) from, 0., the line 12 x − 5 y − 13 =, , Write each of the following equations in, ax + by + c =, 0 form., a) y = 2x-4, c), , 13) Find the distance between parallel lines, 4x − 3y + 5 =, 0 and 4x − 3y + 7 = 0, , b) y = 4, , x, y, x y, =0, + =, 1 d), 3, 2, 2 4, , 3), , 0 and, Show that lines x − 2 y − 7 =, 2x − 4y + 15 = 0 are parallel to each other., , 4), , 0 and, Show that lines x − 2 y − 7 =, 2x + y + 1 = 0 are perpendicular to each, other. Find their point of intersection., , 14) Find the distance between parallel lines, 9x + 6 y − 7 =, 0. and 3x + 2y + 6 = 0, , 0 which, 15) Find points on the line x + y − 4 =, are at one unit distance from the line, , x+ y−2=, 0, 16) Find the equation of the line parallel to the, X-axis and passing through the point of, 0 and, intersection of lines x + y − 2 =, 4x + 3y = 10., , 5), , p makes a triangle, If the line 3x + 4 y =, of area 24 square unit with the co-ordinate, axes then find the value of p ., , 6), , Find the co-ordinates of the foot of the, perpendicular drawn from the point A(-2,3), to the line 3x − y − 1 =, 0 ., , 17) Find the equation of the line passing, through the point of intersection of lines, x+ y−2=, 0 and 2 x − 3 y + 4 =, 0 and, making intercept 3 on the X-axis., , 7), , Find the co-ordinates of the circumcenter, of the triangle whose vertices are, A( −2,3), B(6, −1),C(4,3)., , 18) If A(4,3), B(0,0), and C(2,3) are the, vertices of ∆ ABC then find the equation of, , Find the co-ordinates of the orthocenter of, the triangle whose vertices are A(3,-2),, B(7,6), C (-1,2), , 19) D(-1,8), E(4,-2), F(−5,-3) are midpoints of, sides BC, CA and AB of ∆ ABC . Find, , 8), , 9), , bisector of angle BAC., , (i) equations of sides of ∆ ABC ., , Show that lines, 3x − 4 y +=, 5 0,7 x − 8 y +=, 5 0,and 4 x + 5 y − 45, = 0, , are concurrent. Find their point of, concurrence., , (ii) co-ordinates of the circumcenter of, ∆ ABC ., 20) O(0, 0), A(6,0) and B(0,8) are vertices of a, triangle. Find the co-ordinates of the incenter, of ∆ OAB., , 122

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• Angle between intersecting lines : If θ is, the acute angle between lines having slopes, , Let's Remember, , m1 and m2 then tan θ =, , • Locus : A set of points in a plane which, satisfy certain geometrical condition (or, conditions) is called a locus., , • Equations of line in different forms :, , • Equation of Locus : Every point in, XY plane has Cartesian co-ordinates. An, equation which is satisfied by co-ordinates, of all points on the locus and which is not, satisfied by the co-ordinates of any point, which does not lie on the locus is called, the equation of the locus., , • Slope point form :, , • Inclination of a line : The smallest angle, θ made by a line with the positive direction, of the X-axis, measured in anticlockwise, sense, is called the inclination of the line., Clearly 0° ≤ θ ≤ 180°., , • Normal form :, , • Two points form :, , • General form, , x y, + =1, a b, , x cos α + y sin α = p, : ax + by + c = 0, , • Distance of a point from a line :, • The distance of the origin from the line, , ax + by + c = 0 is given by p =, , c, a 2 + b2, , • The distance of the point P ( x1 ,y1 ) from line, ax + by + c = 0 is given by, p=, , (if x1 ≠ x2), , • Perpendicular and parallel lines : Lines, having slopes m1 and m2 are perpendicular, to each other if and only if, , x − x1, y − y1, =, x2 − x1 y2 − y1, , • Double intercept form :, , If A ( x1 , y1 ) , B ( x2 , y2 ) be any two points, on the line whose inclination is θ then, y2 − y1, x2 − x1, , ( y − y1 )= m( x − x1 ), , • Slope intercept form : y = mx + c, , • Slope of a line : If θ is the inclination of, a line then tanθ (if it exist) is called the, slope of the line., , tanθ =, , m1 − m2, 1 + m1m2, , ax1 + by1 + c, a 2 + b2, , • The distance between the Parallel lines, : The distance between the parallel lines, is, ax + by + c2 = 0, ax + by + c1 = 0 and, , m1m2 = −1 ., , Two lines are parallel if and only if they, have the same slope., , give by, , 123, , p=, , c1 − c2, a 2 + b2

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(A), , MISCELLANEOUS EXERCISE - 5, , (C) x+y=6, , (I) Select the correct option from the given, alternatives., , 7), , 1) If A is (5,−3) and B is a point on the, x−axis such that the slope of line AB is, −2 then B ≡, 7, (A) (7,2), (B) ( ,0), 2, 2, 7, (C) (0, ), (D) ( ,0), 7, 2, , 8), , (A) −1, , (B) 0, , (C) 1, , (C) 7x−13y+5 =0, , (D) 13x−7y−5=0, , 9), , (C) 3, , The angle between the line, , (D), , 4, 3, , 3x −y−2=0, , (B) 30°, , (C) 45°, , (D) 60°, , If kx+2y−1=0 and 6x−4y+2=0 are identical, lines, then determine k., , (A), , (B) −, , 1, 3, , (C), , 1, 3, , (D) 3, , 2, 5, , (B), , 1, 5, , (C), , 5, 2, , (D), , 2, 5, , (II) Answer the following questions., 1), , Find the value of k, a) if the slope of the line passing through, the points P ( 3,4 ) , Q ( 5,k ) is 9., b) the points, collinear, , (D) x+y=3, , (B) 2, , (C) 1, , 10) Distance between the two parallel lines, y=2x+7 and y=2x+5 is, , A (1,3) ,B ( 4,1) ,C ( 3,k ), , are, , c) the point P(1,k) lies on the line passing, through the points A(2, 2) and B(3, 3)., , 5) If the line kx+4y=6 passes through the, point of intersection of the two lines, 2x+3y=4 and 3x+4y=5, then k =, (A) 1, , 2, 3, , (B), , (A) −3, , 4) The equation of the line through (1,2),, which makes equal intercepts on the axes,, is, (A) x+y=1, (B) x+y=2, (C) x+y=4, , 1, 3, , (A) 15°, , 3) If A(1,−2), B(−2,3) and C(2,−5) are the, vertices of ∆ABC, then the equation of, the median BE is, (B) 13x+7y+5=0, , (D) x+y=−6, , and x− 3y +1=0 is, , 1, (D), ab, , (A) 7x+13y+47=0, , 3x +y±6=0, , A line passes through (2,2) and is, perpendicular to the line 3x+y=3. Its, y−interecpt is, (A), , 2) If the point (1,1) lies on the line passing, through the points (a,0) and (0,b), then, 1 1, + =, a b, , (B), , 3x ±y+6=0, , 2), , Reduce the equation 6 x + 3 y + 8 = 0 into, slope-intercept form. Hence find its slope., , 3), , Find the distance of the origin from the, line x=−2., , 4), , Does point A ( 2, 3) lie on the line, 3x + 2 y − 6 = 0 ? Give reason., , (D) 4, , 6) The equation of a line, having inclination, 120° with positive direction of X−axis,, which is at a distance of 3 units from, the origin is, 124

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5) Which of the following lines passes, through the origin ?, (a) x = 2, (c) y = x + 2, , 12) Find the equation of the line which, contains the point A(3,5) and makes equal, intercepts on the co−ordinates axes., , (b) y = 3, (d) 2 x − y = 0, , 13) The vertices of a triangle are A(1,4),, B(2,3) and C(1,6). Find equations of, , 6) Obtain the equation of the line which is :, , (a) the sides, , a) parallel to the X−axis and 3 unit below, it., , (c) Perpendicular bisectors of sides, (d) altitudes of ∆ ABC., , b) parallel to the Y−axis and 2 unit to the, left of it., , 14) Find the equation of the line which passes, through the point of intersection of lines, x + y − 3 = 0 , 2 x − y + 1 = 0 and which is, parallel X- axis., , c) parallel to the X−axis and making an, intercept of 5 on the Y−axis., d) parallel to the Y−axis and making an, intercept of 3 on the X−axis., 7), , 15) Find the equation of the line which passes, through the point of intersection of lines, x + y + 9 = 0 , 2 x + 3 y + 1 = 0 and which, , Obtain the equation of the line containing, the point, , makes X-intercept 1., , (i) (2,3) and parallel to the X−axis., , 16) Find the equation of the line through A, , (ii) (2,4) and perpendicular to the Y−axis., 8), , ( −2, 3), , 17) Find the X-intercept of the line whose, slope is 3 and which makes intercept 4, on the Y−axis., , point, , A ( −1, 2 ) ., , b) containing the point T(7,3) and having, , 18) Find the distance of P ( −1,1) from the line, , inclination 900 ., , 12 ( x + 6 ) = 5 ( y − 2 ) ., , c) through the origin which bisects, the portion of the line 3x + 2 y = 2, intercepted between the co−ordinate, axes., 9), , and perpendicular to the line, , through S (1, 2 ) and T ( 2, 5 ) ., , Find the equation of the line :, a) having slope 5 and containing, , (b) the medians, , 19) Line through A ( h,3) and B ( 4,1) intersect, the line 7 x − 9 y − 19 = 0 at right angle., Find the value of h., , Find the equation of the line passing, through the points S(2,1)and T(2,3), , 10) Find the distance of the origin from the, line 12,  x + 5 y + 78= 0, , 20) Two lines passing through M ( 2, 3) intersect, each other at an angle of 45°. If slope, of one line is 2, find the equation of the, other line., , 11) Find the distance between the parallel, lines 3x +4 y + 3 = 0 and 3x +4 y + 15 = 0, , 21) Find the Y-intercept of the line whose, slope is 4 and which has X intercept 5., 125

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22) Find the equations of the diagonals of the, rectangle whose sides are contained in the, x 8=, , x 10, y = 11 and y = 12 ., lines=, , 27) Find points on the X-axis whose distance, x y, + = 1 is 4 unit., from the line, 3 4, , 23) A(1, 4), B(2,3) and C (1, 6) are vertices, of ∆ABC. Find the equation of the altitude, through B and hence find the co-ordinates, of the point where this altitude cuts the, side AC of ∆ABC., , 28) The perpendicular from the origin to a line, meets it at, the line., , Find the equation of, , 29) P (a ,b ) is the mid point of a line segment, between axes. Show that the equation of, x y, the line is, + =2., a b, , 24) The vertices of ∆ PQR are P(2,1),, Q(-2,3) and R(4,5). Find the equation of, the median through R., , 30) Find the distance of the line 4 x − y = 0, , from the point P ( 4,1) measured along, the line making an angle of 1350 with the, positive X-axis., , 25) A line perpendicular to segment joining A, , B ( 2, 3) divides it internally in, the ration 1:2. Find the equation of the, line., , (1, 0 ) and, , 31) Show that there are two lines which pass, through A(3,4) and the sum of whose, intercepts is zero., , 26) Find the co-ordinates of the foot of the, perpendicular drawn from the point P, , ( −1, 3), , ( −2, 9 ) ., , the line 3x-4y-16 = 0 ., , 32) Show that there is only one line which, passes through B(5,5) and the sum of, whose intercept is zero., , 126

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6, , Circle, \ x2 + y2 = r2., , Let's Study, , , , , , , This is the standard equation of a circle., , Equation of a circle and its different forms, , (2) Centre-radius form :, In Fig. 6.2, C(h, k), is the centre and r, is the radius of the, circle. P(x, y) is any, point on the circle., , Equation of Tangent to a circle, Condition for tangency, Director circle, Let's Recall, , ∴ CP = r, , , , , Properties of chords and tangents of a circle., , , , , Product of slopes of perpendicular lines is -1., , CP = ( x − h )2 + ( y − k )2 \ r2 = (x - h)2 + (y - k)2, , Slopes of parallel lines are equal., , (x − h)2 + (y − k)2 = r2 is the centre-radius form of, equation of a circle., , Also,, , The angle inscribed in a semicircle is a right, angle., , Fig. 6.2, , A circle is a set of all points in a plane, which are equidistant from a fixed point in the, plane., , (3) Diameter Form : In the Fig. 6.3, C is the, centre of the circle., A(x1 , y1), B(x2 , y2), are the end points, of a diameter of the, circle. P(x, y) is any, point on the circle., Angle inscribed in a, Fig. 6.3, semi circle is a right, , The fixed point is called the centre of the, circle and the distance from the centre to any, point on the circle is called the radius of the circle., 6.1 Different forms of equation of a circle, (1) Standard form : In Fig. 6.1, the origin, O, is the centre of the, circle. P(x, y) is any, point on the circle., , Fig. 6.1, , ο, angle; hence, m∠ APB = 90 ; that is AP ⊥ BP., , Slope of AP =, , The radius of circle, is r., ∴ OP = r., By distance formula, OP2 = ( x − 0 )2 + ( y − 0 )2, ∴ we get r2 = x2 + y2, , y − y1, x − x1, , and slope of BP =, , As AP ⊥ BP, product of their slopes is –1., y − y1, , y − y2, , ∴ x − x × x − x = –1, 1, 2, , (y – y1) (y – y2) = –(x – x1)(x – x2), (x – x1) (x – x2) + (y – y1)(y – y2) = 0, , 127, , y − y2, x − x2, , ,

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Ex.4 Find the equation of circle touching the, , That is, (x - x1)(x - x2) + (y - y1) (y - y2) = 0, , Y-axis at point (0, 3) and whose Centre is, , This is called the diameter form of the equation, of circle, where (x1, y1) and (x2, y2) are endpoints, , at (–3, 3)., Solution :, , of diameter of the circle., , The circle touches, SOLVED EXAMPLES, , the Y-axis at point, C(−3, 3), , Ex.1 Find the equation of a circle with centre at, origin and radius 3., , (0,3), , is (−3,3) we get, radius r = 3, , Solution : Standard equation of a circle is, , x2 + y2 = r2, ∴ x2 + y2 = 32, , x2 + y2 = 9, , here r = 3, , using, , centre, , radius form;, , is the equation of circle, , (x − h)2 + (y − k)2 = r2, (x + 3)2 + (y – 3)2 = 9, x2 + 6x + 9 + y2 − 6y + 9 = 9, , Solution : Centre C = (−3, 1),, Circle passes through the point P(5, 2)., By distance formula,, , ∴ x2 + y2 + 6x – 6y + 9 = 0 is the equation, of the circle., , Ex.5 Find the equation of the circle whose centre, is at (3,−4) and the line 3x−4y−5 = 0 cuts the circle, at A and B; where l(AB) = 6., , , r2 = CP2 = (5 + 3)2 + (2 − 1)2, = 82 + 12 = 64 + 1 = 65, , ∴ the equation of the circle is, , Solution : centre of the, circle C(h,k) = C(3,−4), , (x + 3) + (y − 1) = 65 (centre-radius form), 2, , By, , Fig. 6.4, , Ex.2 Find the equation of a circle whose centre is, (−3, 1) and which pass through the point (5, 2)., , , , (0,3), and the centre, , 2, , x2 + 6x + 9 + y2 − 2y + 1 = 65, x2 + y2 + 6x − 2y + 10 − 65 = 0, , 3x − 4y − 5 = 0 cuts the, circle at A and B., , x2 + y2 + 6x − 2y − 55 = 0, , l(AB) = 6, CM ⊥ AB, , Ex.3 Find the equation of the circle with A(2, −3), and B(−3, 5) as end points of its diameter., Solution : By using the diameter form;, A (2, –3) ≡ (x1, y1) and B(–3, 5) ≡ (x2, y2) are, the co-ordinates of the end points of a diameter of, the circle., , Fig. 6.5, , ∴ AM = BM = 3, , CM = Length of perpendicular from centre on, the line, =, , ∴ by the diameter form, equation of the circle is, (x − x1) (x − x2) + (y − y1) (y − y2) = 0, ∴ (x – 2) (x + 3) + (y + 3) (y – 5) = 0, ∴ x2 + x – 6 + y2 – 2y – 15 = 0, , ∴ x2 + y2 + x – 2y – 21 = 0, 128, , 3(3) − 4(−4) − 5, (3) 2 + (−4) 2, , =, , 9 + 16 − 5, 9 + 16, , =, , 20, =4, 5

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5. If y = 2x is a chord of circle x2 + y2 −10x = 0,, find the equation of circle with this chord as, diametre., , From right angled triangle AMC, CA2 = CM2 + AM2, = (4)2 + (3)2, = 16 + 9 = 25, CA, , 6. Find the equation of a circle with radius 4, units and touching both the co-ordinate axes, having centre in third quadrant., , = radius of the circle = 5, By centre radius form equation of the circle., (x −3)2 + (y+4)2 = 52, , 7) Find the equation of circle (a) passing through, the origin and having intercepts 4 and −5 on, the co-ordinate axes., , x2−6x+9+y2+8y+16 = 25, x2+y2−6x+8y = 0, , 8) Find the equation of a circle passing through, the points (1,−4), (5,2) and having its centre, , EXERCISE 6.1, , on the line x−2y+9 = 0 ., , 1. Find the equation of the circle with, (i) Centre at origin and radius 4., , Activity :, (1) Construct a circle in fourth quadrant having, radius 3 and touching Y-axis. How many, such circles can be drawn?, , (ii) Centre at (−3, −2) and radius 6., (iii) Centre at (2, −3) and radius 5., (iv) Centre at (−3, −3) passing through point, (−3, −6), , (2), , x2 + y2 − 4x + 6y − 12 = 0. Find the area of, , 2. Find the centre and radius of the circle., (i) x2 + y2 = 25, , the circle., , (ii) (x − 5)2 + (y − 3)2 = 20, , 2, , 6.2 General equation of a circle :, , 2, , 1, 1, 1, (iii)  x −  +  y +  =, 2 , 3, 36, , , The general equation of a circle is of the, form x2 + y2 + 2gx + 2fy + c = 0, if g2 + f 2−c>0, , 3. Find the equation of the circle with centre, (i), , The centre-radius form of equation of a circle, , At (a, b) touching the Y-axis, , is, , (ii) At (-2, 3) touching the X- axis, , (x − h)2 + (y − k)2 = r2, , (iii) on the X-axis and passing through the, , i.e x2 − 2hx + h2 + y2 − 2ky + k2 = r2, , origin having radius 4., (iv) at, , (3,1), , and, , touching, , Construct a circle whose equation is, , i.e x2 + y2 − 2hx − 2ky + (h2 + k2 − r2) = 0, the, , line, , If this is the same as equation x2 + y2 + 2gx +, 2fy + c = 0, then comparing the coefficients, , 8x − 15y + 25 = 0, , 2g = – 2h, 2f = – 2k and c = h2 + k2 – r2., , 4. Find the equation circle if the equations of, two diameters are 2x + y = 6 and 3x + 2y = 4., , ∴ (h, k) ≡ (–g, –f) is the center and, , When radius of circle is 9., , r2 = h2 + k2 – c i.e. r =, , 129, , g 2 +f 2 − c is the radius.

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Lets Note :, , Thus,, , The general equation of a circle is a second degree, equation in x and y, coefficient of xy is zero,, , the general equation of a circle is, x2 + y2 + 2g x + 2 fy + c = 0 whose centre is, (–g, –f) and radius is, , coefficient of x2 = coefficient of y2, , g +f − c ., 2, , 2, , Activity :, , SOLVED EXAMPLES, , Consider the general equation of a circle, , Ex. 1) Prove that 3x2 + 3y2 − 6x + 4y − 1 = 0,, , x2 + y2 + 2gx + 2 fy + c = 0, + y2 + 2 fy +, , x2 + 2gx +, ∴ (x +, , )2 + (y +, , ∴ [x – (, , )] + [y – (, 2, , =, , +, , represents a circle. Find its centre and radius., , –c, , Solution : Given equation is, , )2 = g2 + f 2 – c., , 3x2 + 3y2 − 6x + 4y − 1 = 0, , )] = (   ), 2, , 2, , dividing by 3, we get, , (use centre radius form of equation of the circle)., Therefore centre of the circle is (–g, –f ) and radius, is, , 4y, 1, −, = 0 comparing, 3, 3, , x2 + y2 − 2x +, this with, , g 2 +f 2 − c ., , x2 + y2 + 2gx + 2fy + c = 0, ∴ g = –1, , we get 2g = −2, , Let's Remember, , 2f =, , 4, 3, , ∴f=, , 2, −1, and c =, 3, 3, , (1) If g2 + f 2 − c > 0, the equation, x2 + y2 + 2gx + 2fy + c = 0 represents a circle, in the xy plane., , g + f − c = (−1) +, , (2) If g2 + f 2 − c = 0, then the equation, x2 + y2 + 2gx + 2fy + c = 0 represents a point, which is true degenerate conic and is the, limiting position (radius is 0)., , As, , 2, , 2, , =1+, , 2, , 2, 3,  , , 2, , −, , −1, 3, , 4 1 16, + =, 9, 9 3, , 16, >0, 9, , g2 + f2 − c > 0, , ∴ 3x2 + 3y2 − 6x + 4y − 1 = 0 represents a, , , , (3) If g2 + f 2 − c < 0, then the equation, x2 + y2 + 2gx + 2fy + c = 0 does not represent, , circle., , ∴ Centre of the circle = (−g, −f) = (1,, , any point in the xy plane., , Radius of circle r = ( − g ), , Activity :, Check whether the following equations, represent a circle. If, so then find its centre and, , =, , radius., , 2, , + (− f, , ), , 2, , −2, ), 3, , −c, , 16 4, =, 3, 9, , b) x2 + y2 − 8x + 6y + 29 = 0, , Ex. 2) Find the equation of the circle passing, through the points (5,−6), (1,2) and, (3, −4)., , c) x2 + y2 + 7x – 5y + 15 = 0, , Solution : Conside P = (5, −6), Q = (1, 2),, , a) x + y – 6x – 4y + 9 = 0, 2, , 2, , 130

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Ex. 3) Show that the points (5, 5), (6, 4), (−2, 4), and (7, 1) are on the same circle; i.e. these, points are concyclic., , R = (3, −4), Let the centre of the, circle be at C(h, k), ∴ r = CP = CQ = CR, , Q, R, , C (h, k), , P, = radii of the same circle, , Solution :, Let, the equation of the circle be, x2 + y2 + 2gx + 2fy + c = 0 , , Fig. 6.6, , Consider CP = CQ , , ....... (I), , The circle passing through (5, 5), (6, 4),, , ∴ CP2 = CQ2, , (−2, 4), , By using distance formula,, (h − 5)2 + (k + 6)2 = (h − 1)2 + (k − 2)2, , ∴, , 50 + 10g + 10f + c = 0, , ....... (II), , ∴ h2 − 10h + 25 + k2 + 12k + 36, , ∴, , 52 + 12g + 8f + c = 0 , , ....... (III), , ∴, , 20 − 4g + 8f + c = 0 , , ....... (IV), , = h2 − 2h + 1 + k2 − 4k + 4, i.e −8h + 16k = −56, , Now (III) − (IV) gives, , h – 2k = 7 , Now consider, CQ = CR , , ........ (I), , ∴ 16g = −32, , ∴ CQ2 = CR2, , 32 + 16g = 0, , ∴ g = –2, , Subtracting (II) from (III), , (h − 1)2 + (k − 2)2 = (h − 3)2 + (k + 4)2, , we get 2 + 2g − 2f = 0 , , ∴ h2 − 2h + 1 + k2 − 4k + 4, , substitute g = –2 in equation (V), 2 + (–4) – 2f = 0, , = h2 − 6h + 9 + k2 + 8k + 16, , 20 + c = 4(–2) – 8(1) = −8 + 8, , Now, subtracting (II) from I, we get,, , 20 + c = 0, , k=2, , equation (I) we get,, , ∴ C = (11, 2), , x2 + y2 − 4x −2y – 20 = 0 .... (VI) is the, , Radius of the circle is, , equation of the circle passing through the, , r = CP = (11 − 5) 2 + ( 2 + 6) 2 =, , 100, , points (5,5), (6,4) and (−2,4)., , = 10;, , If (7,1) satisfies equation (VI), the four points, , By using centre-radius form, equation of, , are concyclic., , the circle is (x − 11)2 + (y − 2)2 = 100, , L.H.S. = (7)2 + (1)2 − 28 − 2 − 20, , ∴ x − 22x + 121 + y − 4y + 4 = 100, 2, , ∴, , 2, , = 49 + 1 − 50 = 0 = R.H.S., , x + y − 22x – 4y + 25 = 0, 2, , ∴ c = –20, , Now substituting g = −2, f = –1 and c = –20 in, , Substituting k = 2 in (II) we get, h = 11, , 2 - 4 = 2f ∴ f = –1, , Now substitute g = −2 and f = −1 in eqn. (IV), , ∴ h − 3k = 5 ......... (II), , i.e. 4h − 12k = 20, , ..... (V), , 2, , ∴ Thus, the point (7, 1) satisfies the equation of, , , , circle., , ∴ The given points are concyclic ., , 131

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x = h + r cos θ and y = k + r sin θ., , EXERCISE 6.2, , Hence co-ordinates of any point on the circle, , (1) Find the centre and radius of each of the, , are (h + r cos θ , k + r sin θ)., , following., , (2) Sometimes parametric form is more convenient, for calculation as it contains only one variable., , (i) x + y − 2x + 4y − 4 = 0, 2, , 2, , (ii) x2 + y2 − 6x − 8y − 24 = 0, , 6.3.1 Tangent : When a line intersects a circle in, coincident points, then that line is called as a, tangent of the circle and the point of intersection, is called point of contact., , (iii) 4x2 + 4y2 − 24x − 8y − 24 = 0, (2) Show that the equation, 3x2 + 3y2 + 12x + 18y − 11 = 0 represents a, , The equation of tangent to a standard circle, x + y2 = r2 at point P(x1, y1) on it., , circle., , 2, , (3) Find the equation of the circle passing through, , Given equation of a circle is x2 + y2 = r2. The, , the points (5, 7), (6, 6) and (2, −2)., , centre of the circle is at origin O(0, 0) and radius is r., , (4) Show that the points (3, −2), (1, 0), (−1, −2) and, , Let P (x1, y1) be any point on the circle., , (1, −4) are concyclic., , y1, y1 − 0, Slope of OP = x − 0 = , if x1 ≠ 0, y1 ≠ 0., x1, 1, , 6.3 Parametric Form of a circle :, Let P(x, y) be any, , A tangent is drawn to the circle at point P., , point on a circle, , Since OP is perpendicular, , with centre at O, , to tangent at point P., , and radius r., , θ, , slope of the tangent, , As shown in, Fig. 6.7, OP makes, , m=, , an angle θ with the, positive, , direction, , of X-axis. Draw, , −x1, y1, , O(0, 0), , ∴ equation of the tangent, , Fig. 6.7, , in slope point form is, , PM ⊥ X-axis from P., , , y – y1 = m(x – x1), , ∆ OMP is a right angled triangle,, , , y – y1 =, , OM, , x, , ∴ cos θ = OP = ;, r, ∴ x = r cos θ, , PM, , P(x1, y1), , y, , Fig. 6.8, , −x1, (x – x1), y1, , , yy1 – y12 = –xx1 + x12, , sin θ = OP = r, ∴ y = r sin θ, , , xx1 + yy1 = x12 + y12 .............(I), , x = r cos θ and y = r sin θ is the parametric form, , As (x1, y1) lies on the circle, x12 + y12 = r2,, , of circle x2 + y2 = r2. θ is called parameter., , therefore equation (I) becomes xx1 + yy1 = r2, Thus, In general, equation of the tangent to a, , Note that:, , circle x2 + y2 = r2 at point P(x1, y1) is xx1+yy1= r2, , (1) The parametric form of circle, (x − h)2 + (y − k)2 = r2 is given by, 132

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Ex. 3) Find the equation of the tangent to the circle, , ∴ 16x2 + 9x2 + 90x + 225 = 144, ∴ 25x2 + 90x + 81 = 0, , x2 + y2 − 4x − 6y − 12 = 0 at (−1, −1), , ∴ (5x + 9)2 = 0, , Solution : The equation of circle is, , −9, x=, ; The roots of equation are equal., 5, , x2 + y2 − 4x − 6y − 12 = 0, It is of the type x2 + y2 + 2gx + 2fy + c = 0, , \ line (2) is tangent to given circle (1), \y=, , 1, 4, , (3x + 15), , \ g = −2, f = −3, c = −12, Let P (−1, −1) = (x1, y1), , = 1 (3  − 9  + 12), 4  5, 12, =, 5, , We know that the equation of a tangent to, a circle, ,  9 12  is the only point of intersection of, − , ,  3 5 , , x2 + y2 + 2gx + 2fy + c = 0 at (x1, y1) is, , the line and circle., , xx1 + yy1 + g(x + x1) + f(y+y1) + c = 0, , \ The line 3x − 4y +15 = 0 touches the circle,  9 12 , at  − , ,  3 5 , , −x −y −2x + 2 −3y + 3−12 = 0, , x(−1) + y(−1) + 2(x−1) −3 (y−1) −12 = 0, 3x + 4y + 7 = 0, ,  9 12 , \ point of contact =  − , ,  3 5 , , Activity :, , EXERCISE 6.3, , Equation of a circle is x + y = 9., Its centre is at ( , ) and radius is, Equation of a line is 3x – 4y + 15 = 0, 2, , ∴y=, , 2, , (1) Write the parametric equations of the circles, (i) x2 + y2 = 9 (ii) x2 + y2 + 2x − 4y − 4 = 0, , x+, , (iii) (x − 3)2 + (y + 4)2 = 25, , Comparing it with y = mx + c, m=, , (2) Find the parametric representation of the, , and c =, , circle 3x2 + 3y2 − 4x + 6y − 4 = 0., , We know that, if the line y = mx + c is a tangent, to x2 + y2 = a2 then c2 = a2m2 + a2, Hence c2 = ( )2, 225, , =, .... (I), 16, , Also, c2 = a2m2 + a2, =9, , +9, , , , =9, , 9, +1, 16, , , , 9(25), =, 16, , , , =, , (3) Find the equation of a tangent to the circle, x2 + y2 − 3x + 2y = 0 at the origin., (4) Show that the line 7x − 3y − 1 = 0 touches the, circle x2 + y2 + 5x − 7y + 4 = 0 at point (1, 2), (5) Find the equation of tangent to the circle, x2 + y2 − 4x + 3y + 2 = 0 at the point (4, −2), , 225, ...........(II), 16, , From equations (I) and (II) we conclude that, 135

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(II) Answer the following :, , (6) Area of the circle centre at (1, 2) and passing, through (4, 6) is, (A) 5π , , (B) 10π , , C) 25π , , (D) 100π, , Q. 1 Find the centre and radius of the circle, x2 + y2 − x +2y − 3 = 0, Q. 2 Find the centre and radius of the circle, x = 3 – 4 sinq, y = 2 – 4cosq, , (7) If a circle passes through the point (0, 0),, (a, 0) and (0, b) then find the co-ordinates of, its centre., (A)  , ,  2 2 , ,  a −b , (B)  , , 2 2 , , (C)  −a , b , , a b, (D)  , ,  2 2, ,  −a −b , ,  2 2, , Q. 3 Find the equation of circle passing through, the point of intersection of the lines, x + 3y = 0 and 2x − 7y = 0 whose centre, is the point of intersection of lines, x + y + 1 = 0 and x − 2y + 4 = 0., Q. 4 Find the equation of circle which passes, through the origin and cuts of chords of, length 4 and 6 on the positive side of, x - axis and y axis respectively., , (8) The equation of a circle with origin as centre, and passing through the vertices of an, equilateral triangle whose median is of, length 3a is, (A) x2 + y2 = 9a2 , , (B) x2 + y2 = 16a2, , (C) x2 + y2 = 4a2 , , (D) x2 + y2 = a2, , Q. 5 Show that the points (9, 1), (7, 9), (−2, 12), and (6, 10) are concyclic., Q. 6 The line 2x − y + 6 = 0 meets the circle, x2 + y2 + 10x + 9 = 0 at A and B. Find the, equation of circle on AB as diameter., , (9) A pair of tangents are drawn to a unit circle, with cnetre at the origin and these tangents, intersect at A enclosing an angle of 60 . The, area enclosed by these tangetns and the area, of the circle is, (A), (C), , 2, 3, , −, , π, (B), 6, , π, 3, (D), −, 3 6, , 3−, , Q. 7 Show that x = −1 is a tangent to circle, x2 + y2 − 2y = 0 at (−1, 1)., Q. 8 Find the equation of tangent to the circle, , π, 3, ,  2π , , x2 + y2 = 64 at the point P  ,  3 , ,  π, 3 1 − , 6, , , Q.9 Find the equation of locus of the point of, intersection of perpendicular tangents, drawn to the circle x = 5cosq and y = 5sinq., , (10) The parametric equations of the circle, x2 + y2 + mx + my = 0 are, (A) x =, , −m m, −m m, +, sin θ, +, cos θ , y =, 2, 2, 2, 2, , (B) x =, , +m m, −m m, +, sin θ, +, cos θ , y =, 2, 2, 2, 2, , Q.10 Find the equation of the circle concentric with, x2 + y2 – 4x + 6y = 1 and having radius, 4 units., Q.11 Find the lengths of the intercepts made on, the co-ordinate axes, by the circle., (i) x2 + y2 – 8x + y – 20 = 0, (ii) x2 + y2 – 5x + 13y – 14 = 0, , (C) x = 0 , y = 0, (D) x = mcosθ ; y = msinθ, 137

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4), , 2), , Fig. 6.13, Fig. 6.11, , Circles intersecting each other., Line joining the point of intersection is the, common chord also called as the radical axis., , Circles touching each other internally., d(c1 c2) = | r1 − r2 |, Exactly one common tangent can be drawn., , (seg AB), , 3), , Exactly two common tangent can be drawn., 5), , Fig. 6.12, Disjoint circles., | r1 + r2 | < d(c1 c2), Exactly four common tangents can be drawn., , Fig. 6.14, Concentric circles, No common tangent can be drawn., , 139

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7, , Conic sections, a wide range of applications such as planetary, motions, in designs of telescopes and antennas,, reflection in flash light, automobile headlights,, construction of bridges, navigation, projectiles, etc., A straight line, a circle, parabola, ellipse and, hyperbola are all conic sections. Of these we have, studied circle and straight line., , Let's Learn, , , , , , , Conic sections : parabola, ellipse, hyperbola, Standard equation of conics, Equation of tangent to the conics, Condition for tangency, , Earlier we have studied different forms of, equations of line, circle, and their properties. In, this chapter we shall study some more curves,, namely parabola, ellipse and hyperbola which are, conic sections., , Let's Recall, , , , Section formulae : Let A (x1,y1), B(x2,y2) be two points in a plane., and Q divide seg AB in the ratio, internally and externally respectively, , and, If P, m:n, then, , Let's Learn, ,  mx + nx1 my2 + ny1 , ,, P=  2,  and, m+n ,  m+n, , 7.1.1 Double cone:, Let l be a fixed, line and m another line, intersecting it at a fixed, point V and inclined, at an acute angle θ, (fig 7.1.) Suppose we, rotate the line m around, the line l in such a way, that the angle θ remains, ,  mx − nx1 my2 − ny1 , ,, Q=  2, , m−n ,  m−n, Introduction:, The Greek mathematician Archimedes, and Apollonius studied the curves named conic, sections. These curves are intersections of a plane, with right circular cone. Conic sections have, , Fig.7.1, , V, , Fig.7.2, , Fig.7.3(a), , Fig.7.3(b), 140, , Fig.7.4

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This constant ratio is, called the eccentricity of the, conic section, denoted by e., , constant. Then the surface generated is a doublenapped right circular cone., The point V is called the vertex, the line l, is called axis, and the rotating line m is called a, generator of the cone. The vertex V separates the, cone in to two parts called nappes (Fig.7.2)., , Hence we write, , or SP = e PM. This is called, Focus - Directrix property, of, the, conic, section., The nature of the conic section, depends upon the value of e., , 7.1.2. Conic sections :, Let’s construct, Take a carrot or a cone made of drawing paper, and cut it with a plane satisfying the following, conditions., i), , The plane is perpendicular to the axis and, does not contain vertex, the intersection is a, circle (studied earlier Fig. 7.3(a))., , ii), , The plane is parallel to one position of the, generator but does not pass through the, vertex, we get a parabola (Fig. 7.3(a))., , iii), , The plane is oblique to the axis and not, parallel to the generator we get an ellipse, (Fig. 7.3(a))., , iv), , If a double cone is cut by a plane parallel to, axis, we get parts of the curve at two ends, called hyperbola (Fig. 7.3(b))., , v), , A plane containing a generator and tangent, to the cone, intersects the cone in that, generator. We get pair to straight lines, (Fig. 7.4)., , SP, = e, PM, , Fig. 7.5, , i), , If e = 1, the conic section is called parabola., , ii), , If 0 < e < 1, the conic section is called an, ellipse., , iii), , If e > 1, the conic section is called, hyperbola., , 7.1.4. Some useful terms of conic sections:, 1), , Axis: A line about which a conic section, is symmetric is called an axis of the conic, section., , 2), , Vertex : The point of intersection of a conic, section with its axis of symmetry is called a, vertex., , 3), , Focal Distance : The distance of a point on, a conic section from the focus is called the, focal distance of the point., , 4), , Focal chord : A chord of a conic section, passing through its focus is called a focal, chord., , 5), , The fixed point is called the focus of the conic, section, denoted by S. The fixed straight line is, called the directrix of conic section, denoted by d., , Latus-Rectum: A focal chord of a conic, section which is perpendicular to the axis of, symmetry is called the latus-rectum., , 6), , If S is the focus, P is any point on the, conic section and segment PM is the length of, perpendicular from P on the directrix, then by, , Centre of a conic : The point which bisects, every chord of the conic passing through it,, is called the centre of the conic., , 7), , Double ordinate : A chord passing through, any point on the conic and perpendicular to, the axis is called double ordinate., , 7.1.3. Definition of a conic section and its, equation:, A conic section or conic can be defined as, the locus of the point P in a plane such that the, ratio of the distance of P from a fixed point to its, distance from a fixed line is constant., , definition, , SP, = constant. (fig. 7.5), PM, 141

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7.1.5. Parabola, , ( x − a ) + ( y − 0), 2, , Definition: A parabola is the locus of the point, in plane equidistant from a fixed point and a, fixed line in that plane. The fixed point is called, the focus and the fixed straight line is called, the directrix., , 2, , =, , ( x + a) + ( y − y), 2, , 2, , Squaring both sides (x – a )2 + y2 = (x + a)2, that is x2 – 2ax + a2 + y2 = x2 + 2ax + a2, , Standard equation of the parabola:, , that is y2 = 4 ax (a > 0), , Equation of the parabola in the standard form, y2 = 4ax., , This is the equation of parabola in standard form., , Let S be the focus and d be the directrix of the, parabola., , Activity :, Trace the parabola using focus directrix, , Let, SZ, be, perpendicular, to the directrix., Bisect, SZ, at, the point O. By, the, definition, of parabola the, midpoint O is, on the parabola., Take O as the, Fig. 7.6, origin, line OS as, the X - axis and the line through O perpendicular, to OS as the Y - axis., Let SZ = 2 a, a > 0., , property., 1), , find the equation of parabola with focus at, (2, 0) and directrix x + 2 = 0., , 2), , Find the equation of parabola with focus at, (– 4 , 0) and directrix x = 4., , 7.1.6. Tracing of the parabola y2 = 4 ax (a>0), 1), , Symmetry : Equation of the parabola can, be written as y = ± 2 ax that is for every, value of x, there are two values of y which, are negatives of each other. Hence parabola, , Then the coordinates of the focus S are (a, 0) and, the coordinates of Z are (–a, 0)., , is symmetric about X- axis., , The equation of the directrix d is, , 2), , Region : For every x < 0,the value of y is, , x = –a, i.e. x + a = 0, , imaginary therefore entire part of the curve, , Let P (x, y) be any point on the parabola. Draw, segment PM perpendicular to the directrix d., , lies to the right of Y-axis., 3), , ∴ M= (– a , y), , y = 0, therefore the curve meets the co, , By using distance formula we have, SP =, , ( x − a ) 2 + ( y − 0) 2 ,, , PM =, , ( x + a)2 + ( y − y)2, , Intersection with the axes: For x = 0 we have, ordinate axes at the origin O(0, 0), , 4), , Shape of parabola: As x → ∞, y → ∞., Therefore the curve extends to infinity as x, grows large and opens in the right half plane, , By focus – directrix property of the parabola, SP = PM, , Shape of the parabola y2 = 4 ax (a > 0) is as, shown in figure 7.6., 142

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7.1.7 Some results, , LS = L'S = l (say)., , 1) Focal, distance : Let, P (x1,y1) be, any point on, the parabola, y2 = 4 ax, , So the coordinates of L are (a , l), Since L lies on y2 = 4 ax, l2 = 4a (a), l2 = 4a2, l = ±2a, As L is in the first quadrant,, , Let segment, PM, is, perpendicular, to the directrix, d, then M is (−a , y1), SP=PM=, , l = 2a, Fig.7.7, , 2, , ( x1 + a ) + ( y1 − y1 ), , Length of latus rectum LSL' = 2l = 2(2a), = 4a, The co-ordinates of ends points of the latus, rectum are L (a, 2a) and L' (a,−2a), , 2, , = x1 + a, Activity :, , ∴ focal distance SP = x1 + a, = a + abscissa of point P., , 2), , 1), , Find the length and end points of latus, rectum of the parabola x2= 8y, , Length of latus-Rectum:, In figure 7.7 LSL' is the latus-rectum of the, , 2), , parabola y2 = 4 ax. By symmetry of the curve, , Find the length and end points of latus, rectum of the parabola 5y2= 16x, , 7.1.8 Some other standard forms of parabola, y2 = −4ax, , Fig.7.8, , l>0, , x2 = 4by, , Fig.7.9, , We summarize the properties of parabola in four standard forms, , 143, , x2 = −4by, , Fig.7.10

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Equation of the, parabola y2 = 4ax, Terms, , 7.1.10 General forms of the equation of a, parabola, , x2 = 4by, , 1 Focus, , (a, 0), , ( 0, b), , 2 Equation of, directrix, , x+a=0, , y+b = 0, , 3 Vertex, , 0(0,0), , 0(0,0), , 4 End points of latus, rectum, , (a, ±2a), , (±2b, b), , Length of latus, 5 rectum, , |4a|, , |4b|, , 6 Axis of symmetry, , X-axis, , Y-axis, , 7 Equation of axis, , y=0, , x=0, , 8 Tangent at vertex, , Y-axis, , X-axis, , 9 Focal distance of a, point P(x1,y1), , |x1 + a|, , | y1 + b|, , If the vertex is shifted to the point (h, k) we, get the following form, 1), , (y – k)2 = 4a (x – h), This represents a parabola whose axis of, , symmetry is y – k = 0 which is parallel to the, X-axis, vertex is at (h, k) and focus is at (h + a , k), and directrix is x = h – a., It can be reduced to the form x = Ay2 + By + C., OR, Y2 = 4a X, where X = x – h , Y= y – k, Activity :, 1), , Obtain the equation of the parabola with its, axis parallel to Y– axis and vertex at (h, k), , Parameter : If the co-ordinates of a point on, 2), , the curve are expressed as functions of a variable,, , Find the coordinates of the vertex, focus, and equation of the directrix of the parabola, , that variable is called the parameter for the curve., , y2 = 4x + 4y, , 7.1.9 Parametric expressions of standard, parabola y2= 4ax, , SOLVED EXAMPLES, , x = at , y = 2at are the expressions which, 2, , Ex. 1) Find the coordinates of the focus, equation, of the directrix, length of latus rectum and, coordinates of end points of latus rectum of, each of the following parabolas., , satisfies given equation y2 = 4ax for any real value, of t that is y2 = (2at)2 = 4 a2 t2 = 4a (at2) = 4ax, where t is a parameter, P(x1, y) ≡ (at2, 2at) describes the parabola, , i) y2 = 28x , , y2 = 4ax, where t is the parameter., , Solution:, , Activity :, 1), , i) y2 = 28x, , For the parabola y2= 12x , find the parameter, , Equation of the parabola is y2 = 28x, , for the point a) (3, –6) b) (27,18), 2), , ii) 3x2 = 8y, , Find the parameter for the point (9, –12), , comparing this equation with y2 = 4ax, we get, 4a = 28 ∴ a = 7, , of the parabola y2 = 16x, , Coordinates of the focus are S(a,0) = (7,0), 144

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Equation of the directrix is x + a = 0 that is, x+7=0, , ∴ (6) 2 = 4 b × – 3), , ∴ –12 b = 36 ∴ b = – 3, , Length of latus rectum = 4a = 4 × 7 = 28, , ∴ equation of parabola is x2 = 4(–3)y, , End points of latus rectum are (a,2a) and, (a,–2a). that is (7,14) and (7, –14), , x2 = –12y that is x2 + 12y = 0., , Ex. 3) Find the equation of the parabola whose, diretrix is x + 3 = 0, , ii) 3x2 = 8y, Equation of the parabola is 3x2 = 8y that is, , Solution:, , 8, y, 3, comparing this equation with x2 = 4by, we, 8, 2, get 4b = ∴ b =, 3, 3, , x2 =, , Here equation of diretrixs is x + a = 0 that, is x + 3 = 0 comparing we get a = 3., ∴ Equation of the parabola y2 = 4ax that is, y2 = 12x., Ex. 4) Calculate the focal distance of point P on, the parabola y2 = 20x whose ordinate is 10, , Co-ordinates of the focus are S (0, b) =, (0,, , 2, ), 3, , Solution : Equation of parabola is y2 = 20x, comparing this with y2 = 4ax, , Equation of the directrix is y + b = 0 that is, , we get, , 2, y + = 0 that is 3y + 2 = 0, 3, , 4a = 20 ∴ a = 5, , Here ordinate = y – coordinate = 10, ∴ (10)2 = 20 x, , 2 8, Length of latus rectum = 4b = 4 ×   =, 3 3, Coordinates of end points of latus rectum are, , ∴ 20x = 100, , 100, =5, 20, Now focal distance = a + x, , ∴x=, , 4 2, (2b, b) and (−2b, b). that is ( , ) and, 3 3, 4 2, (– , ), 3 3, , , , = a + abscissa of point, , , , = 5 + 5 = 10 units, , Ex. 5) Find the equation of the parabola having, (4,−8) as one of extremities of porabola., , Ex. 2) Find the equation of the parabola with, vertex at the origin, axis along Y-axis and, passing through the point (6,–3), , Solution : Given that, one of the extrimities, of the latus rectum of the parabola is, (4, −8) therefore other must be (4,8)., End-coordinates of latus - rectum (a, ± 2a) =, (4, ±8)., , Solution:, The vertex of the parabola is at the origin,, it’s axis is along Y-axis. Hence equation of the, parabola is of the form x2 = 4by., , ∴a=4, , Now the point (6,−3) lies on this parabola., Hence the coordinates of the points satisfy the, equation of the parabola., , Equation of parabola is y2 = 4ax, , 145, , y2 = 4(4)x ∴ y2 = 16x

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Ex. 6) For the parabola 3y2 =16x, find the, parameter of the point (3,−4), , Equation of axis is X = 0 that is x + 2 = 0, Equation of diretrix is Y + b = 0 that is, y + 3 – 1 = 0 that is y + 2 = 0, Equation of tangent at vertex is Y = 0 that, is y + 3 = 0, , Solution : Equation of parabola is 3y2 =16x, 16, x comparing this with y2 = 4ax we get, ∴ y2 =, 3, 16, 4, 4a =, ∴ a =, Parametric equations of the, 3, 3, 4 2 8 , parabola y2 = 4ax are (at2, 2at) =  t , t , 3 3 , , 7.1.11 Tangent :, A straight line which intersects the parabola in, coinsident point is called a tangent of the parabola, , 4 2 8 ,  t , t  = (3,–4), 3 3 , , 8, Equating second components we get, t, 3, =–4, 3, 3, ∴t=–4× =–, 8, 2, ∴ Parameter t = –, , 3, 2, , Ex. 7) Find the coordinates of the vertex and, focus, the equation of the axis of symmetry,, diretrix and tangent at the vertex of the, parabola x2 +4x + 4y + 16 = 0, , Fig. 7.11, Point Q moves along the curve to the point, P. The limiting position of secant PQ is the tangent, at P., A tangent to the curve is the limiting, position of a secant intersecting the curve in, two points and moving so that those points of, intersection come closer and finally coincide., , Solution : Equation of parabola is, x2 + 4x + 4y + 16 = 0, x2 + 4x = – 4y – 16, x2 + 4x + 4 = – 4y – 12, (x+2)2 = – 4 (y+3), , Tangent at a point on a parabola., Let us find the equation of tangent to the, parabola at a point on it in cartesion form and in, parametrics form., , Comparing this equation with X2 = –4bY, We get X = x + 2 , Y = y + 3 and 4b = –4, ∴ b = –1, , We find the equation of tangent to the, parabola y2 = 4ax at the point P(x1, y1) on it. Hence,, obtain the equation of tangent at P(t)., , Coordinates of the vertex are X = 0 and, Y = 0 that is x + 2 = 0 and y + 3 = 0, ∴ x = –2 and y = –3, , Equation of the tangent to the curve y = f(x), at point (x1 y1) on it is,, , ∴ Vertex = (x, y) = (–2,–3), , y – y1 = [f (x)](x, , Coordinates of focus are given by X = 0 and, Y=+b, that is x + 2 = 0 and y + 3 = – 1, ∴ x = – 2 and y = –4, ∴ Focus = (–2, –4), , 1, y1), , (x – x1) [f (x)](x, , 1, y1), , We need to know the slope of the tangent at, P(x1, y1). From the theory of derivative of a function,, dy, the slope of the tangent is, at (x1, y1), dx, 146

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Ex. 2) Find the equation to tangent to the parabola, y2 = 12x from the point (2,5)., Solution :, Equation of the parabola is y2 = 12x, comparing it with y2 = 4ax ⇒ 4a = 12, ∴a=3, Tangents are drawn to the parabola from the, point (2,5)., We know, equation of tangent to the, a, parabola y2 = 4ax having slopes m is y = mx + ., m, (3), (5) = m(2) +, m, 2, 5m = 2m + 3, 2m2 − 5m + 3 = 0, 2m2 − 2m − 3m + 3 = 0, (2m − 3) (m − 1) = 0, 3, m=, or m = 1, 2, These are the slopes of tangents., Therefore the equations of tangents by slope, - point form are, 3, (y − 5) = (x − 2) and (y − 5) = 1 (x − 2), 2, ∴ 2y − 10 = 3x − 6 and y − 5 = x − 2, ∴ 3x − 2y + 4 = 0 and x − y + 3 = 0, , This is quadratic equation in m and in general it, has two roots say, m1 and m2 which are the slopes, of two tangents., Thus, in general, two tangents can been drawn to, a parabola from a given point in its plane., If the tangent drawn from P are mutually, perpendicular we have, m1 m2 = –1, From equation (1) m1 m2 =, ∴, , a, = –1, x1, , a, (product of roots), x1, , ∴ x1 = –a, , which is the equation of directrix., Thus, the locus of the point, the tangents from, which to the parabola are perpendicular to each, other is the directrix of the parabola., SOLVED EXAMPLES, Ex. 1) Find the equation of tangent to the parabola, y2 = 9x at (1,−3)., , Solution :, Equation of the parabola is y2 = 9x;, , Ex. 3) Show that the tangents drawn from the, point (−4,−9) to the parabola y2 = 16x are, perpendicular to each other., , comparing it with y2 = 4ax, 9, 4a = 9 ⇒ a =, 4, Tangent is drawn to the parabola at (1,−3) = (x1,y1), , Solution :, Equation of the parabola is y2 = 16x., comparing it with y2 = 4ax ⇒ 4a = 16, ∴a=4, Tangents are drawn to the parabola from, point (−4,−9)., Equation of tangent to the parabola y2 = 4ax, , Equation of tangent to the parabola y = 4ax at, (x1,y1) is yy1 = 2a(x + x1), 2, , ∴ Equation of tangent to the parabola, 9, y2 = 4x at (1,−3) is y(−3) = 2   (x + 1), 4, 9, i.e. −3y =   (x + 1), 2, , having slope m is y = mx +, , i.e. −6y = 9x + 9, i.e. 3x + 2y + 3 = 0, , ∴ (−9) = m(−4) +, 148, , 4, m, , a, m

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∴ −9m = −4m2 + 4, ∴ 4m2 − 9m + 4 = 0, m1 and m2 be the slopes (roots), constant, (m1.m2) = +, co-efficient of m 2, , 8), , Find coordinate of the point on the parabola., Also find focal distance. i) y2= 12x whose, parameter is 1/3 ii) 2y2 = 7x whose parameter, is –2, , 9), , 4, ∴ m1.m2 = −1, 4, hence tangents are perpendicular to each, , For the parabola y2 = 4x, find the coordinate, of the point whose focal distance is 17., , 10), , Find length of latus rectum of the parabola, y2 = 4ax passing through the point (2.–6)., , 11), , Find the area of the triangle formed by, the line joining the vertex of the parabola, x2 = 12y to the end points of latus rectum., , 12), , If a parabolic reflector is 20cm in diameter, and 5 cm deep, find its focus., , 13), , Find coordinate of focus, vertex and, equation of directrix and the axis of the, parabola y = x2 – 2x + 3, , 14), , Find the equation of tangent to the parabola, i) y2 = 12x from the point (2,5) , ii) y2 = 36x from the point (2,9), , 15), , If the tangent drawn from the point (–6,9), to the parabola y2 = kx are perpendicular to, each other, find k., , m1.m2 = −, other., , Activity :, 1), , Find the equation of tangent to the parabola, y2 = 9x at the point (4,-6), , 2), , Find the equation of tangent to the parabola, y2 = 24x having slope 3/2, , 3), , Show that the line y = x + 2 touches the, parabola y2 = 8x. Find the coordinates of, point of contact., EXERCISE 7.1, , 1), , Find co-ordinate of focus, equation of, directrix, length of latus rectum and the, co ordinate of end points of latus rectum, of the parabola i) 5y2 = 24x ii) y2 = –20x, iii) 3x2 = 8y iv) x2 = –8y v) 3y2 = –16x, , 16), , Two tangents to the parabola y2 = 8x meet, the tangents at the vertex in the point P and, Q. If PQ = 4,prove that the equation of the, locus of the point of intersection of two, tangent is y2 = 8(x + 2)., , 2), , Find the equation of the parabola with, vertex at the origin, axis along Y-axis and, passing through the point (–10,–5), , 17), , Find the equation of common tangent to the, parabola y2 = 4x and x2 = 32y., , 3), , Find the equation of the parabola with, vertex at the origin, axis along X-axis and, passing through the point (3,4), , 18), , Find the equation of the locus of a point, the, tangents from which to the parabola y2 =, 18x are such that some of their slopes is -3, , 4), , Find the equation of the parabola whose, vertex is O (0,0) and focus at(–7,0)., , 19), , 5), , Find the equation of the parabola with, vertex at the origin, axis along X-axis and, passing through the point i) (1,–6) ii) (2,3), , The tower of a bridge, hung in the form of, a parabola have their tops 30 meters above, the road way and are 200 meters apart. If the, cable is 5meters above the road way at the, centre of the bridge, find the length of the, vertical supporting cable from the centre., , 6), , For the parabola 3y2 =16x, find the parameter, of the point, a) (3,–4), b) (27,–12), , 20), , 7), , Find the focal distance of a point on the, parabola y2 = 16x whose ordinate is 2 times, the abscissa., , A circle whose centre is (4,–1), passes through the focus of the parabola, x2 + 16y = 0., Show that the circle touches the diretrixs of, the parabola., , 149

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7.2 Ellipse, , If S is a fixed point is called focus and, directrix d is a fixed line not containing the focus, , Let's Study, •, •, •, •, , then by definition PS = e and PS = e PM , where, PM, PM is the perpendicular on the directrix and e is, the real number with 0 < e < 1 called eccentricity, of the ellipse., , Standard equation of the ellipse., Equation of tangent to the ellipse., Condition for tangency., Auxilary circle and director circle of the, ellipse, , Fig. 7.12, , Fig. 7.13, , The ellipse is the intersection of double, napped cone with an oblique plane., , 7.2.1 Standard equation of ellipse, Let’s derive the standard equation of the ellipse, x2, y2, +, = 1, a > b, a2, b2, , Fig. 7.14, , Fig. 7.15, , An ellipse is the locus of a point in a plane, which moves so that its distance from a fixed, point bears a constant ratio e ( 0 < e < 1 ) to its, distance from a fixed line. The fixed point is, called the focus S and the fixed line is called, the directrix d., , Let S be the focus, d be the directrix and e be the, eccentricity of an ellipse., Draw SZ perpendicular to directrix. let A and, A' divide the segment SZ internally as well as, externally in the ratio e : 1., 150

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Yaxis, , d directrix, , d, , 1, , B, , L, , 1, M, M', , A, A'1, , zz'1, , O, , S1, S', , P, , M, , A, , S, , Z, , X axis (y = 0), , LL'1, , B, B', , 1, , latus rectum (LSL1), , Fig. 7.16, let AA' = 2 a , midpoint O of segment AA' be the, origin. Then O ≡ (0,0) , A ≡ (a, 0) and A' ≡ (– a ,0), , a, That is x – � = 0., e, , By definition of ellipse A and A’ lie on ellipse., SA e, =, AZ 1, , SP = focal distance, , SA ' −e, =, A'Z, 1, , =, , Let P(x , y) be any point on the ellipse., , ( x − ae ) + ( y − 0 ), 2, , 2, , ….. ( 4 ), , PM = distance of point P from directrix, , Since P is on the ellipse SP = e PM . . . (1), therefore SA = e AZ ., , =, , Let Z ≡ (k,0) and S ≡ (h,0), , a, e = x − a ……. ( 5 ), e, 12 + 02, x−, , By section formula, e k + 1h, a=, e +1, , From (1) , (4) and (5), , e k -1h, also – a =, e -1, , ae+a=ek+h, , ( x − ae) 2 + ( y − 0) 2 = e x −, , . . . (2), , a, e, , ( x − ae) 2 + ( y − 0) 2 = ex − a, , – a e + a = e k – h . . . (3), Solving these equations , we get, , Squaring both sides, , k = a/e and, , (x – ae)2 + (y – 0) = e2x2 – 2aex + a2, , h=ae, , Focus S ≡ (ae , 0) and Z ≡ (a/e , 0), Equation of the directrix is x =, , x2 – 2aex + a2e2 + y2 = e2x2 – 2aex + a2, , a, �, e, , x2 + a2e2 + y2 = e2x2 + a2, (1 – e2) x2 + y2 = a2 (1 – e2), 151

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Since (1 – e2) > 0 , Dividing both sides by a2, (1 – e2), 2, , Thus for the ellipse ( a e , 0 ) and ( - a e , 0 ) are, two foci and x =, , 2, , x, y, + 2, =1, 2, a, a (1 − e 2 ), 2, , two directrices., ∴ Standard equation of Ellipse is, , 2, , x, y, + 2 = 1 where b2 = a2 (1 – e2) and a > b, 2, a, b, , Note:, , This is the standard equation of ellipse ., , Equation of the ellipse, , We also get for a point P(x,y) on the locus PS' = e, PM' . . . (6) where PM' is the perpendicular on, directrix x = –, , ( x + ae ) + ( y − 0 ), 2, , 2, , ii), , The line segment through the foci of the, ellipse is called the major axis and the line, segment through centre and perpendicular, to major axis is the minor axis. The major, axis and minor axis together are called, principal axis of the ellipse. In the standard, form X axis is the major axis and Y axis is, the minor axis., , iii), , The segment AA' of length 2a is called the, major axis and the segment BB' of length 2b, is called the minor axis. Ellipse is symmetric, about both the axes., , iv), , The origin O bisects every chord through it, therefore origin O is called the centre of the, ellipse., , v), , latus rectum is the chord through focus, which is perpendicular to major axis. It is, bisected at the focus. There are two latera, recta as there are two foci., , PM' = distance of point P from directrix, x+, , =, , From (6) , (7) and (8), ( x + ae) 2 + ( y − 0) 2 = e x +, , a, e, , ( x + ae) 2 + ( y − 0) 2 = ex + a, Squaring both sides, (x + ae)2 + (y – 0) = e2x2 + 2aex + a2, x2 + 2aex + a2e2 + y2 = e2x2 + 2aex + a2, x2 + a2e2 + y2 = e2x2 + a2, , 7.2.2 Some Results :, , (1 – e ) x + y = a (1 – e ), 2, , 2, , 2, , 2, , x2 y 2, +, = 1 (a > b), a 2 b2, , The ellipse Intersects x-axis at A(a, 0), A', (-a, 0) and y-axis at B(0, b), B'(0, -b), these, are the vertices of the ellipse., , …..(7), , a, e = x + a …….(8), 2, e, 1 + 02, , x2 y 2, +, =1, a 2 b2, , i), , a, , from point P., e, , S'P = focal distance, =, , a, a, and x = – are corresponding, e, e, , 2, , 1), , Since (1 – e ) > 0 ,, 2, , d (dd') is the same that of distance ZZ', ie. d (ZZ'), , Dividing both sides by a (1 – e ), 2, , Distance between directrices, , 2, , Z (a/e, 0) and Z' (− a/e , 0), , x2, y2, +, =1, a 2 a 2 (1 − e 2 ), , ⇒ d (dd') = d (zz') = |, , x2 y 2, + 2 = 1 where b2 = a2 (1 – e2) and a > b, 2, a, b, , =2, 152, , a, e, , a  a, − − |, e  e

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2), , End co-ordinates of latera recta, Let LSL′ be the latus rectum of the ellipse., , SP = e PM and S'P = e PM', , x2 y 2, +, =1 a > b, (refer fig. 7.16), a 2 b2, (SL and SL′ are the semi latus rectum), Let : L ≡ (ae, l), , SP + S'P = e PM + e PM', , Sum of focal distances of point P, = e (PM + PM'), = e (MM'), = e (distance between the directrix),  a, = e  2×  = 2a, e, , SP + S'P = 2a = constant, , (ae) 2 (l 2 ), + 2 =1, a2, b, e2 +, , l2, =1, b2, , = length of major axis., , l2, = 1 − e2, 2, b, , Sum of focal distances of point on the ellipse, is the length of major axis which is a constant., , l2 = b2 (1 – e2),  b2 , 2, 2, l =b  2, a , , Using this property one can define and draw, an ellipse. If S1 and S2 are two fixed points and a, point P moves in the plane such that PS1 + PS2 is, equal to constant K , where K > d(S1S2) , then the, locus of P is an ellipse with S1 and S2 as foci. Here, in the standard form K = 2a., , [∴ b2 = a2 (1 – e2)], , b4, l = 2, a, 2, , l=±, , b2, a, , , , b2 , −b 2 , ae, ,, L, ae, ,, and, ≡, , , , L= , a , a , , , These are the co-ordinates of end points of latus, rectum., 3), , 4), , A circle drawn with the major axis AA′ as a, diameter is called an auxiliary circle of the, ellipse., , 6), , Parametric form of an ellipse, , P(x, y) be any point on the ellipse. Let Q, , Length of latus rectum, l (LSL′) = l (SL) + l (SL′) =, , 5), , be a point on the on the auxiliary circle such that, 2, , 2, , b b, 2b, + =, a a, a, , 2, , QPN ⊥ to the major axis., Let m ∠ XOQ = θ ∴ Q = (a cosθ, a sinθ), , SP and S'P are the focal distances of the point, P on the ellipse. (ref. Fig.7.17), , Let P(x , y) ≡ (a cos θ, y)., , ( a cosθ ), a2, cos 2θ +, , 2, , +, , y2, =1, b2, , y2, =1, b2, , y2 = b2 (1 – cos2θ) , y2 = b2 sin2θ, y = ± bsinθ, , Fig. 7.18, , P(x, y) ≡ (a cosθ , b sinθ) ≡ P(θ), , Fig. 7.17, 153

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Thus x = a cosθ and y = b sinθ is the parametric, 2, , is θ = tan-1, , 2, , x, y, + 2 = 1 (a > b) where θ, 2, a, b, is the parameter which is called as an eccentric, angle of the point P., , form of the ellipse, , note that θ is not the angle made by OP with X, axis., , of the ellipse., , If θ is the eccentric angle of P , we know that, y, a� y, x = a cosθ , y = b sinθ then tanθ = b =, , that, x, b� x, a, Standard equation, , 2, , Centre, , 3, , Axes of symmetry, , 4, , Vertices, , 5, , Major axis and minor axis, , 6, , Length of major axis, , 7, , Length of minor axis, , 8, , Relation between a b and c, , 9, , Foci, , 10, , Distance between foci, , 11, , Equation of directrix, , 12, , Distance between the directrix, , 13, , End points of latus rectum, , 14, , Length of latus rectum, , 15, , Parametric form, , 16, , Equation of tangent at vertex, , 17, , Sum of Focal distance of a point, P(x1, y1) is the length of it's, , x2, y2, a2 + b2 = 1 ( b > a) is other standard form, , 7), , To find the eccentric angle of a point P(x,y) on, the ellipse in terms of x and y., , 1, , a� y, b� x, , It is called vertical ellipse. (Ref. figure 7.22), , x2, y2, 2 +, a, b2 = 1, a > b, 0(0,0), , x2, y2, 2 +, a, b2 = 1, b > a, 0(0,0), , Both x axis and y axis, , Both x axis and y axis, , A(a,0) A'(−a,0), B(0,b) B'(0,−b), , A(a,0) A'(−a,0), B(0,b) B'(0,−b), , X axis and Y axis, , X axis and Y axis, , 2a, , 2b, , 2b, , 2a, , b = a (1 – e ), , a = b (1 – e2), , S(ae, 0) S(−ae,0), , S(0,be) S(0, −be), , 2 ae, , 2 be, , a, a, x = e , and x = e, 2a, e, , b, b, y = e and y = - e ., 2a, e., , b2, L = ae, a and, b2, L' ≡ ae, a, , a2, , be, b, a2, L' b , be, , 2b2, a, , 2a2, b, , x = a cosq and y = b sinq, , x = a cosq and y = b sinq, , x = a, x = − a and, y = b, y = − b, , x = a, x = − a and, y = b, y = − b, , 2a, major axis, , 2b, major axis, , 2, , 2, , 154, , 2, , 2, , 2, , L

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Centre O (0, 0), Vertices A(± a, 0) ≡ (± 4, 0) , B (0, ± b) ≡ (0, ± 3), Length of major axis (2a) = 2(4) = 9, Length of minor axis (2b) = 2(3) = 6, By relation between a, b and e., b2 = a2 (1 – e2), 9 = 16 (1 – e2), 9, = 1 – e2, 16, 9, e2 = 1 –, 16, e2 =, , Fig. 7.19, , Foci, , Ex. 1) Find the coordinates of the foci, the vertices,, the length of major axis, the eccentricity and the, length of the latus rectum of the ellipse, , , 7 , , 0  =, S(ae, 0) ≡  4., 4, , , , (, , 7,0, , , 7 , , 0  = − 7 , 0, S′(−ae, 0) =  −4., 4, , , , (, , x2 y 2, +, =1, i), 16 9, , ), , ), , Distance between foc = 2 ae = 2 7,, , ii) 4x2 + 3y2 = 1, , Equation of directrix x = ±, , iii) 3x + 4y = 1, 2, , a, e, , 16, 4, that is x = ±, 7, 7, 4, a,  16  32, distance between directrix = 2 = 2 , =, e, 7,  7, End coordinates of latus rectum, , iv) 4x2 + 9y2 – 16 x + 54 y + 61 = 0, , x=±, , Solution :, x2 y 2, +, =1, i) Given equation of an ellipse is, 16 9, Comparing with standard equation, , 7, 4, , but 0 < e < 1 therefore e =, , SOLVED EXAMPLES, , 2, , 7, 7, that is e = ±, 16, 4, , 9, b2, , L ae, a =  7 , , 4, , , x2 y 2, +, =1, a 2 b2, , a = 4 ; b = 3 (a > b), , -b2, 9, , L' ae, a =  7 , − , 4, , , X-axis (y = 0) is the major axis and y-axis, (x = 0 ) the minor axis., , 2� b2, 9, 9, =2( )=, Length of latus rectum =, 2, a, 4, , a2 = 16 ; b2 = 9, , 155

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Parametric form x = a cosθ , y = b sinθ, , Length of Latus rectum =, , That is x = 4 cosθ , y = 3 sinθ, ii), , Parametric form x =, , 3x2 + 4 y2 = 1, , x2 y 2, +, =1, 1, 1, 3, 4, x2 y 2, Comparing with 2 + 2 = 1, a, b, 1, 1, , b2 =, a2 =, 3, 4, 1, 1, a=, b =, (a > b), 2, 3, , x2, y2, =1, +, 1 1,    ,  4 3, Comparing with, x2 y 2, +, =1, a 2 b2, 1, 1, , b2 =, 4, 3, 1, 1, b =, b>a, a=, 2, 3, a2 =, , \ Major axis is, By the relation between a, b & e, (1 – e2), , therefore y-axis is major axis, Y-axis (ie x = 0) is the major axis, , (1 – e2) =, \ e2 =, , ae=, , centre is at 0(0, 0), , 1, 2, , ∵0<e<1, , x-axis (ie. y = 0) is the minor axis, 1, Length of major axis 2b = 2, 3, 1, =1, length of minor axis 2a = 2, 2, Centre is at O(0 ,0 ), 1, Vertices A(± a, 0) ≡ (± , 0) ,, 2, 1, B (0, ± b) ≡ (0, ±, ), 3, , 1, vertices (±a, 0) = (± 3 , 0) and, 1, (0, ± b) = (0, ± 2 ), 1 1, foci (± ae, 0) = (± 3 2 , 0), 1, , , = ±, ,0,  2 3 , distance between foci =, a, is x = ± e, , Equation of, i.e. x = ±, , ; y=, , iii) 4x2 + 3y2 = 1, , =, , =, , =, , 1, = 3, , Relation between a, b, e, a2 = b2 (1 – e2), 1, 1, = (1 – e2), 4, 3, , a, , a, Distance between directrices = 2 e, , =, b2, End points of Latera recta = (ae; ± a ), =(, ,±, , 3, = 1 – e2, 4, 3, 1, ∴ e2 = 1 –, e2 =, 4, 4, 1, ∴ e =, (∵ 0 < e < 1), 2, , ∴, , ), 156, , 3, 2

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foci S (0, + be) and S′ (0, – be) ≡ (0, ± be) =, , a2 = 9 ; b2 = 4, , 1 1 , 1 , , .  =  0, ±, ,  0, ±, 3 2 , 2 3, , , a = 3 ; b = 2 (a > b), ,  1  1  1, distance between foci = 2 be = 2 ,   =, 3,  3  2 , , X = 0 ie. x – 2 = 0 is the minor axis., , Y = 0 ie. y + 3 = 0 is the major axis and, , equation of directrices is y = ±, , Centre (X = 0,Y= 0) ≡ (x – 2= 0, y + 3 = 0) ≡, (2, – 3), , b, e, , Vertices A(x – 2 = ±3, y + 3 = 0) ≡ (2 ± 3, – 3) i.e., A(5,−3) and A'(−1,−3), ,  1 , , , 2, y= ±  3 y=±, 3, 1,  , 2, , B (x – 2=0, y + 3 = ± 2) ≡ (2 , –3 ± 2), i.e. B(2,−1) and B'(2,−5), A(5 , – 3), A’(– 1 , – 3). B(2 , – 1) B' (2 , – 5 ), , 4, b, Distance between directrices 2 =, 3, e, and coordinates of latus rectum., , Length of major axis (2a) = 2(3) = 6, Length of minor axis (2b) = 2(4) = 8, b2 = a2 (1 – e2), , a2, LL' ≡ (±, , be), b, , ∴ 4 = 9 (1 – e2), , ,  1, ,   , 4  1  1 , =  ±   ,,  ,   1   3   2  , , ,  , ,   3, , e2 =, , 4, = 1 – e2, 9, , Foci, , 2, , 2a, 2, =, 3, b, , ∴ e2 = 1 –, , 5, 3, , , , 5, , y + 3 = 0  ≡ (2 +, S  x − 2 = 3, 3, , , , Parametric form, x = a cosq, y = b sinq, , , , 5, , y + 3 = 0  = (2 –, S'  x − 2 = −3, 3, , , , 1, 1, sin θ, x = 2 cosq , y =, 3, , Distance between foci = 2 5, Equation of directrix x – 2 = ±, , iv) 4x2 + 9y2 – 16 x + 54 y + 61 = 0, By the method of completing square the above, equation becomes 4 ( x – 2 )2 + 9 (y+3)2 = 36, That is, , that is x = 2 ±, , ( x − 2) 2 ( y + 3) 2, +, =1, 9, 4, , Comparing with standard equation, , 9, 5, , distance between directrix = 2, x2 y 2, +, =1, a 2 b2, , 5 , –3), , 5 , –3), , 3,  5, , ,  3 , , a,  9  18, = 2, =, e, 5,  5, , coordinates of end point of latera recta, , 157, , 4, 9, , 5, 5, that is e = ±, 9, 3, , but 0 < e < 1 therefore e =, , , 3 1 , ,, =  ±, ,  4 2 3, Length of latus rectum =, , ∴

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, b2 , L  ae ,  =, a , , , Ex. 3) Find the eccentricity of an ellipse whose, length of the latus rectum is one third of its, minor axis., , , 5 (2) 2  , 4, ,, 3, ×,  =  5 , , , 3, 3  , 3, , , Solution :, , , 4, b2  , ae, ,, −, L' ,  =  5, − , a  , 3, , , 1, (minor axis), 3, , Length of latus rectum =, , 4 9, 2� b2, Length of latus rectum =, = 2  =, 3 2, a, , 1, 2 b2 1, = ( 2 b ) that is b = a, 3, a, 3, , Parametric form X = 3 cosθ , Y = 2 sinθ, , We know that, , That is x – 2 = 3 cos , y + 3 = 2 sinθ, , b2 = a2 (1 – e2), , 1 2, a = a2 (1 – e2), 9, , i.e. x = 2 + 3cosθ, y = −3 + 2sinθ, , 1, = 1 – e2, 9, e2 =, , ∴ e2 = 1 −, , 1, 9, , 8, 2 2, that is e = ±, 9, 3, , but 0 < e < 1, , ∴, , 2 2, 3, , e=, , Activity :, Find the equation of an ellipse whose major axis, is on the X-axis and passes through the points, (4, 3) and (6, 2), , Fig. 7.20, Ex. 2) Find the equation of an ellipse having, vertices (± 13, 0) and foci (± 5, 0), , Solution :, Let equation an ellipse, , Solution : Since vertices and foci are on the, x-axis, the equation of an ellipse will be of the, , a > b , since major axis is the X-axis., , x2 y 2, form 2 + 2 = 1 (a > b), a, b, , Also ellipse passes through points (4, 3) and (6, 2), , Vertices (± 13, 0) = (± a, 0) ⇒ a = 13, , ∴, , Foci (± 5, 0) = (±ae, 0) ⇒ ae = 5, , ( 4), a2, , 2, , ( 3), +, b2, , 2, , = 1 and, , (6), , 2, , a2, , ( 2), +, b2, , 2, , =1, , Solve these equations simultaneous to set a2 and, b2., , 5, ∴ e=, 13, , 7.2.3 Special cases of an ellipse:, , We know b2 = a2 (1 – e2) = a2 – a2e2, = (13)2 – (5)2 = 169 – 25 = 144, Equation of the ellipse is, , x2 y 2, +, =1,, a 2 b2, , Consider the standard ellipse, , x2, y2, +, =1., 169 144, , b2 = a2 (1 – e2) and a > b ., 158, , x2 y 2, +, = 1 where, a 2 b2

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equation of the tangent at P(x1,y1) to the ellipse, xx yy, x2 y 2, + 2 = 1 is a21 + b21 = 1 ., 2, a, b, that is, , The equation of tangent to the ellipse in, terms of slope is, 2, 2, y = m x ± a 2 m 2 + b 2 , P  −a m , b , c ,  c, , y1, x1, y – 1 = 0 . . . (2), 2 x +, a, b2, , Thus the line y = m x + c is tangent to the ellipse, , If the line given by equation (1) is a tangent to the, ellipse at P(x1,y1)., , x2 y 2, +, = 1 if c = ± a 2 m 2 + b 2 and the point, a 2 b2, , Comparing coefficients of like terms in equation, (1) and (2), , of contact is (–, ,  x1   y1 ,  2  2, we get,  a  =  b  = − 1, m, 1, c, , 7.2.6 Tangents from a point to the ellipse, Two tangents can be drawn to the ellipse from any, point outside the ellipse., ,  x1 ,  y1 ,  2  −1,  2, ∴a  =, and  b  = −1, m, c, −1, c, ∴, , Let P(x1,y1) be any point in plane of the ellipse., , –1, 1, x1, y, =, and 1 =, 2, 2, c, c, am, b, , The equation of tangent, with slope m to the, ellipse is, , b2, a2m, and y1 =, c, c, , ∴ x1 = –, , y = m x ± a 2 m2 + b2 ., This pass through (x1,y1), , x2 y 2, P(x1,y1) lies on the ellipse 2 + 2 = 1, a, b, ∴, , ∴, , ∴ y1 = m x1 ± a 2 m 2 + b 2, , x12 y12, + 2 =1, a2, b, (−, , 4, , 2 2, 2, ∴ y1 − m x1 = ± a m + b , we solve it for m., , Squaring on both sides and simplifying we get the, quadratic equation in m., , a2m 2, b2, ) ( )2, c, + c2 = 1, a2, b, 2, , (, , a m, b, ( 2), 2, ∴ c 2 + c2 = 1, a, b, , it has two roots say, m1 and m2 which are the, slopes of two tangents., Thus, in general, two tangents can be drawn to a, ellipse from a given point in its plane., , a 2 m2 b2, ∴, + 2 =1, c2, c, ∴ a2 m2 + b2 =c2, i.e. c2 = a2 m2 + b2, ∴ c = ± a m + b is the condition for tangency., 2, , ), , ( x12 − a 2 )m 2 − 2 x1 y1m + y12 − b 2 = 0, , 4, , 2, , 2, a2m b, ,, )., c, c, , 2, , 160, , Sum of the roots = m1 + m2 =, , - (-2 x1 y1 ), ( x12 - a 2 ), , =, , (2 x1 y1 ), ( x12 - a 2 )

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Product of roots = m1 m2, , =, , (y, , 2, 1, , − b2, , 2, 1, , 2, , ), , (x − a ), , 7.2.7 Locus of point of intersection of, perpendicular tangents, If the tangent drawn from P are mutually, perpendicular then we have m1 m2 = –1, y12 – b2, \ 2 2 =1, x1 – a, Fig. 7.23, SOLVED EXAMPLE, Ex. 1) Find the equation of tangent to the ellipse, , (, , ∴ y −b, , 2, , ) = − (x, , 2, 1, , x2 y 2, +, = 1 at the point (2,, 8 6, , ii), , x2 y 2, +, = 1 at the point whose eccentric, 25 9, , 3 )., , angle is π/4., , Fig. 7.22, 2, 1, , i), , Solution :, , 2, , −a ), , i), , ∴ x12 + y12 = a 2 + b 2, , , , This is the equation of standard circle with centre at, origin and radius a 2 + b 2 which is called the, director circle of the ellipse., , , , 7.2.8 Auxiliary circle, director circle of the, ellipse, x2, y2, For the standard ellipse a2 + b2 = 1(a > b) the, circle drawn with major axis as a diameter is, called the auxillary circle of the ellipse and it's, equation is x2 + y2 = a2 ., The locus of point of intersection of perpendicular, x2, y2, tangents to the ellipse a2 + b2 = 1 is called the, director circle of the ellipse and its equation is, x2 + y2 = a2 + b2 ., 161, , Equation of the ellipse is, , x2 y 2, +, =1, 8 6, , x2, y2, comparing it with a2 + b2 = 1, a2 = 8 and b2 = 6., , , , Tangent is drawn to the ellipse at point, (2, 3 ) on it. Say (x1, y1) ≡ (2, 3 )., , , , We know that,, , , , the equation of tangent to the ellipse, , , , x2, y2, a2 + b2 = 1 at point (x1 y1) on it is, , , , xx1 yy1, a2 + b2 = 1, , ∴, , x(2) y( 3 ), 8 + 6 =1

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2, and c = 4, 3, We know that,, , x, 3y, +, =1, 4, 6, , m=–, , i.e. 6x + 4 3 y = 24, , if the line y = m x + c is tangent to the ellipse, , i.e. 3x + 2 3 y = 12, , x2 y 2, + 2 = 1 then c2 = a2 m2 + b2., 2, a, b, Here c2 = (4)2 = 16 and, , Thus required equation of tangent is, 3x + 2 3 y = 12., ii), , Equation of ellipse is, compairing it with, a2 = 25 and b2 = 9., , 2, , 4, 2, a2 m2 + b2 = (18) (– )2 + (8) = (18) ( ) + 8, 9, 3, = (2) (4) + 8 = 16, , 2, , x, y, +, =1, 25 9, , hence the given line is tangent to the given, ellipse., , x2 y 2, +, =1, a 2 b2, , Ex. 3) Find the equations of tangents to the ellipse, 4x2 + 9y2 = 36 passing through the point, (2, –2)., , p, eccentric angle θ = ., 4, By parametric form equation of tangent is, , Solution : Equation of the ellipse is 4x2 + 9y2 = 36, , x ⋅ cos θ, y ⋅ sin θ, +, =1, a, b, π, π, x ⋅ cos, y ⋅ sin, 4 +, 4 =1, i.e., 5, 3, 1, 1, x⋅, y⋅, 2 +, 2 =1, 5, 3, x, y, +, =1, 5 2 3 2, , i.e., , compairing it with, , x2 y 2, +, =1, a 2 b2, , a2 = 9 and b2 = 4, Equation of tangent in terms of slope m, to, the ellipse is, y = m x ± a 2 m2 + b2, Point (2, –2) lies on the tangent, ∴ (–2) = m (2) ±, , 3x + 5y = 15 2, , ∴ –2 m –2 = ±, , 9m 2 + 4, , 9m 2 + 4, squaring both sides, 4m2 + 8m + 4 = 9m2 + 4, – 5m2 + 8m = 0, m (–5m + 8) = 0 ⇒ m = 0 or m = 8/5, Equation of tangent line having slope m and, passing through pt (2, –2) is y + 2 = m (x – 2), 8, i.e. y + 2 = 0(x – 2) or y + 2 = (x – 2), 5, y+2=0, 5y + 10 = 8x – 16, , 8x – 5y – 26 = 0, Thus equation of tangents are y + 2 = 0 and, 8x – 5y – 26 = 0, , Ex. 2) Show that the line 2x + 3y = 12 is tangent, to the ellipse 4x2 + 9y2 = 72., Solution : Equation of the ellipse is 4x2 + 9y2 = 72, i.e., , x2, y2, +, =1, 9, 4, , x2, y2, +, =1, 18, 8, , x2 y 2, compairing it with 2 + 2 = 1, a, b, a2 = 18 and b2 = 8, Equation of line is 2x + 3y = 12, 2, i.e. y = – x + 4, 3, compairing it with y = m x + c, 162

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5., , Show that the product of the lengths of, ht perpendicular segments drawn from, the foci to any tangent line to the ellipse, x2 /25 + y2/16 = 1 is equal to 16., , 6., , A tangent having slop –1/2 to the ellipse, 3x2 + 4y2 = 12 intersects the X and Y axes in, the points A and B respectively. If O is the, origin, find the area of the triangle., , 7., , Show that the line x – y = 5 is a tangent to, the ellipse 9x2 + 16y2 = 144. Find the point, of contact., , 8., , Show that the line 8y + x = 17 touches, the ellipse x2 + 4y2 = 17. Find the point of, contact., , 9., , Determine whether the line x + 3y√2 = 9 is, a tangent to the ellipse x2/9 + y2/4 = 1. If so,, find the co-ordinates of the pt of contact., , the distance between foci is 6 and the, distance between directrix is 50/3., , 10., , Find k, if the line 3x + 4y + k = 0 touches 9x2, + 16y2 = 144., , vi) The latus rectum has length 6 and foci, are (+2, 0)., , 11., , Find the equation of the tangent to the, ellipse (i) x2/5 + y2/4 = 1 passing through, the point (2, -2)., , EXERCISE 7.2, 1., , Find the (i) lengths of the principal axes., (ii) co-ordinates of the focii (iii) equations, of directrics (iv) length of the latus rectum, (v) distance between focii (vi) distance, between directrices of the ellipse:, (a) x2/25 + y2/9 = 1, (c) 2x2 + 6y2 = 6, , 2., , (b) 3x2 + 4y2 = 12, (d) 3x2 + 4y2 = 1., , Find the equation of the ellipse in standard, form if, i), , eccentricity = 3/8 and distance between, its focii = 6., , ii), , the length of major axis 10 and the, distance between focii is 8., , iii) distance between directrix is 18 and, eccentricity is 1/3., iv) minor axis is 16 and eccentricity is 1/3., v), , vii) passing through the points (−3, 1) and, (2, −2), , ii), , viii) the dist. between its directrix is 10 and, which passes through (−√5, 2), , iii) 2x2 + y2 = 6 from the point (2, 1)., iv) x2 + 4y2 = 9 which are parallel to the line, 2x + 3y – 5 = 0 ., , ix) eccentricity is 2/3 and passes through, (2, −5/3)., 3., , 4., , 4x2 + 7y2 = 28 from the pt (3, -2)., , v), , Find the eccentricity of an ellipse, if the, length of its latus rectum is one third of its, minor axis., , x2/25 + y2/4 = 1 which are parallel to, the line x + y + 1= 0., , vi) 5x2 + 9y2 = 45 which are ⊥ to the line, 3x + 2y + y = 0., , Find the eccentricity of an ellipse if the, distance between its directrix is three times, the distance between its focii., , vii) x2 + 4y2 = 20 , ⊥ to the line 4x+3y = 7., 12., , 163, , Find the equation of the locus of a point the, tangents form which to the ellipse 3x2 + 5y2, = 15 are at right angles.

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13., , Tangents are drawn through a point P to the, ellipse 4x2 + 5y2 = 20 having inclinations θ1, and θ2 such that tan θ1 + tan θ2 = 2. Find the, equation of the locus of P., , 14., , Show that the locus of the point of, intersection of tangents at two points on an, ellipse, whose eccentric angles differ by a, constant, is an ellipse., , 15., , P and Q are two points on the ellipse, x2 y 2, +, = 1 with eccentric angles θ1 and θ2., a 2 b2, Find the equation of the locus of the point, of intersection of the tangents at P and Q if, θ1 + θ2 = π/2., , 16., , The eccentric angles of two points P and Q, the ellipse 4x2 + y2 = 4 differ by 2π/3. Show, that the locus of the point of intersection, of the tangents at P and Q is the ellipse, 4x2 + y2 = 16., , 17., , Find the equations of the tangents to the, ellipse x2/16 + y2/9 = 1, making equal, intercepts on co-ordinate axes., , 18., , A tangent having slope – ½ to the ellipse, 3x2 + 4y2 =12 intersects the X and Y axes in, the points A and B respectively. If O is the, origin, find the area of the triangle., , a fixed point bears a constant ratio e (e > 1) to, its distance from a fixed line. The fixed point is, called the focus S and the fixed line is called the, directrix d., , Fig. 7.24, If S is the focus and d is the directrix not, containing the focus and P is the moving point,, PS, then, = e, where PM is the perpendicular, PM, on the directrix. e > 1 called eccentricity of the, hyperbola. (Fig. 7.24), 7.3.1 Standard equation of the hyperbola, Let S be the focus, d be the directrix and e be the, eccentricity of a hyperbola., Draw SZ perpendicular to directrix. Let A and A', divide the segment SZ internally and externally in, the ratio e : 1. By definition of hyperbola A and A', lie on hyperbola., , 7.3 Hyperbola, Let's Study, •, , Standard equation of the hyperbola., , •, , Equation of tangent to the hyperbola., , •, , condition for tangency., , •, , auxillary circle and director circle of the, hyperbola., , Let AA' = 2 a , midpoint O of segment AA' be the, origin. Then O ≡ (0,0) , A≡ (a,0) and A' ≡ (− a,0), SA e, =, AZ 1, , (, , SA ' −e, ), =, A'Z, 1, , therefore SA = e AZ ., Let Z ≡ (k,0) and S ≡ (h,0), By section formula for internal and external division., , The hyperbola is the intersection of double, napped cone with plane parallel to the axis., , a=, , The hyperbola is the locus of a point in, a plane which moves so that its distance from, 164, , e k + 1h, e +1, , also − a =, , e k −1h, e −1

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ae+a=ek+h, , This is the standard equation of hyperbola ., a, Let S' be (− ae,0) and d' be the line x = − ., e, , . . . (2), , − a e + a = e k – h . . . (3), Solving these equations , we get, k = a/e and, , For any point P on the hyperbola, PM' is perpendicular on d' then it can be verified that, PS '= e PM'., , h=ae, , Focus S ≡ (ae , 0) and Z ≡ (a/e , 0), a, Equation of the directrix is x =, e, a, That is x – = 0, e, , Thus for the hyperbola ( a e , 0 ) and ( − a e , 0 ), a, a, are two focii and x =, and x = −, are, e, e, corresponding two directrices., , Let P(x,y) be a point on the hyperbola., SP = focal distance, =, , ( x − ae ) + ( y − 0 ), 2, , 2, , …..(4), , PM = distance of point P from the directrix, , a, e = x − a …….(5), e, 12 + 02, x−, , =, , Fig. 7.25, , From (1) , (4) and (5), , 7.3.2 Some useful terms of the hyperbola, , a, ( x − ae) + ( y − 0) = e x −, e, 2, , 2, , Equation of the hyperbola, , = ( x − ae) 2 + ( y − 0) 2 = ex − a, , i), , The hyperbola intersects x-axis at A(a, 0),, A′ (−a, 0)., , (x – ae)2 + (y – 0)2 = e2x2 – 2 aex + a2, , ii), , It does not intersects the y-axis., , x2 – 2 aex + a2e2 + y2 = e2x2 – 2 aex + a2, , iii) The segment AA′ of length 2a is called the, transverse axis and the segment BB′ of length, 2b is called the conjugate axis., , Squaring both sides, , x + ae +y =ex + a, 2, , 2 2, , 2, , 2 2, , 2, , (1 – e2) x2 + y2 = a2 (1 – e2), , iii) The line segment through the foci of the, hyperbola is called the transverse axis and the, line segment through centre and perpendicular, to transverse axis is the conjugate axis. The, transverse axis and conjugate axis together, are called principal axes of the hyperbola., In the standard form X axis is the transverse, axis and Y axis is the conjugate axis., , Since e > 1, ( e2 – 1) x2 – y2 = a2 ( e2 – 1), Dividing both sides by a2 ( e2 – 1), , x2, y2, −, =1, ∴ 2, a, a 2 (e 2 − 1), where b2 = a2 ( e2 – 1), , 165

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iv) latus rectum is the chord passing through the, focus which is perpendicular to transverse, axis. It is bisected at the focus. There are two, latera recta as there are two focii., , 3), , b 2 b 2 2b 2, + =, = l (SL) + l (SL′) =, a a, a, 4), , 7.3.3 Some Results :, 1), , same that of distance ZZ′ ie. d (ZZ′), , Z ≡ (a/e, 0) and Z′ (− a/e , 0), , =2, 2), , a, e, , a a, + |, e e, , Let LSL′ be the latus rectum of the, hyperbola., , Difference between the focal distances of point, on the hyperbola is the length of transverse, axis which is a constant., , (SL and SL′ are the semi latus rectum), Let L ≡ (ae, l), , 5), , A circle drawn with the transverse axis AA′ as, a diameter is called an auxiliary circle of the, hyperbola and its equation is x2 + y2 = a2 ., , ∙, , Parametric Equation of the Hyperbola, , (ae), l, a2 + b2 = 1, 2, , 2, , l2, e2 + b2 = 1, l2, 2, b2 = 1–e, , :, Taking the transverse axis AA' as diameter., Draw a circle with centre at origin and radius, , l = b (1 – e ), 2, , 2, , 2, , 'a' so that its equation is x2+y2 = a2. It is called the, ,  b2 , l =b  2, a , 2, , l2 =, , SP and S′P are the focal distances of the point, P on the hyperbola., , SP = e PM and S′P = e PM′, Difference between the focal distances of, point P, SP – S′P = e PM – e PM′, = e (PM – PM′), = e (MM′), = e (distance between the directrices), = e (2 a/e), SP – S′P = 2a = constant, = length of major axis ie. transverse axis., , Distance between directrices i.e. d (dd′) is the, , ⇒ d (dd1) = d (zz′) = |, , Length of latus rectum = l (LL′), , 2, , auxiliary circle of the hyperbola, , 4, , b, a2, , l=±, , b2, a, 2, ,  b , L =  ae,  and L' ≡, a, , , 2, , −b , ,  ae,, , a , , , These are the co-ordinates of end points of latus, rectum., Fig. 7.26, , 166, , ...(I)

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∴ for any value of θ, the point (a secθ, b tanθ), always lies on the hyperbola., , Let P(x1 y) be any point on the hyperbola. Draw, PM perpendicular to OX. Draw the tangent MQ, touching auxiliary circle at Q. Point Q is called the, corresponding point of P on the auxiliary circle., , Let us denote this point by P(θ) = P (a secθ,, b tanθ) where θ is called parameter also called the, eccentric angle of point P., , Let m∠XOQ = θ. Then by trigonometry, co-ordinates of Q are (a cos θ, a sinθ)., , The equations x = a secθ, y = b tan θ are called, parametric equations of the hyperbola., , Further,, x = OM =, , OM, . OQ = secθ. a = a sec θ, OQ, , 7), , Other standard form of hyperbola., y 2 x2, −, = 1 is called the conjugate hyperbola, b2 a 2, , The point P(x,y) = P(a sec θ, y) ∴ P lies on the, hyperbola-(I) therefore., , of the hyperbola, , a 2 sec 2 θ y 2, − 2 =1, a2, b, y2, = sec2θ − 1 = tan2θ, 2, b, y, ∴ = ± tanθ, b, Since P lies in the first quadrant and angle, θ < 90°, y and tanθ both are positive., ∴, , ∴ y = b tanθ, ∴ P ≡ P (a secθ, b tanθ), Substituting these co-ordinates in the LHS of, equation of hyperbola (I), we get, , , 5, , y + 3 = 0  = sec2θ − tan2θ = 1,  x − 2 = 3, 3, , , , 1, , y 2 x2, −, =1, b2 a 2, , Standard equation, , 2, , Centre, , 3, , Axes of symmetry, , 4, , Fig. 2.27, , Vertices, , O(0,0), , O(0,0), , Both x axis and y axis, , Both x axis and y axis, , A(a,0) A′(−a,0), B(0,b) B′(0,−b), , A(a,0) A′(−a,0), B(0,b) B′(0,−b), , X axis and Y axis, , Y axis and X axis, , 5, , Major axis ie. transverse axis and, minor axis ie. conjugate axis, , 6, , Length of transverse axis, , 2a, , 2b, , 7, , Length of conjugate axis, , 2b, , 2a, , 167

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8, , Foci, , 9, , Distance between foci, , 10 Equation of directrix, , S(ae, 0) S(− ae,0), , S(0,be) S(0, − be), , 2 ae, , 2 be, , x=, , a, a, . and x = −, e, e, , y=+, , b, b, and y = − ., e, e, 2b, ., e, , 2a ., e, , 11 Distance between the directrics, , , b2 , L =  ae,  and, a , , , 12 end points of latus rectum, , a2, , be) and, b, –a2, , be), L' ≡ (, b, , L≡(, , , −b 2 , ae, ,, L' ≡ , , a , , 2b 2, a, , 2a 2, b, , x = a secθ and y = b tanθ, , x = a tanθ and y = b secθ, , 13 Length of latus rectum, 14 Parametric form, , x=a,x=−a, , 15 Equation of tangent at vertex, , y=b, y=−b, , 2a, , 2b, , 16 Sum of Focal distance of a point (length of major axis ie. transP(x1,y1), verse axis), , Eccentricity b2 = a2 (e2 – 1), , SOLVED EXAMPLE, , ∴e=, , Ex. 1) Find the length of transverse axis, length of, conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the, length of latus rectum of the hyperbola, x2 y 2, − =1, (i), 4 12, , (ii), , (i) The equation of the hyperbola is, , 2, , ∴ foci (± ae, 0) are (± 4,0), a, 2, = =1, e, 2, , a, are, ∴ the equations of directrices x = ±, e, x = ±1., 2b 2 2(12), Length of latus rectum = = = 12, a, 2, , 2, , x, y, − =1 ., 4 12, , Comparing this with equation, , y 2 x2, − =1, (ii) The equation of the hyperbola is, 9 16, Comparing this with the equation, , We have a2 = 4, b2 = 12, ∴ a = 2, b = 2 3, , y 2 x2, −, =1, b2 a 2, , Length of transverse axis = 2a = 2(2) = 4, Length of conjugate axis = 2b = 2(2 3 ), = 4 3, , a 2+ b2, 4 + 12, 16 4, =, =, = =2, a, 2, 2, 2, (Q e > 0), , ae = 2(2) = 4, , y 2 x2, − =1, 9 16, , Solution :, , (length of minor axis ie., conjugate axis), , We have a2 = 16, b2 = 9, 168

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Ex. 3) Find the equation of the hyperbola referred, to its principal axes whose distance between, 5, 18, directrices is, and eccentricity is ., 3, 5, Solution :, The equation of the hyperbola referred to its principal axes be, , ∴ a = 4, b = 3, Length of transverse axis = 2b = 2(3) = 6, Length of conjugate axis = 2a = 2(4) = 8, Eccentricity a2 = b2 (e2 – 1), a 2 + b2, 16 + 9, 25 5, =, =, =, b, 3, 3, 3, , ∴e=, , (Q e > 0), , .................(1), , 5, be = 3   = 5, 3, , Since eccentricity = e =, , ∴ focii (0, ± be) are (0,+5), , directrices =, , b, 3 9, = =, e 5/3 5, ∴ the equations of directrices y = ±, 9, y=± ., 5, , 2a 18, =, e, 5, , 5, and distance between, 3, , a 9, =, e 5, 9, 9 5, ∴a= e = × ,a=3, 5, 5 3, ∴, , b, are, e, , ∴ a2 = 9, , 2a 2 2(16) 32, Length of latus rectum = = =, b, 3, 3, ,  25 , Now b2 = a2 (e2 – 1) = 9  −1 =, ,  9, 16, =9×, = 16, 9, Then from (1), the equation of the required, , Ex. 2) Find the equation of the hyperbola with the, centre at the origin, transverse axis 12 and, one of the foci at (3 5 ,0), Solution :, , hyperbola is, , Let the equation of the required hyperbola be, , x2 y 2, − =1, 9 16, , 7.3.4 Tangent to a hyperbola:, , ...................(1), , A straight line which intersects the curve hyperbola, in two coincident points is called a tangent of the, hyperbola, , Length of transverse axis = 2a = 12, ∴ a = 6 ∴ a2 = 36, Since focus (ae, 0) is (3 5 , 0), ∴ ae = 3 5, ∴ a2e2 = 45, ∴ a2 + b2 = 45, ∴ 36 + b2 = 45, ∴ b2 = 9, Then from (1), the equation of the required, hyperbola is, x2 y 2, − =1, 36 9, , Fig. 7.28, 169

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∴, , -1, x1, =, 2, c, am, , and, , If the tangent passes through (x1,y1)., , 1, y1, 2 = –, c, b, , ∴ y1 = m x1 ±, , a2m, -b 2, and y1 =, ∴ x1 = –, c, c, , ∴ y1 – m x1 = ±, , x12 y12, −, =1, a 2 b2, , (x12 – a2)m2 – 2x1 y1 m + (y12 + b2) = 0, , 2, , it has two roots say, m1 and m2 which are the, slopes of two tangents., , 2, ,  a 2m   b2 ,  − , −, ∴  c  −  c  =1, a2, b2, , Thus, in general, two tangents can be drawn to a, hyperbola from a given point in its plane., ,  a 4 m2   b4 ,  2   2, ∴  c  −  c  =1, a2, b2, , ∴, , a 2 m2 - b2 ., , Squaring on both sides and simplifying we get the, quadratic equation in m which is found to be,, , P(x1,y1) lies on the hyperbola, ∴, , a 2 m2 - b2 ., , Sum of the roots = m1 + m2 =, , a 2 m2 b2, − 2 =1, c2, c, , ∴ c2 = a2m2 – b2, , c = ±, , 2, , 2, , a m -b, , =, , 2, , Product of roots = m1 m2, , is called the condition of, , tangency., , 2, 1, , + b2, , 2, 1, , − a2, , ), ), , ∴ (y12 + b2) = – (x12 − a2), , ∴ x12 + y12 = a2 – b2, , Which is the equation of standard circle with Centre at origin and radius a 2 -� b 2 (a > b). This is, called the director circle of the hyperbola., , a 2 m2 - b2 ,, , 7.3.6 Tangents from a point to the hyperbola, , 7.3.8 Auxiliary Circle, Director Circle, , Two tangents can be drawn to a hyperbola from, any point outside the hyperbola in its plane., , The director circle of the given hyperbola is, the locus of a point, the tangents from which to, the hyperbola are at right angles., , Let P(x1,y1) be any point in plane of the hyperbola., The equation of tangent to the hyperbola is, y=mx±, , (y, =, (x, , If the tangent drawn from P are mutually perpendicular then we have m1 m2 = –1, , a 2 m 2 - b 2 and the point of, , 2, −b 2 ,  am, , −, contact is  −, ., c ,  c, The equation of tangent in terms slope is, , y=mx±, , (2 x1 y1 ), ( x12 - a 2 ), , 7.3.7 Locus of point of intersection of perpendicular tangents :, , Thus the line y = m x + c is tangent to the, hyperbola, if c = ±, , - (-2 x1 y1 ), ( x12 - a 2 ), , a 2 m2 - b2 ., 171

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Consider the point P moving along the line so that, distance OP goes on increasing, then the distance, between P and hyperbola goes on decreasing but, does not become zero. Here the distance between, the point P and hyperbola is tending to zero,, such a straight line is called an asymptote for the, hyperbola, , Fig. 7.29, , Fig. 7.31, , SOLVED EXAMPLES, Ex.1 : Find the equation of the tangent to the, hyperbola 2x2 – 3y2 = 5 at a point in the third, quadrant whose abscissa is –2., , Fig. 7.30, x2, y2, –, = 1, the, a2 b2, circle drawn with transverse axis as a diameter is, called the auxillary circle of the hyperbola and its, equation is x2 + y2 = a2 ., For the standard hyperbola, , Solution : Let P(–2, y1) be the point on the, hyperbola, ∴ 2(–2)2 – 3y2 = 5, ∴ 8 – 5 = 3y2, , The locus of point of intersection of perpenx2, y2, dicular tangents to the hyperbola 2 – 2 = 1 is, a, b, called the director circle of the hyperbola and its, equation is x2 + y2 = a2 – b2 . (a > b) ., , ∴ 3 = 3y2, ∴ y2 = 1, ∴ y = ±1, But P lies in the third quadrant., , auxillary circle x2 + y2 = 25;, director circle x2 + y2 = 16 of the, hyperbola 9x2 − 16y2 = 144. (Refer fig 7.30), , ∴ P ≡ (–2, –1), The equation of the hyperbola is, x2, y2, −, =1 . Comparing this with the, 5/ 2 5/3, , 7.3.9 Asymptote:, , x, y, Consider the lines, = ± , they pass through, a, b, origin O., 172

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equation, a2 =, , Let P(x1, y1) be the point of contact., , , we have, , 4, −25  , −a m,  3  = 25, =, ∴ x1 =, c, 4,  16 , − ,  3, , 5, 5, , b2 =, 2, 3, , 2, , The equation of tangent at, P(x1, y1) ≡ P(–2. –1) is, , –16, –b2, and y1 = c = – 16/3 = 3, , , xx1 yy1, 5 (2) 2  , 4, −, =, 1, ,, 3, ×, ∴,  =  5 , , , 2, 2, , a, b, 3, 3  , 3, , , ∴ the point of contact is P, , Ex. 3 : If the line 2x + y + k = 0 is tangent to the, x2, y2, hyperbola 6 − 8 = 1 then find the value of k., , ∴ 2xx1 – 3yy1 = 5, ∴ 2x(–2) – 3y(–1) = 5, ∴ –4x + 3y = 5, , Solution : The equation of the hyperbola is, x2, y2, 6 − 8 = 1., , ∴ 4x – 3y + 5 = 0, Ex. 2 : Show that the line 4x – 3y = 16 touches, the hyperbola 16x2 – 25y2 = 400. Find the co-ordinates of the point of contact., , The equation of the line is 2x + y + k = 0, ∴ y = –2x – k, Putting this value of y in the equation of hyperbola, we get, , Solution : The equation of the hyperbola is, x2 y 2, − =1 ., 25 16, Comparing it with the equation, , x2 (–2x – k)2, =1, 6 –, 8, , ., , ∴ 4x2 – 3(4x2 + 4kx + k2) = 24, , we have a = 25, b = 16, 2, , 2, , ∴ 4x2 – 12(4x2 – 12kx – 3k2) = 24, , The equation of the line is 3y = 4x – 16, , ∴ 8x2 + 12kx + (3k2 + 24) = 0 .............(1), , 4, 16, ∴ y= x−, 3, 3, ∴ Comparing it with the equation, , Since given line touches the hyperbola, ∴ the quadratic equation (1) in x has equal roots., , y = mx + c, , ∴ its discriminant = 0 i.e. b2 − 4ac = 0, , 4, -16, We get m = , c =, 3, 3, Now, , ∴ (12k)2 – 4(8) (3k2 + 24) = 0, ∴ 144k2 – 32(3k2 + 24) = 0, ,  16 , a2m2 – b2 = 25   – 16, 9, , ∴ 9k2 – 2(3k2 + 24) = 0, ∴ 9k2 – 6k2 – 48 = 0, , 2, , =, , 25, ,3, 4, , 256  −16 , 2, =,  =c, 9  3 , , ∴ 3k2 = 48, , Thus the condition of tangency is satisfied., , ∴ k2 = 16, , ∴ the given line touches the hyperbola., , ∴ k2 = ±4, 173

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Another Method :, , EXERCISE 7.3, , Here c = – k, m = – 2, a2 = b, b2 = 8, , 1), , ∴ c2 = a2m2 – b2, ∴ k2 = [(–2)2 , 6] = 8, ∴ k2 = 16, ∴ k2 = ± 4, , x2 y 2, i) − =1, 25 16, x2 y 2, ii) − = 1–1, 25 16, , Ex.4 : The line x – y + 3 = 0 touches the hyperbola, whose foci are (± 41, 0) . Find the equation of, the hyperbola., , iii) 16x2 – 9y2 = 144, iv) 21x2 – 4y2 = 84, v) 3x2 – y2 = 4, vi) x2 – y2 = 16, , Solution : Let the equation of the hyperbola be, .................(1), , y 2 x2, − =1, 25 9, y 2 x2, −, =1, viii) , 25 144, , Its foci (±ae, 0) are (± 41, 0), , vii) , , ∴ ae =, , 41, ∴ a2 e2 = 41, ∴ a2 + b2 = 41..............(2) (a2e2 = a2+b2), , x2 y 2, − = +1, 100 25, (x) x = 2 sec θ, y = 2 3 tan θ, , ix) , , The given line is y = x + 3, Comparing this with y = mx + c, we get, m = 1, c = 3, , 2), , Find the equation of the hyperbola with centre at the origin, length of conjugate axis 10, and one of the foci (–7,0)., , 3), , Find the eccentricity of the hyperbola, which, is conjugate to the hyperbola x2 – 3y2 = 3., , 4), , If e and e' are the eccentricities of a hyperbola and its conjugate hyperbola respectively,, 1, 1, prove that 2 +, =1, e (e ') 2, , Since the given line touches the hyperbola, ∴ It satisfies the condition of tangency., ∴am –b =c, 2, , 2, , 2, , Find the length of transverse axis, length of, conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the, length of latus rectum of the hyperbola., , 2, , ∴ a2(1)2 – b2 = (3)2, ∴ a2 – b2 = 9...................(3), By adding (2) and (3), we get, , 5), , Find the equation of the hyperbola referred to, its principal axes., i), whose distance between foci is 10 and, 5, eccentricity ., 2, ii) whose distance between foci is 10 and, length of conjugate axis 6., iii) whose distance between directrices is, 8, 3, and eccentricity is ., 3, 2, , 2a2 = 50 ∴ a2 = 25, from (2), we get, 25 + b2 = 41 ∴ b2 = 16, From (1), the equation of the required hyperbola is, x2 y 2, − =1, 25 16, , 174

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2), , The length of latus rectum of the parabola, x2 – 4x – 8y + 12 = 0 is………, A) 4, B) 6, C) 8, D) 10, , 3), , If the focus of the parabola is (0,–3) its, directrix is y = 3 then its equation is, A) x2 = –12y, B) x2 = 12y, C) y2 = 12x, D) y2 = −12x, , 4), , The coordinates of a point on the parabola, y2 = 8x whose focal distance is 4 are …….., , Let's Remember, •, , A conic section or a conic can be defined as, the locus of the point P in a plane such that, the ratio of the distance of P, from a fixed, point to its distance from a fixed line is constant., The constant ratio is called the eccentricity of, the conic section, denoted by 'e'., , •, , If e = 1 the conic section is called parabola if, 0 < e < 1 the conic section is called ellipse. if, e > 1 the conic section is called hyperbola., , •, , eccentricity of rectangular hyperbola is, , •, , standard equations of curve., , A) (1/2, ±2), C) (2, ±4) , , 2., , 5), , x2 y 2, x2 y2, ellipse a2 + b2 = 1, (a>b) a2 + b2 =1 (b>a), , 6), , x2 y2, hyperbola a2 – b2 = 1 a > b ,, , •, , focal distance of a point P on the parabola, y2 = 4ax is a abscissa of point P., , •, , sum of focal distances of point on the ellipse, is the length of major axis., , •, , Difference between the focal distances, of point on the hyperbola is the length of, transverse axis., , 7), , MISCELLANEOUS EXERCISE - 7, (I), , Select the correct option from the given, alternatives., , 1), , The line y = mx + 1 is tangent to the parabola, y2 = 4x if m is …………., B) 2, , C) 3, , B) 14, , 176, , C) 17, , D) 19, , The equation of the parabola having (2, 4), and (2 , –4) as end points of its latus rectum, is…………….…………….., A) y2 = 4x, C) y2 = –16x, , D) 4, , B) 20 sq.units, D) 14 sq.units, , π , If P   is any point on he ellipse, 4, 2, 9x + 25y2 = 225. S and S1 are its foci then, SP.S1P =, A) 13, , 9), , B) y2 = 32x, D) x2 = 32y, , The area of the triangle formed by the, line joining the vertex of the parabola, x2 = 12y to the end points of its latus rectum, is…………, A) 22 sq.units, C) 18 sq.units, , 8), , B) (12, ±6), D) none of these, , Equation of the parabola with vertex at the, origin and diretrix x + 8 = 0 is……………, A) y2 = 8x, C) y2 = 16x, , y2 x2, b2 – a2 = 1 b > a, , A) 1, , The end points of latus rectum of the, parabola y2= 24x are…………, A) (6, ± 12), C) (6, ± 6), , parabola y2 = 4ax , x2 = 4by, , B) (1, ±2√2), D) none of these, , B) y2 = 8x, D) x2 = 8y

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10), , If the parabola y2 = 4ax passes through, ( 3, 2) then the length of its latus rectum, is…………….……………., A), , 11), , 12), , 2, 3, , B), , 4, 3, , C), , 1, 3, , D) 4, , The eccentricity of rectangular hyperbola is, A) ½ , B) 1 / (2 ½ ), C) 2 ½ , D) 1 / (3 ½ ), The equation of the ellipse having foci, 1, (+ 4, 0) and eccentricity is,, 3, 2, 2, A) 9x + 16y = 144, , 18), , Centre of the ellipse 9x2 + 5y2 − 36x − 50y, − 164 = 0 is at, A) (2, 5) B) (1, −2) C) (−2, 1) D) (0, 0), , 19), , If the line 2x − y = 4 touches the hyperbola, 4x2 − 3y2 = 24, the point of contact is, A) (1, 2) B) (2, 3) C) (3, 2) D)(−2, −3), , 20), , The foci of hyperbola 4x2 − 9y2 − 36 = 0 are, , D) 144x2 + 128y2 = 18432, equation, , eccentricity, , of, 3, 2, , the, , ellipse, , 14), , 15), , 16), , 17), , and passing through, , B) x2 + 4y2 = 100, C) x2 + 4y2 = 4, , If the line 4x − 3y + k = 0 touches the ellipse, 5x2 + 9y2 = 45 then the value of k is, A) + 21 B) ± 3 21 C) + 3 D) + 3 (21), The equation of the ellipse is 16 x2 + 25 y2 =, 400. The equations of the tangents making, an angle of 1800 with the major axis are, A) x = 4 B) y = ± 4 C) x = −4 D) x = ±5, The equation of the tangent to the ellipse, 4x2 + 9y2 = 36 which is perpendicular to the, 3x + 4y = 17 is,, A) y = 4x + 6, B) 3y + 4x = 6, C) 3y = 4x + 6 5, C) 3y = x + 25, , 2), , Find the Cartesian co-ordinates of the points, on the parabola y2 = 12x whose parameters, are (i) 2, (ii) −3., , 3), , Find the co-ordinates of a point of the, parabola y2 = 8x having focal distance 10., , 4), , Find the equation of the tangent to the, parabola y2 = 9x at the point (4, −6) on it., , 5), , Find the equation of the tangent to the, parabola y2 = 8x at t = 1 on it., , 6), , Find the equations of the tangents to the, parabola y2 = 9x through the point (4,10)., , 7), , Show that the two tangents drawn to the, parabola y2 = 24x from the point (−6,9) are, at the right angle., , 8), , Find the equation of the tangent to the, parabola y2 = 8x which is parallel to the line, 2x + 2y + 5 = 0. Find its point of contact., , 9), , A line touches the circle x2 + y2 = 2 and the, parabola y2 = 8x. Show that its equation is, y = ± (x+2)., , 10) Two tangents to the parabola y2 = 8x, meet the tangent at the vertex in P and Q., If PQ = 4, prove that the locus of the point, of intersection of the two tangents is, y2 = 8(x + 2)., , Eccentricity of the hyperbola 16x2 − 3y2 −, 32x − 12y − 44 = 0 is, A), , 17, 3, , B), , 19, 3, , C), , 19, 3, , D), , S) (0 , ± 12 ), , For each of the following parabolas, find, focus, equation of the directrix, length of the, latus rectum, and ends of the latus rectum., (i) 2y2 = 17x (ii) 5x2 = 24y., , having, , (− 8 , 3 ) is, A) 4x2 + y2 = 4, C) 4x2 + y2 = 100, , C) ( ± 12 , 0), , 1), , C) 128x2 + 144y2 = 18432, The, , B) ( ± 11 , 0), , (II) Answer the following., , B) 144x2 + 9y2 = 1296, , 13), , A) ( ± 13 , 0), , 17, 3, 177

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19) Show that the line 8y + x = 17 touches the, ellipse x2 + 4y2 = 17. Find the point of contact., , 11) The slopes of the tangents drawn, from P to the parabola y2 = 4ax are, m1 and m2, show that (i) m1 − m2 = k, (ii) (m1/m2) = k, where k is a constant., , 20) Tangents are drawn through a point P to the, ellipse 4x2 + 5y2 = 20 having inclinations θ1, and θ2 such that tan θ1 + tan θ2 = 2. Find the, equation of the locus of P., , 12) The tangent at point P on the parabola, y2 = 4ax meets the y − axis in Q. If S is the, focus, show that SP subtends a right angle, at Q., , 21) Show that the product of the lengths of its, perpendicular segments drawn from the foci, to any tangent line to the ellipse x2/25 + y2/16, = 1 is equal to 16., , 13) Find the (i) lengths of the principal axes, (ii) co-ordinates of the foci (iii) equations, of directrices (iv) length of the latus rectum, (v) Distance between foci (vi) distance, between, directrices, of, the, curve, (a) x2/25 + y2/9 = 1 (b) 16x2 + 25y2 = 400, (c) x2/144 − y2/25 = 1 (d) x2 − y2 = 16, , 22) Find the equation of the hyperbola in the, standard form if (i) Length of conjugate, axis is 5 and distance between foci is 13., (ii) eccentricity is 3/2 and distance between, foci is 12. (iii) length of the conjugate axis is, 3 and distance between the foci is 5., , 14) Find the equation of the ellipse in standard, form if (i) eccentricity = 3/8 and distance, between its foci=6. (ii) the length of major, axis 10 and the distance between foci is 8., (iii) passing through the points (−3, 1) and, (2, −2)., , 23) Find the equation of the tangent to, the hyperbola, (i) 7x2 − 3y2 = 51 at (−3, −2), (ii) x = 3 secθ, y = 5 tanθ at θ = π/3 (iii) x2/25, −y2/16 = 1 at P(30°)., , 15) Find the eccentricity of an ellipse if the, distance between its directrices is three times, the distance between its foci., , 24) Show that the line 2x − y = 4 touches the hyperbola 4x2 − 3y2 = 24. Find the point of contact., , 16) For the hyperbola x2/100−y2/25 = 1, prove, that SA. S'A = 25, where S and S' are the foci, and A is the vertex., , 25) Find the equations of the tangents to the hyperbola 3x2 − y2 = 48 which are perpendicular, to the line x + 2y − 7 = 0, , 17) Find the equation of the tangent to the ellipse, x2/5−y2/4 = 1 passing through the point, (2,−2)., , 26) Two tangents to the hyperbola, make angles θ1, θ2, with the transverse axis., Find the locus of their point of intersection if, tanθ1 + tanθ2 = k., , 18) Find the equation of the tangent to the ellipse, x2 + 4y2 = 100 at (8,3)., , 178

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Measures of Dispersion, , 8, , All the above series have the same size, (n=5) and the same mean (50), but they are, different in composition. Thus, to decide who, is more dependable, the measure of mean is not, sufficient. The spread of data or variation is a, factor which needs our attention. To understand, more about this we need some other measure., One such measure is Dispersion., , Let's Study, ∙, , Meaning and Definition of Dispersion, , ∙, , Range of data., , ∙, , Variance and Standard Deviation, , ∙, , Coefficient of Variation, , In the above example, observations from, series X and series Z are more scattered as, compared to those in series Y. So Y is more, consistent. The extent of scatter in observations, which deviate from mean is called dispersion., , Let's Recall, ∙, , Concept of Constant and Variable, , ∙, , Concept of an Average, , ∙, , Computation of Mean for Ungrouped and, Grouped Data, , Activity :, Given two different series-, , “An average does not tell the full story., It is hardly fully representative of a mass unless, we know the manner in which the individual items, scatter around it. A further description of the series is necessary if we are to gauge how representative the, average is.”, , - George Simpson and Fritz Kafka, , Let’s Observe, , Runs scored, 90, 17, 104, 33, 6, 40, 60, 55, 50, 45, 112, 8, 96, 29, 5, , A : 0.5, 1, 1.5, 3, 4, 8, , , , B : 2, 2.2, 2.6, 3.4, 3.8,, , Find arithmetic means of the two series., Plot the two series on the number line., Observe the scatter of the data in each series, and decide which series is more scattered., , Let's Learn, , In the earlier classes we have learnt about, the measures of central tendency mean, median, and mode. Such an average tells us only about, the central part of the data. But it does not give, any information about the spread of the data. For, example, consider the runs scored by 3 batsmen, in a series of 5 One Day International matches., Batsman, X, Y, Z, , , , Total, 250, 250, 250, , According to Spiegel:, “The degree to which numerical data tend, to spread about an average value is called the, variation or dispersion of the data.”, 8.1 Measures of Dispersion :, , Mean, 50, 50, 50, , Following measures of dispersion are the, commonly used –, (i) Range, (ii) Variance, (iii) Standard deviation, 179

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8.1.1 Range :, Range is the simplest measure of dispersion., It is defined as the difference between the largest, value and the smallest value in the data., Thus, Range = Largest Value – Smallest Value, , = L –S, , 2., , Find range of the following data:, 575, 609, 335, 280, 729, 544, 852, 427, 967,, 250, , 3., , The following data gives number of typing, mistakes done by Radha during a week. Find, the range of the data., , Where, L = Largest Value and S = Smallest, Value., Uses of Range:, 1), , It is used in stock market., , 2), , It is used in calculations of mean temperature, of a certain place., , , 4, , SOLVED EXAMPLES, , Wednesday, , No. of, mistakes, , 15, , 20, , 21, , Thurs- Friday, day, 12, , 17, , Saturday, 10, , Following results were obtained by rolling a, die 25 times. Find the range of the data., 1, 4, , 2, 6, , 3, 2, , 4, 7, , 5, 3, , 6, 3, , Find range for the following data., Frequency, , 62-64 64-66 66-68 68-70 70-72, 5, , 3, , 4, , 5, , 3, , , , Solution : Smallest Value = S = 36, , Let's Learn, , Largest Value = L = 73, , 8.2 VARIANCE and STANDARD DEVIATION:, , Range = L – S = 73 – 36 = 37, , The main drawback of the range is that it is, based on only two values, and does not consider, all the observations. The variance and standard, deviation overcome this drawback as they are based, on the deviations taken from the mean., , Ex.2. Calculate range for the following data., Salary, 30- 5 0 - 70- 9 0 - 110- 130(00’s, 50, 70, 90 110 130 150, Rs.), No. of, Employ- 7, 15, 30 24, 18, 11, ees, , 8.2.1 Variance:, The variance of a variable X is defined as the, arithmetic mean of the squares of all deviations of, X taken from its arithmetic mean., , Solution :, L = Upper limit of highest class = 150, S = Lower limit of lowest class = 30, , It is denoted by Var(X) or σ2 ., , ∴ Range = L – S = 150 – 30 = 120, , Var (X) = σ2 =, , EXERCISE 8.1, 1., , Tuesday, , Classes, , 70, 62, 38, 55, 43, 73, 36, 58, 65, 47, , , Monday, , Score, Frequency, 5., , Ex.1) Following data gives weights of 10, students (in kgs) in a certain school. Find, the range of the data., , Day, , 1 n, ∑ (x – x )2, n i =1 i, , Note :, , Find range of the following data:, 19, 27, 15, 21, 33, 45, 7, 12, 20, 26, , We have,, Var (X), , 180, , 1 n, = ∑ (xi – x )2, n i =1

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1 n, = ∑ (xi2 – 2xi x + x 2), n i =1, , 1 n, Var (X) = σ = ∑ fi (xi – x )2, n i =1, 2, , =, , 1 n 2, 1 n, 1 n, 2, x, –, 2, •, x, +, 1, x, ∑, ∑ x n∑, n i =1 i, n i =1 i, i =1, , =, , 1, 1, x i2 – 2 x . x + x 2 × n n, ∑, n i =1, , n, , =, , , , n, , n, , Standard Deviation is defined as the positive, square root of the variance., It is denoted by σ (sigma) and σ =, , =, , Var ( X ), , , , Where , x =, , =, , i =1, , n, , n, , ∑x, i =1, , i, , i i, , , and ∑  fi = N = Total, , i =1, , i i, , 2, , N, , – x 2,, , n, , x =, , ∑fx, i =1, , i i, , N, , S. D. = σ =, , n, , , and, , ∑f, i =1, , i, , = N = Total frequency, , Var ( X ), , 8.2.3 Change of origin and scale:, 1. The variance and consequently the standard, deviation are independent of change of, origin., , 1 n, ∑ (x – x )2, n i =1 i, 2, i2, 1, , ∑fx, , Where,, , (i) Variance and Standard Deviation for raw, data :, Let the variable X takes the values x1, x2, x3,, … xn. Let x be the arithmetic mean., , n, , i =1, , (iii) Variance and Standard Deviation for, grouped frequency distribution :, Let x1, x2,…, xn be the mid points of the, intervals. and f1, f2, …, fn are corresponding, class frequencies, then the variance is, defined as :, 1 n, Var (X) = σ2 = ∑ fi (xi – x )2, n i =1, , 1 n, ∑ (x – x )2, n i =1 i, , ∑ xx, , – x 2,, , N, frequency S. D. = σ = Var ( X ), , 1 n 2, , = ∑ xi – x 2, n i =1, 8.2.2 Standard Deviation :, , Var (X) = σ2 =, , 2, , N, , ∑fx, , Where, x =, , 1 n 2, = ∑ xi – x 2, n i =1, , Then,, , i =1, , i i, , n, , 1 n 2, = ∑ xi – 2 x 2 + x 2, n i =1, , Therefore, Var (X) = σ2 =, , ∑fx, , That is if d = x – A where A the class mark of, middle class interval if the number of classes, is odd. If the number of classes is even, then, there will be two middle class intervals and, A is the class mark of the one having greater, frequency., , − x2, , S. D. = σ = Var ( X ), , (ii) Variance and Standard Deviation for, ungrouped frequency distribution :, , then σ d2 = σ x2 and σd = σx, , Let x1, x2,…, xn be the values of variable X, with corresponding frequencies f1, f2, …,, fn respectively, then the variance of X is, defined as, , It means that if σx is standard deviation of the, values x1, x2, x3, ... xn. The standard deviation, σd, of x1 − A, x2−A, x3 − A, ......, xn − A is also, same as that of σx., 181

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2., , Ex.2) Given below are the marks out of 25, of 5 students in mathematics test. Calculate, the variance and standard deviation of these, observations., , The variance and consequently the standard, deviation are not independent of change of, scale., x– A, Let u =, where h is width of the class, h, interval if given. If the class intervals are not, given, then h is the difference (or distance), between the two consecutive value of xi., , Marks : 10, 13, 17, 20, 23, Solution : We use alternate method to solve this, problem., Calculation of variance :, , and h ≠ 0, then σx = h σu, , xi, 10, 13, 17, 20, 23, 83, , and σx2 = h2 σu2, It means that if σx is standard deviation of the, values x1, x2, ... xn. Then standard deviation, σu of, , x −A, x1 − A x2 − A x3 − A, is, ,, ,, , ...., n, h, h, h, h, , 1, times of σx., h, , n, , Here, n = 5, , SOLVED EXAMPLES, , Therefore, Var (X) = σ2 =, , 9, 12, 15, 18, 21, 24, 27, Solution :, , xi – x, −9, −6, −3, 0, 3, 6, 9, , 9, 12, 15, 18, 21, 24, 27, 126, Here, n = 7,, , (xi – x )2, 81, 36, 9, 0, 9, 36, 81, 252, , n, , ∑x, , 2, i, , i =1, , n, , 83, = 16.6, 5, – ( x )2, , 1487, – (16.6)2, 5, , , , = 297.4 – 275.56, , , , = 21.84, S.D. = σ = Var ( X ) =, , 21.84, , = 4.67, , Ex.3) A die is rolled 30 times and the following, distribution is obtained. Find the variance and, S.D., Score (X), Frequency (f), , ∑ x1 126, x =, n = 7 = 18, , Var ( X ) =, , =, , =, , , , 1, 2, , 2, 6, , 3, 2, , 4, 5, , 5, 10, , Solution :, , 1, 252, Var (X) = σ = •∑(xi – x )2 =, = 36, n, 7, =σ=, , i, , i =1, , , , , , 2, , S.D., , and x =, , ∑x, n, , Ex.1) Compute variance and standard deviation, of the following data observations., , xi, , xi2, 100, 169, 289, 400, 529, 1487, , X, 1, 2, , 36 = 6, 182, , f, 2, 6, , f.x, 2, 12, , f.x2, 2, 24, , 6, 5

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3, 4, 5, 6, Total, , 2, 5, 10, 5, 30, , 6, 20, 50, 30, 120, , =, , ∑ fx, 120, =, =4, N, 30, , x, , We get,, , Now, σ, , 2, x, , = 3.72 – 0.0016, , , , 18, 80, 250, 180, 554, , = 3.7184, , , , Therefore, σu = 3.1784 = 1.92, σx = h σu, , , ∴ σx = 9.6, Ex.5. Compute variance and standard deviation, for the following data., , ∑ fx2, = N – x2, =, , C.I. 45- 55- 65- 7555 65 75 85, f, 7 20 27 23, , 554, – 42 = 18.47 – 16 = 2.47, 30, , Therefore, σ x =, , 15, 13, , 2.47 = 1.57, , 20, 12, , 25, 15, , 30, 18, , Let u =, , 35, 17, , 40, 10, , Class-intervals, , 45, 15, , 45-55, 55-65, 65-75, 75-85, 85-95, 95-105, 105-115, 115-125, Total, , x − 30, 5, , X, 15, 20, 25, 30, 35, 40, 45, Total, We get,, Now,, , u, −3, −2, −1, 0, 1, 2, 3, , u, , σ, , =, 2, u, , f, 13, 12, 15, 18, 17, 10, 15, 100, , f.u, −39, −24, −15, 0, 17, 20, 45, 4, , f.u2, 117, 48, 15, 0, 17, 40, 135, 372, , Now,, , X − 90, ., 10, , u =, , Mid, value, ( xi ), 50, 60, 70, 80, 90, 100, 110, 120, , fi, , ui, , fiui, , fiui2, , 7, 20, 27, 23, 13, 6, 3, 1, 100, , −4, −3, −2, −1, 0, 1, 2, 3, , −28, −60, −54, −23, 0, 6, 6, 3, −150, , 112, 180, 108, 23, 0, 6, 12, 9, 450, , ∑ fiui −150, N = 100 = – 1.5, 2, , Var(u) = σ u =, , 4, ∑ f.u, =, = 0.04, N, 100, , ∑fiui2, 2, N –u, , 450, – (–1.5)2 = 4.5 – 2.25, 100, = 2.25, , Var(u) =, , , ∑ f.u2, 2, =, N – u, , =, , 95- 105- 115105 115 125, 6, 3, 1, , Calculation of variance of u :, , Solution:, Let u =, , 8595, 13, , Solution :, , Ex. 4) Compute variance and standard deviation, for the following data:, x, f, , = 5(1.92), , Thus, Var(u) = 2.25, , 372, – 0.042, 100, , ∴ Var(X)= h2.Var(u) = 102 ×2.25 = 225, , 183, , ∴ S.D. = σ x = Var ( X ) =, , 255 = 15

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Ex.6) Find the standard deviation of the following, frequency distribution which gives distribution of, heights of 500 plants in centimeters., Height of, plants, (in cm), No. of, plants, , 2025, , 2530, , 3035, , 3540, , 4045, , 4550, , 145, , 125, , 90, , 40, , 45, , 55, , 3., X, F, , 4., , 5., , 6., , 7., , ∑ fiui −120, N = 500 = – 0.24, ∑fu, 2, 2, Var(u) = σ u = Ni i – u, , , , = 2.84 – 0.0576 = 2.7824, , 8., , Thus, Var(u) = 2.7824, , Q. Find variance and S.D. for the following set, of numbers., 65, 77, 81, 98, 100, 80, 129, , 34, 8, , 35, 9, , 36, 10, , 37, 6, , Following data gives age of 100 students in, a college. Calculate variance and S.D., 18, 11, , 19, 17, , 20, 30, , 21, 15, , Find mean, variance and S.D. of the, following data., , Find the variance and S.D. of the following, frequency distribution which gives the, distribution of 200 plants according to their, height., 1418, 5, , 1923, 18, , 2428, 44, , 2933, 70, , 3438, 36, , 3943, 22, , 4448, 5, , The mean of 5 observations is 4.8 and the, variance is 6.56. If three of the five observations, are 1, 3 and 8, find the other two observations., , If σ1, σ2 are standard deviations and x 1, x 2 are, the arithmetic means of two data sets of sizes n1, and n2 respectively ,then the mean for the combined, data is :, , EXERCISE 8.2, , 2., , 33, 10, , 8.3 Standard Deviation for Combined data :, , ∴ S.D. = σ x = Var ( X ) = 69.56 = 8.34cm, , 7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6, , 31 32, 15 12, , Let's :Learn, , ∴ Var(X)= h2.Var(u) = 52 ×2.7824 = 69.56 cm2, , 1., , 10 12 14 16 18 20, 6 4 3 4 2 6, , Compute the variance and S.D., , Height, (in cm), No. of, plants, , 1420, – (–0.24)2, 500, , =, , 8, 7, , Class- 10- 20- 30- 40- 50- 60- 70- 80- 90es, 20 30 40 50 60 70 80 90 100, Freq., 7 14 6 13, 9, 15 11 10 15, , Now, u =, , , , 4 6, 10 10, , Age (In years) 16 17, No. of Students 20 7, , Mid value fi, ui, fiui, fiui2, ( xi ), 22.5, 145 −2 −290 580, 27.5, 125 −1 −125 125, 32.5, 90, 0, 0, 0, 37.5, 40, 1, 40, 40, 42.5, 45, 2, 90, 180, 47.5, 55 3, 165, 495, −120 1420, , 20-25, 25-30, 30-35, 35-40, 40-45, 45-50, Total, , 2, 8, , X, Frequency, , Solution : Let u = X − 32.5 ., 5, Calculation of variance of u :, Class, , Compute variance and standard deviation, for the following data:, , xc =, , n1 x1 + n2 x2, n1 + n2, , And the Standard Deviation for the combined, series is :, , 184

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Solution :, , n (σ 2 + d12 ) + n2 (σ 22 + d 22 ), σc = 1 1, n1 + n2, , Given, n1 = 100,, , 2, x 1 = 45, σ 1 = 49, , For combined group,, σ c2 = 130,, , Where, d1= x 1 – x c and d2= x 2 – x c, , n = 250,, , x c = 51,, , To find : x 2 and σ c, , SOLVED EXAMPLES, , n = n1 + n2 a 250 = 100 + n2, , Ex.1) The means of two samples of sizes 10 and, 20 are 24 and 45 respectively and the standard, deviations are 6 and 11. Obtain the standard, deviation of the sample of size 30 obtained by, combining the two samples., , n2 = 150, , n1 x1 + n2 x2, n1 + n2, , We have, xc =, , 100 × 45 + n2 x2, 100 + 150, , 51 =, , Solution : , , a, , 12750 = 4500 + 150 x 2, , Let n1 = 10, n2 = 20, x 1 = 24, x 2 = 45,, σ1 = 6, σ2 = 11, , Therefore, x 2 = 55, , Combined mean is :, , Combined standard deviation is given by,, , xc =, , n1 x1 + n2 x2 10 × 24 + 20 × 45, =, n1 + n2, 10 + 20, , σ c2 =, , 1140, = 38, 30, Combined standard deviation is given by,, , n1 (σ 12 + d12 ) + n2 (σ 22 + d 22 ), n1 + n2, , Where, d1= x 1 – x c = 45-51 = – 6, , =, , and d2= x 2 – x c = 55-51 = 4, , 100(49 + 36) + 150(σ 22 + 16), 130 =, 100 + 150, , n1 (σ 12 + d12 ) + n2 (σ 22 + d 22 ), σc =, n1 + n2, , 2, , 32500 = 150 σ 2 + 10900, , Where, d1= x 1 – x c and d2= x 2 – x c, , σ 22 = 144, , ∴ d1= x 1 – x c = 24 – 38 = – 14 and, , Therefore, σ 2 = 144 = 12, ∴ S.D. of second group = 12, , d2= x 2 – x c = 45 – 38 = – 7, σc, , =, , 10(62 + (−14) 2 ) + 20(112 + 7 2 ), 10 + 20, , =, , 2320 + 3400, =, 30, , =, , 190.67 = 13.4, , Let's :Learn, , 5720, 30, , 8.3.1 Coefficient of Variation :, Standard deviation depends upon the unit, of measurement. Therefore it cannot be used to, compare two or more series expressed in different, units. For this purpose coefficient of variation, (C.V.) is used and is defined as,, , Ex.2) The first group has 100 items with mean, 45 and variance 49. If the combined group has, 250 items with mean 51 and variance 130, find, the mean and standard deviation of second group., , C. V. = 100 ×, 185, , σ, x

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(a) Based on consistency :, Since C.V. of Varad is smallest, he is more, consistent and hence is to be selected., , Coefficient of Variation is used to compare, the variability of two distributions. A distribution, with smaller C.V. is said to be more homogenous, or compact and the series with larger C.V. is said, to be more heterogeneous. The distribution with, smaller C.V. is said to be more consistent., , (b) Based on expected score :, If the player with highest expected score, (mean) is to be selected, then Viraj will be, selected., , SOLVED EXAMPLES, , Ex.2. The following values are calculated in, respect of prices of shares of companies X and Y., State the share of which company is more stable, in value., , Share of Y Share of Y, , Ex.1) The arithmetic mean of runs scored by, 3 batsmen Varad, Viraj and Akhilesh in the, same series are 50, 52 and 21 respectively. The, standard deviation of their runs are 11, 16 and 5, respectively. Who is the most consistent of the, three? If one of the three is to be selected, who, will be selected?, , Mean, Variance, , 50, 7, , 105, 4, , Solution :, , Solution :, Let x1 , x 2 , x 3 and σ1, σ2, σ3 be the means, and standard deviations of the three batsmen Varad,, Viraj and Akhilesh respectively., , Here, σ x2 = 7, σ y2 = 4, x = 50, y = 105, , Therefore, x1 = 50, x 2 = 52,, σ1 = 11, σ2 = 16, σ3 = 5, , C.V.( X) =100 ×, , Therefore σ x =, , σ1, Now, C. V. of runs scored by Varad = 100 × x, 1, 11, = 22, = 100 ×, 50, , = 100 ×, , C. V. of runs scored by Akhilesh = 100 ×, 5, = 100 ×, = 23.81, 21, (i), , Since C.V.(Y) < C.V.(X),, The shares of company Y are more stable in, value., , σ2, x2, , 16, = 30.76, 52, , 4 =2, , σx, 2.64, =100 ×, = 5.28, x, 50, σy, 2, C.V.( Y ) = 100 ×, =100 ×, = 1.90, y, 105, , x3 = 21 and, , C. V. of runs scored by Viraj = 100 ×, , 7 = 2.64, σ y =, , Activity : Construct the table showing the, frequencies of words with different number, of letters occurring in the following passage,, omitting punctuation marks. Take the number of, letters in each word as one variable and obtain, the mean, S.D. and the coefficient of variation of, its distribution., , σ3, x3, , Since the C. V. of the runs is smaller for Varad,, he is the most consistent player., , “Take up one idea. Make that one idea your, life – think of it, dream of it, live on that idea., Let the brain, muscles, nerves, every part of your, body, be full of that idea, and just leave every, other idea alone. This is way to success.”, , (ii) To take decision regarding the selection, let us, consider both the C.V.s and means., , 186

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(ii) Which worker seems to be faster in, completing the job?, , EXERCISE 8.3, 1., , 2., , 4., 5., , 6., , A company has two departments with 42 and, 60 employees respectively. Their average, weekly wages are Rs. 750 and Rs. 400. The, standard deviations are 8 and 10 respectively., (i) Which department has a larger bill?, (ii) Which department has larger variability, in wages?, , For a certain data, following information is, available., X, 13, 3, 20, , Mean, S. D., Size, 3., , 7., , The means of two samples of sizes 60 and, 120 respectively are 35.4 and 30.9 and the, standard deviations 4 and 5. Obtain the, standard deviation of the sample of size 180, obtained by combining the two sample., , 8., , Y, 17, 2, 30, , Obtain the combined standard deviation., Calculate coefficient of variation of marks, secured by a student in the exam, where the, marks are: 85, 91, 96, 88, 98, 82, , The following table gives weights of, the students of two classes. Calculate, the coefficient of variation of the two, distributions. Which series is more variable?, Weight (in kg), Class A, Class B, 30-40, 22, 13, 40-50, 16, 10, 50-60, 12, 17, Compute coefficient of variation for team A, and team B., , 9., , Find the coefficient of variation of a sample, which has mean equal to 25 and standard, deviation of 5., , No. of goals, No. of matches played by, team A, No. of matches played by, team B, , A group of 65 students of class XI have their, average height is 150.4 cm with coefficient, of variance 2.5%. What is the standard, deviation of their height?, , 0, 19, , 1, 6, , 2, 5, , 3, 16, , 4, 14, , 16, , 14, , 10, , 14, , 16, , Which team is more consistent?, , Two workers on the same job show the, following results:, , Worker P Worker Q, Mean time for, completing the job, 33, 21, (hours), Standard Deviation, 9, 7, (hours), , 10. Given below is the information about marks, obtained in Mathematics and Statistics by, 100 students in a class. Which subject shows, the highest variability in marks?, Mathematics, 20, 2, , Mean, S.D., , (i) Regarding the time required to complete, the job, which worker is more consistent?, , Statistics, 25, 3, , Range :, Activity 1 :, The daily sale of wheat in a certain shop is given below., Day, , Monday, , Tuesday, , Wednesday, , Thursday, , Friday, , Saturday, , Sale in Kg, , 135, , 39, , 142.5, , 78, , 120.5, , 93, , Then L = ––––––––––––, , S = ––––––––––––, , Range L – S = ––––––––––––, 187

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Activity 2 :, Neeraj Chopra is an Indian track and field athlete, who competes in the Javelin throw., The following data reveals his record of throws in Asian Championships (A.C.) and World Championships, (W.C.), Championship, Throw, (Meters), , 2015, , 2016, , 2017, , 2018, , 2013, , 2016, , 2017, , (A.C.), , (A.C.), , (A.C.), , (A.C.), , (W.C.), , (W.C.), , (W.C.), , 70.50, , 77.60, , 85.23, , 88.06, , 66.75, , 86.48, , 82.26, , Then L = ––––––––––––, , S = ––––––––––––, , Range L – S = ––––––––––––, , Variance and Standard Deviation :, Activity 1 :, The marks scored in a test by seven randomly selected students are, 3, , 4, , 6, , 2, , 8, , 8, , 5, , Find the Variance and Standard Deviation of these seven students., Solution :, Mean x =, , 3 + 4 + + + + + 5 36, =, 7, 7, , The deviation from mean for each observation is ( x – x ), 36, 3– 7, , 36, 8– 7, , 15, – 7, , 5– 7, , 20, 7, , The deviations squared are ( x – x )2, 225, 49, , Variance =, , 400, 49, , ∑, , ( x − x )2, n, , Standard Deviation = Variance =, , 188, , =

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Activity 2 :, The number of centuries scored in a year by seven randomly selected batsmen are, 3, , 5, , 6, , 3, , 7, , 6, , 4, , Find the Variance and Standard Deviation of these seven batsmen., Solution :, x, , 3, , x2, , 9, , ∑x, , 3, , +, , +, , +, , +, , ∑x, , 81, , +, , +, , +, , +, , Variance, , 5, , 6, , =σ, , 2, , Standard, Deviation, , σ, , 3, , 7, , ∑ x2, ∑x, = n – n, , 6, , 49, , +, , =, , +, , +, , =, =, , 2, , 2, , = 7 –, , = Variance =, , 7, , Variance and Standard Deviation for, frequency distribution :, , =, , Variance and Standard Deviation for, raw data :, , , , n, , 1, = ∑ ( xi − x ) 2 =, n i =1, , ∑x, Where , x = n i, And, , i =1, , x12, n, , S. D. =, , , – x, , ∑, , n, i =1, , fi xi2, , – x2, , and ∑ fi = N = Total frequency, , Let, x be the arithmetic mean. Then,, , ∑, , 1 n, ∑ f (x – x )2, n i =1 i i, , N, ∑ f ix i, x = N, , Where ,, , Let the variable X takes the values x1, x2,, x3, … xn., , n, , )–, 49, , =, , Var (X) σ2 =, , Range = Largest Value – Smallest, Value = L –S, , Var (X) = σ 2, , 7(, , =, , Let's Remember, , , , Result, , +, , , , , , 4, , Var ( X ), , Change of origin and scale method :, , 2, , Let u =, , , , X −A, h, , Then Var (u) = σ u2 =, , S. D. = σ = Var ( X ), , n, , ∑, i =1, , fi ui 2, N, , And Var (X) = h2. Var (u), 189, , 2, –u

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5), , i.e. σ x2 = h2 σu2, S.D. is σ x = h σ u, , , A) 5, C) 7, , Standard Deviation for Combined, data :, , 6), , n (σ 2 + d12 ) + n2 (σ 22 + d 22 ), σc = 1 1, n1 + n2, , Coefficient of Variation :, C. V. = 100 ×, , 7), , σ, x, , 8), (I) Select the correct option from the given, alternatives :, 1), , 2), , 3), , B) 20, D) 10, , B) 4 patients, D) 15 patients, , B) 40, D) 48, , For any two numbers SD is always, , Given the heights (in cm) of two groups of, students:, , 10) Standard deviation of data is 12 and mean is, 72 then coefficient of variation is, , The standard deviation of a distribution, divided by the mean of the distribution and, expressing in percentage is called:, A), B), C), D), , B) 8, D) 8 − 5 = 3, , Group A : 131 cm, 150 cm, 147 cm, 138 cm,, 144 cm, Group B : 139 cm, 148 cm, 132 cm, 151 cm,, 140 cm, Which of the following is / are the true?, I) The ranges of the heights of the two, groups of students are the same., II) The means of the heights of the two, groups of students are the same., A) I only B) II only C) Both I and II, D) None, , If the observations of a variable X are, −4,, −20, −30, −44 and −36, then the value of the, range will be:, A) -48, C) –40, , 4), , 9), , Number of patients who visited cardiologists, are 13, 17, 11, 15 in four days then standard, deviation (approximately) is, A) 5 patients, C) 10 patients, , B) Range, D) C.V., , A) Twice the range, B) Half of the range, C) Square of the range D) None of these, , If there are 10 values each equal to 10, then, S.D. of these values is : ––––––––––., A) 100, C) 0, , B) 4, D) 6, , The variance of 19, 21, 23, 25 and 27 is 8., The variance of 14, 16, 18, 20 & 22 is:, A) Greater than 8, C) Less than 8, , MISCELLANEOUS EXERCISE - 8, , 2,, , The positive square root of the mean of the, squares of the deviations of observations, from their mean is called:, A) Variance, C) S.D., , Where, d1= x1 – x c and d2= x 2 – x c, , , , If the S.D. of first n natural numbers is, then the value of n must be ––––––., , A) 13.67%, C) 14.67%, , B) 16.67%, D) 15.67%, , (II) Answer the following :, , Coefficient of Standard deviation, Coefficient of skewness, Coefficient of quartile deviation, Coefficient of variation, , 190, , 1., , 76, 57, 80, 103, 61, 63, 89, 96, 105, 72, Find the range for the following data., , 2., , 116, 124, 164, 150, 149, 114, 195, 128, 138,, 203, 144

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3., , Given below the frequency distribution, of weekly wages of 400 workers. Find the, range., , Weekly wages 10, (in ’00 Rs.), No. of workers 45, , 4., , 15, , 20, , 25, , 30 35 40, , 63, , 102, , 55, , 74 36 25, , misread as 30. Find the corrected mean and, standard deviation., 12. The mean and S.D. of a group of 50, observation are 40 and 5 respectively. If two, more observations 60 and 72 are added to the, set, find the mean and S.D. of 52 items., 13. The mean height of 200 students is 65 inches., The mean heights of boys and girls are, 70 inches and 62 inches respectively and the, standard deviations are 8 and 10 respectively., Find the number of boys and the combined, S.D., , Find the range of the following data, , Classes 115- 125125 135, Fre1, 4, quency, , 135145, 6, , 145155, 1, , 155165, 3, , 165175, 5, , Find variance and S.D. for the following set of, numbers., 5., , 25, 21, 23, 29, 27, 22, 28, 23, 21, 25, , 6., , 125, 130, 150, 165, 190, 195, 210, 230, 245,, 260, , 14. From the following data available for 5 pairs, of observations of two variables x and y, obtain, the combined S.D. for all 10 observations., Where,, , 8., X, F, , 9., , 0, , 1, , 2, , 3, , 4, , 5, , 5, , 20, , 25, , 15, , 20, , 5, , n, , ∑y, i =1, , 30, 8, , 64, 3, , 47, 4, , 63, 5, , 46, 7, , 35, 8, , 28, 3, , 2029, 65, , 3039, 100, , 4049, 55, , 60, 7, , No. of students, , 4049, 4, , 5059, 12, , 7079, 28, , ∑x, i =1, , i, , 2, , = 220,, , Boys, 60, 16, , Girls, 47, 9, , Which of the two is more variable?, , 50- 60- 70- 8059 69 79 89, 87 42 38 13, , 6069, 25, , i =1, , n, , = 340, , Mean, Variance, , 17. The mean and standard deviations of two, bands of watches are given below :, , 10. Given below is the frequency distribution of, marks obtained by 100 students. Compute, arithmetic mean and S.D., Marks, , 2, , ∑ yi =40,, , 16. Following data relates to the distribution of, weights of 100 boys and 80 girls in a school., , Calculate S.D. from following data., , Age, (In yrs), Freq., , i, , n, , 15. Calculate coefficient of variation of the, following data., 23, 27, 25, 28, 21, 14, 16, 12, 18, 16, , Compute the variance and S.D. for the, following data:, 62, 5, , ∑ xi =30,, i =1, , 7. Following data gives no. of goals scored by a, team in 100 matches. Compute the standard, deviation., No. of Goals, Scored, No. of matches, , n, , Brand-I, 36 months, 8 months, , Mean, S.D., , Brand-II, 48 months, 10 months, , Calculate coefficient of variation of the two, brands and interpret the results., , 80- 9089 99, 26, 5, , 18. Calculate coefficient of variation for the data, given below, , 11. The arithmetic mean and standard deviation, of a series of 20 items were calculated by a, student as 20 cms and 5 cms respectively., But while calculating them, an item 13 was, , Size (cm), No of items, , 191, , 58, 3, , 811, 14, , 1114, 13, , 14- 17- 20- 2317 20 23 26, 16 19 24 11

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19. Calculate coefficient of variation for the data, given below, Income, (Rs.), No. of, families, , 30004000, 24, , 40005000, 13, , 50006000, 15, , 21. There are two companies U and V which, manufacture cars. A sample of 40 cars each, from these companies are taken and the, average running life (in years) is recorded., , 6000- 7000- 8000- 90007000 8000 9000 10000, 28, 12, 8, 10, , Life ( in years), , 20. Compute coefficient of variations for the, following data to show whether the variation, is greater in the yield or in the area of the field., , Year, , Area (in acres), , 2011-12, 2012-13, 2013-14, 2014-15, 2015-16, 2016-17, 2017-18, , 156, 135, 128, 117, 141, 154, 142, , 0-5, 5-10, 10-15, , Yield, (in lakhs), 62, 70, 68, 76, 65, 69, 71, , No of Cars, Company U, Company V, 5, 14, 18, 8, 17, 18, , Which company shows greater consistency?, 22. The means and S.D. of weights and heights of, 100 students of a school are as follows., , Mean, S.D., , Weights, 56.5 kg, 8.76 kg, , Heights, 61 inches, 12.18 inches, , Which shows more variability, weights or, heights?, , 192

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9, , Probability, Types of Events:, , Let's Learn, •, •, •, •, •, •, •, , Elementary Event: An event consisting, of a single outcome is called an elementary, event., , Basic Terminologies, Concept of probability, Addition Theorem, Conditional probability, Multiplication Theorem, Baye's Theorem, Odds, , Certain Event: The sample space is called, the certain event if all possible outcomes are, favourable outcomes. i.e. the event consists of, the whole sample space., Impossible Event: The empty set is called, impossible event as no possible outcome is, favorable., , Let's Recall, , Algebra of Events:, Events are subsets of the sample space., Algebra of events uses operations in set theory, to define new events in terms of known events., , 9.1.1 Basic Terminologies, Random Experiment :, experiment having more than, All possible results are known, result cannot be predicted such, is called a random experiment., , Suppose an, one outcome., but the actual, an experiment, , Union of, two events in, of A and B is, of all possible, one of A and, , Outcome: A possible result of random, experiment is called a possible outcome of the, experiment., , Two Events: Let A and B be, the sample space S. The union, denoted by A∪B and is the set, outcomes that belong to at least, B., , Ex. Let S = Set of all positive integers, not exceeding 50;, , Sample space: The set of all possible, outcomes of a random experiment is called the, sample space. The sample space is denoted by, S or Greek letter omega (Ω). The number of, elements in S is denoted by n(S). A possible, outcome is also called a sample point since it, is an element in the sample space., , Event A = Set of elements of S that are, divisible by 6; and, Event B = Set of elements of S that are, divisible by 9. Find A∪B, Solution :, , A = {6,12,18,24,30,36,42,48}, B = {9,18,27,36,45}, , Event: A subset of the sample space is, called an event., , ∴ A∪B = {6,9,12,18,24,27,30,36,42,45, 48} is, the set of elements of S that are divisible by, 6 or 9., , Favourable Outcome: An outcome that, belongs to the specified event is called a, favourable outcome., , Exhaustive Events: Two events A and B, in the sample space S are said to be exhaustive, if A∪B = S., 193

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Example: Consider the experiment of, throwing a die and noting the number on the, top., , Event B = set of elements of S that are, divisible by 13., Then A = {8,16,24,32,40,48},, , Let S be the sample space, , B = {13,26,39}, , ∴ S = {1,2,3,4,5,6}, , ∴ A∩B = φ because no element of S is, divisible by both 8 and 13., , Let, A be the event that this number does, not exceed 4, and, , Note: If two events A and B are mutually, exclusive and exhaustive, then they are called, complementary events., , B be the event that this number is not, smaller than 3., Then A = {1,2,3,4} B = {3,4,5,6}, , Symbolically, A and B are complementary, events if A∪B = S and A∩B = φ., , and therefore, A∪B = {1,2,3,4,5,6} = S, ∴ Events A and B are exhaustive., , Notation: Complement of an event A is, , Intersection of Two Events: Let A and, B be two events in the sample space S. The, intersection of A and B is the event consisting, of outcomes that belong to both the events A, and B., , denoted by A', A or Ac. The following table, shows how the operations of complement,, union, and intersection can be combined to, define more events., Operation, , Example, Let S = Set of all positive, integers not exceeding 50,, Event A = Set of elements of S that are, divisible by 3, and, Event B = Set of elements of S that are, divisible by 5., Then A = {3,6,9,12,15,18,21,24,27,30,33,, 36,39,42,45,48},, , Interpretation, , A', A or Ac, , Not A., , A∪B, , At least, one of A and B, , A∩B, , Both A and B, , (A'∩B) ∪ (A∩B'), , Exactly one of A and B, , (A'∩B') = (A∪B)', , Neither A nor B, , SOLVED EXAMPLES:, Ex. 1: Describe the sample space of the, experiment when a coin and a die are thrown, simultaneously., , B = {5,10,15,20,25,30,35,40,45,50}, ∴ A∩B = {15,30,45} is the set of, elements of S that are divisible by both 3, and 5., , Solution : Sample space S = {(H,1), (H,2), (H,3),, (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4),, (T,5), (T,6)}, , Mutually Exclusive Events: Event A, and B in the sample space S are said to be, mutually exclusive if they have no outcomes, in common. In other words, the intersection of, mutually exclusive events is empty. Mutually, exclusive events are also called disjoint events., , Ex. 2: Sunita and Samrudhi who live in Mumbai, wish to go on holiday to Delhi together., They can travel to Delhi from Mumbai either, by car or by train or plane and on reaching, Delhi they can go for city-tour either by bus, or Taxi. Describe the sample space, showing, all the combined outcomes of different ways, they could complete city-tour from Mumbai., , Example: Let S = Set of all positive, integers not exceeding 50,, Event A = set of elements of S that are, divisible by 8, and, 194

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Solution : Sample space, S = {(car, bus), (car, taxi), (train, bus), (train,, taxi), (plane, bus), (plane, taxi)}, , 9.1.4 Probability of an Event:, The probability of an event A is defined as, , Ex. 3: Three coins are tossed. Events E1, E2, E3, and E4 are defined as follows., E1 : Occurence of at least two heads., E2 : Occurence of at least two tails., E3 : Occurence of at most one head., E4 : Occurence of two heads., Describe the sample space and events E1, E2,, E3 and E4., Find E1∪E4, E3'. Also check whether, i) E1 and E2 are mutually exclusive, ii) E2 and E3 are equal, , P(A) =, , n(A), n(S), , Where,, n(A) = number of outcomes favorable for, event A,, n(S) = number of all possible outcomes., 9.1.5 Elementary Properties of Probabilty:, 1), , A' is complement of A and therefore, P(A') = 1 − P(A), , 2), , For any event A in S, 0 ≤ P(A) ≤ 1, , Solution : Sample space, S= {HHH, HHT, HTH, HTT, THH, THT, TTH,, TTT}, E1: {HHH, HHT, HTH, THH}, E2: {HTT, THT, TTH, TTT}, E3: {HTT, THT, TTH, TTT}, E4: {HHT, HTH, THH,}, , 3), , For the impossible event φ, P(φ) = 0, , 4), , For the certain event S, P(S) = 1, , 5), , If A1 and A2 two mutually exclusive events, then P(A1 ∪ A2) = P(A1) + P(A2), , 6), , If A ⊆ B, then P(A) ≤ P(B) and, P(A'∩B) = P(B) – P(A), , E1∪E4 = {HTT, HTH, THH, HHH, HHT}, E3 = {HHH, HHT, HTH, THH}, i) E1∩E2 = {} = ϕ, ∴ E1 and E2 are mutually exclusive., ii) E2 and E3 are equal., , 7), , Addition theorem: For any two events A and, B of a sample space S,, , 8), , For any two events A and B,, P(A∩B') = P(A) − P(A∩B), , 9.1.2 Concept of Probability:, A random experiment poses uncertainty, regarding the actual result of the experiment, even, though all possible outcomes are already known., The classical definition of probability is based on, the assumption that all possible outcomes of an, experiment are equally likely., , 9), , For any three events A, B and C of a sample, space S,, , P(A∪B) = P(A) + P(B) − P(A∩B), , P(A∪B∪C) = P(A) + P(B) + P(C) – P(A∩B), – P(B∩C) – (P(A∩C) + P(A∩B∩C), 10) If A1, A2, ......, Am are mutually exclusive, events in S, then P(A1 ∪ A2 ∪, ...... ∪Am, = P(A1) + P(A2) + .... + P(Am), , 9.1.3 Equally likely outcomes:, , Remark: Consider a finite sample space S, with n finite elements., , All possible outcomes of a random, experiment are said to be equally likely if none, of them can be preferred over others., 195

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S = {a1, a2, a3, ... an}. Let A1, A2, A3, ... An be, elementary events given by Ai = {ai} with, probability P(Ai). We have, P(S) = P(A1) + P(A2) +.... + P(An) = 1, , ii) Since, , ...(I), , When all elementary events given by Ai, (i = 1, 2, 3, .... n) are equally likely, that is, P(A1) = P(A2) = ... = P(An), then from (I), we, have P(Ai) = 1/n, i = 1, 2,..... n, , Let event A: Card drawn is King, and event B: Card drawn is Queen., Since pack of 52 cards contains, 4 king cards, from which any one king card can be drawn, in 4C1 = 4 ways. ∴ n(A) = 4, , ∴ P(A) = (Number of favourable outcomes, for the occurrence of event A)/ (Total number, of distinct possible outcomes in the sample, space S), , ∴ P(A) =, , Ex. 1) If A∪B∪C = S (the sample space) and A, B, and C are mutually exclusive events, can the, following represent probability assignment?, , ∴ P(B) =, , ∴ required probability P(king or queen), , ii) P(A) = 0.4, P(B) = 0.6, P(C) = 0.2, Solution:, i) Since P(S) = P(A∪B∪C), , and 0≤ P(A), P(B), P(C) ≤ 1, , 4, n (B), =, 52, n (S), , Since A and B are mutually exclusive events, , i) P(A) = 0.2, P(B) = 0.7, P(C) = 0.1, , , , n (A), 4, =, n (S), 52, , Similarly, a pack of 52 cards contains, 4, queen cards from which any one queen card, can be drawn in 4C1 = 4 ways. ∴ n(B) = 4, , SOLVED EXAMPLES, , = 0.2 + 0.7 + 0.1 = 1, , = 0.4 + 0.6 + 0.2 = 1.2 > 1, , ∴ n(S) = 52C1 = 52., , n (A), m, =, .... (II), n (S), n, , , , , , Random Experiment = One card is drawn, at random from a pack of 52 cards, , 1, 1, 1, ) + ( ) +... +( ) (m times), n, n, n, , = P(A) + P(B) + P(C) [Property 10], , = P(A) + P(B) + P(C) [Property 10], , Solution:, , P(A) = P(A1) + P(A2) + P(A3) +.... + P(Am), , , , , , Ex. 2) One card is drawn at random from a pack, of 52 cards. What is the probability that it is, a King or Queen?, , A = A1∪A2∪A3∪ ... ∪Am, then using property, 10, we have, , ∴ P(A) =, , P(S) = P(A∪B∪C), , ∴ P(A), P(B) and P(C) cannot represent, probability assignment., , If A is any event made up of m such, elementary events, i.e., , =(, , , , = P(A∪B) = P(A) + P(B) =, , 4, 4, 2, +, =, 52, 52 13, , Ex. 3: Five employees in a company of 20 are, graduates. If 3 are selected out of 20 at, random. What is the probability that, i) they are all graduates?, , ∴ The given values can represent the, probability assignment., , ii) there is at least one graduate among them?, , 196

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Solution : Out of 20 employees, any 3 are to, be selected in 20C3 ways., , letters X, S, O and R which can be arranged, among themselves in 4! = 24 different ways., After this is done, two letters T and Y can be, arranged among themselves in 2! = 2 ways., Therefore, by fundamental principle, total, number of arrangements in which T and Y, are always together is 24 × 2 = 48., 48 2, =, ∴ required probability P(A) =, 120 5, , ∴ n(S) = 20C3 where S is the sample space., Let event A: All 3 selected employees are, graduates., Out of 5 graduate any 3 can be selected in, C3 ways., 5, C3, 10, ∴ required probability P(A) = 20, =, C3 1140, 5, , b) Event B: An arrangement begins with T and, ends with Y., , 1, =, 114, , Remaining 3 letters in the middle can be, arranged in 3! = 6 different ways., 6, 1, =, ∴ required probability P(B) =, 120 120, , Let event B: At least one graduate employee, is selected., ∴ B' is the event that no graduate employee, is selected., Since out of 20 employee, 5 are graduates,, therefore from the remaining 15 nongraduate any 3 non-graduates can be selected, in 15C3 ways., 15, , ∴ P(B') =, , 20, , EXERCISE 9.1, 1), , There are four pens: Red, Green, Blue and, Purple in a desk drawer of which two pens, are selected at random one after the other, with replacement. State the sample space, and the following events., a) A : Selecting at least one red pen., b) B : Two pens of the same color are not, selected., , 2), , A coin and a die are tossed simultaneously., Enumerate the sample space and the, following events., a) A : Getting a Tail and an Odd number, b) B : Getting a prime number, c) C : Getting a head and a perfect square., Find n(S) for each of the following random, experiments., a) From an urn containing 5 gold and 3, silver coins, 3 coins are drawn at random, b) 5 letters are to be placed into 5 envelopes, such that no envelop is empty., c) 6 books of different subjects arranged, on a shelf., , C3, 455, 91, =, =, C3, 1140, 228, , ∴ required probability, P(B) = 1 − P(B') = 1 −, , 91 137, =, 228 228, , Ex. 4) The letters of the word STORY be arranged, randomly. Find the probability that, a) T and Y are together., b) arrangment begins with T and end with, Y., , 3), , Solution:, The word STORY consists of 5 different, letters, which can be arranged among, themselves in 5! ways., ∴ n(S) = 5! = 120, a) Event A: T and Y are a together., , d) 3 tickets are drawn from a box containing, 20 lottery tickets., , Let us consider T and Y as a single letter say, X. Therefore, now we have four different, 197

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4), , 5), , 6), , 7), , 8), , Two fair dice are thrown. State the sample, space and write the favorable outcomes for, the following events., a) A : Sum of numbers on two dice is, divisible by 3 or 4., b) B : Sum of numbers on two dice is 7., c) C : Odd number on the first die., d) D : Even number on the first die., e) Check whether events A and B are, mutually exclusive and exhaustive., f) Check whether events C and D are, mutually exclusive and exhaustive., , 9), , From a bag containing 10 red, 4 blue and, 6 black balls, a ball is drawn at random. Find, the probability of drawing, a) a red ball., b) a blue or black ball., c) not a black ball., , 10) A box contains 75 tickets numbered 1 to 75., A ticket is drawn at random from the box., Find the probability that,, a) Number on the ticket is divisible by 6, b) Number on the ticket is a perfect square, c) Number on the ticket is prime, d) Number on the ticket is divisible by 3, and 5, , A bag contains four cards marked as 5, 6, 7, and 8. Find the sample space if two cards are, drawn at random, a) with replacement, b) without replacement, , 11) What is the chance that a leap year, selected, at random, will contain 53 sundays?., 12) Find the probability of getting both red, balls, when from a bag containing 5 red and, 4 black balls, two balls are drawn, i) with, replacement ii) without replacement, , A fair die is thrown two times. Find the, probability that, a) sum of the numbers on them is 5, b) sum of the numbers on them is at least 8, c) first throw gives a multiple of 2 and, second throw gives a multiple of 3., d) product of numbers on them is 12., , 13) A room has three sockets for lamps. From a, collection 10 bulbs of which 6 are defective., At night a person selects 3 bulbs, at random, and puts them in sockets. What is the, probability that i) room is still dark ii) the, room is lit, , Two cards are drawn from a pack of 52, cards. Find the probability that, a) one is a face card and the other is an ace, card, b) one is club and the other is a diamond, c) both are from the same suit., d) both are red cards, e) one is a heart card and the other is a non, heart card, , 14) Letters of the word MOTHER are arranged, at random. Find the probability that in the, arrangement, a) vowels are always together, b) vowels are never together, c) O is at the begining and end with T, , Three cards are drawn from a pack of 52, cards. Find the chance that, a) two are queen cards and one is an ace, card, b) at least one is a diamond card, c) all are from the same suit, d) they are a king, a queen and a jack, , d) starting with a vowel and end with a, consonant, 15) 4 letters are to be posted in 4 post boxes. If, any number of letters can be posted in any of, the 4 post boxes, what is the probability that, each box contains only one letter?, 198

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n(A) = x, number of favourable outcomes, for the occurrence of event A., , 16) 15 professors have been invited for a round, table conference by Vice chancellor of a, university. What is the probability that two, particular professors occupy the seats on, either side of the Vice Chancellor during the, conference., , n(B) = y, number of favourable outcomes for, the occurrence of event B., n(A∩B) = z, the number of favourable, outcomes for the occurrence of both event A and, B., n (A), y, x, n (B), ∴ P(A) = ( ) = , P(B) =, =, nS, n, n, n (S), , 17) A bag contains 7 black and 4 red balls. If 3, balls are drawn at random find the probability, that (i) all are black (ii) one is black and two, are red., , P(A∩B) =, , 9.2.1 Addition theorem for two events, For any two events A and B of a sample, space S, P(A∪B) = P(A) + P(B) − P(A∩B). This, is the property (7) that we had seen earlier. Since, it is very important we give its proof. The other, properties can also be proved in the same way., , n(A I B), n(S), , z, = n, , As all outcomes are equally likely., From Venn diagram, , This can be proved by two methods, a), , Using the definition of probability., , b), , Using Venn diagram., We assume that all outcomes are equally, likely and sample space S contains finite, number of outcomes., , (a) Using the definition of probability:, , Fig. 9.1, , If A and B are any two events, then event, A∪B can be decomposed into two mutually, exclusive events A∩B' and B, , n(A∪B) = (x−z) + z + (y−z), ∴ n(A∪B) = x + y – z, , i.e. A∪B = (A∩B')∪B, ∴ P(A∪B) =, , n(A∪B) = n(A) + n(B) – n(A∩B), , P[(A∩B')∪B], , , =, , , P(A∩B') + P(B), [By property 10], , , =, , , P(A) – P(A∩B) + P(B), [By property 8], , Dividing both sides by n(S), we get, n(A U B), n(S), , =, , n (A), n (B) n(A I B), +, – n(S), n (S), n (S), , ∴ P(A∪B) = P(A) + P(B) – P(A∩B)., , Hence P(A∪B) = P(A) + P(B) − P(A∩B), (b) Using Venn diagram:, Let n(S) = n = Total no. of distinct possible, outcomes in the sample space S., 199

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ii) Let event A: Sum of the numbers is divisible, by 3, , SOLVED EXAMPLES, Ex. 1) Two dice are thrown together. What is the, probability that,, , ∴ P(A) =, , i) Sum of the numbers is divisible by, 3 or 4?, ii) Sum of the numbers is neither divisible by 3, nor 5?, Solution : Let S be the sample space, , Let event Y: Sum of the numbers is divisible, by 5., ∴ possible sums are 5, 10, , ∴ Y ={(1, 4), (2, 3), (3, 2), (4, 1), (4, 6),, (5, 5), (6, 4)}, , Let N1 = N2 = {1,2,3,4,5,6}, , ∴ n(Y) = 7 ∴P(Y) =, , S = N1 × N2 = {(x,y)/x ∈ N1, y ∈ N2}, n(S) = 36, i), , 12, 36, , n(Y) 7, =, n(S) 36, , ∴ Event A∩Y: sum is divisible by 3 and 5, , Let event A: Sum of the numbers is divisible, by 3, ∴ possible sums are 3, 6, 9, 12., , ∴ A∩Y = ϕ, , [X and Y are mutually exclusive events], , ∴ A = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3),, (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3),, (6, 6)}, , ∴ P(A∩Y) =, , n(A ∩ Y), =0, n(S), , ∴ required probability = P(Sum of the, numbers is neither divisible by 3 nor 5), , ∴ n(A) = 12 ∴ P(A) = n(A) / n(S) = 12/36, , P (A'∩Y') = P (A∪Y)' [De'Morgan's law], , Let event B: Sum of the numbers is divisible, by 4., ∴ possible sums are 4, 8, 12, , ∴ B ={(1, 3), (2, 2), (2, 6), (3, 1), (3, 5),, (4, 4), (5, 3), (6, 2), (6, 6)}, 9, ∴ n(B) = 9 ∴ P(B) = n(B) / n(S) =, 36, ∴ Event A∩B: Sum of the numbers is, divisible by 3 and 4 i.e. divisible by 12., , , , = 1 − P(X∪Y), , [Property 1], , , , = 1 – [P(A) + P(Y) – P (A∩Y)], , , , = 1−, , 19 17, =, 36 36, , ∴ possible Sum is 12, , Ex. 2) The probability that a student will solve, problem A is 2/3, and the probability that, he will not solve problem B is 5/9. If the, probability that student solves at least one, problem is 4/5, what is the probability that, he will solve both the problems?, , 1, ∴ P(A∩B) = n(A∩B)/n(S) =, 36, P (Sum of the numbers is divisible by, 3 or 4), , Solution : Let event A: student solves problem A, 2, ∴ P(A) =, 3, , ∴ A∩B = {(6, 6)}, ∴ n(A∩B) = 1, , event B: student solves problem B., , P(A∪B) = P(A) + P(B) − P(A∩B), =, , B., , 12 9, 1 20 5, + −, =, =, 36 36 36 36 9, 200, , ∴ event B': student will not solve problem, ∴ P(B') =, , 5, 9

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5 4, =, 9 9, Probability that student solves at least one, 4, problem = P(A∪B) =, 5, ∴ P(B) = 1 − P(B') = 1 −, , ∴ P(A∪B) = P(A) + P(B) − P(A∩B), , ∴ required probability = P(he will solve both, the problems), = P(A∩B) = P(A) + P(B) − P(A∪B), =, , 5), , The probability that a student will pass in, French is 0.64, will pass in Sociology is, 0.45 and will pass in both is 0.40. What is, the probability that the student will pass in, at least one of the two subjects?, , 6), , Two fair dice are thrown. Find the probability, that number on the upper face of the first die, is 3 or sum of the numbers on their upper, faces is 6., , 7), , For two events A and B of a sample space, , 2 4 4 14, + − =, 3 9 5 45, , S, if P(A) =, , Find the value of the following., a) P(A∩B), b) P(A'∩B'), c) P(A'∪B'), , EXERCISE 9.2, 1), , 2), , 3), , 4), , 3, 5, 1, , P(B) =, and P(A∪B) = ., 8, 8, 2, , First 6 faced die which is numbered 1 through 6, is thrown then a 5 faced die which is numbered, 1 through 5 is thrown. What is the probability, that sum of the numbers on the upper faces, of the dice is divisible by 2 or 3?, , 8), , A card is drawn from a pack of 52 cards., What is the probability that,, i) card is either red or black?, ii) card is either black or a face card?, , For two events A and B of a sample space, 5, 1, S, if P(A∪B) = , P(A∩B) = and P(B') =, 6, 3, 1, , then find P(A)., 3, , 9), , A girl is preparing for National Level, Entrance exam and State Level Entrance, exam for professional courses. The chances, of her cracking National Level exam is 0.42, and that of State Level exam is 0.54. The, probability that she clears both the exams is, 0.11. Find the probability that (i) She cracks, at least one of the two exams (ii) She cracks, only one of the two (iii) She cracks none, , A bag contains 5 red, 4 blue and an unknown, number m of green balls. If the probability, of getting both the balls green, when two, 1, balls are selected at random is , find m., 7, , 10) Form a group of 4 men, 4 women and, 3 children, 4 persons are selected at random., Find the probability that, i) no child is, selected ii) exactly 2 men are selected., 11) A number is drawn at random from the, numbers 1 to 50. Find the probability that it is, divisible by 2 or 3 or 10., , A bag contains 75 tickets numbered from 1, to 75. One ticket is drawn at random. Find, the probability that,, a) number on the ticket is a perfect square, or divisible by 4, b) number on the ticket is a prime number, or greater than 40, , 9.3.1 Conditional Probability:, Let S be a sample space associated with the, given random experiment. Let A and B be any, two events defined on the sample space S. Then, the probability of occurrence of event A under, the condition that event B has already occurred, 201

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9.3.2 Let S be a finite sample space, associated, with the given random experiment, containing, equally likely outcomes. Then we have the, following result., , and P(B) ≠0 is called conditional probability of, event A given B and is denoted by P(A/B)., SOLVED EXAMPLES, , Statement: Conditional probability of event, A given that event B has already occurred is, given by, , Ex.1) A card is drawn from a pack of, 52 cards, given that it is a red card, what is the, probability that it is a face card., , P (A/B) =, , Solution : Let event A: Red card is drawn, and event B: face card is drawn, , P(A ∩ B), , P(B) ≠ 0, P(B), , (Read A/B as A given B), , Let S be a sample space associated with the, given random experiment and n(S) be the number, of sample points in the sample space S. Since, we are given that event B has already occurred,, therefore our sample space reduces to event B, only, which contains n(B) sample points. Event B, is also called reduced or truncated sample space., Now out of n(B) sample points, only n(A∩B), sample points are favourable for occurrence of, event A. Therefore, by definition of probability, , A card is drawn from a pack of 52 cards,, therefore n(S) = 52. But we are given that red, card is drawn, therefore our sample space reduces, to event A only, which contains n(A) = 26 sample, points. Event A is called reduced or truncated, sample space. Out of 26 red cards, 6 cards are, favourable for face cards., ∴ P[card drawn is face card given that it is a, red card] = P[B/A] = 6 /26 = 3/13, Ex. 2) A pair of dice is thrown. If sum of the, numbers is an odd number, what is the probability, that sum is divisible by 3?, , P(A/B) =, , n(A ∩ B), , n(B) ≠0, n(B), , n(A ∩ B), n(S), =, n(B), n(S), , Solution : Let Event A: sum is an odd, number., Event B: Sum is divisible by 3., , ∴ P(A/B) =, , A pair of dice is thrown, therefore n(S) = 36., But we are given that sum is odd, therefore our, sample space reduces to event A only as follows:, , P(A ∩ B), , , P(B), , Similarly P(B/A) =, , A = {(1,2), (1,4), (1,6), (2,1), (2,3), (2,5),, (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4),, (5,6), (6,1), (6,3), (6,5)}, , P(B) ≠0, , P(A ∩ B), , P(A) ≠ 0, P(A), , SOLVED EXAMPLES, , ∴ n(A) = 18, , Ex. 1: Find the probability that a single toss, of a die will result in a number less than 4 if it is, given that the toss resulted in an odd number., Solution : Let event A: toss resulted in an, odd number and, , Out of 18 sample points following 6 sample, points are favourable for occurrence of event B, B = {(1, 2), (2, 1), (3, 6), (4, 5), (5, 4), (6, 3)}, ∴ P[sum is divisible by 3 given that sum is, an odd number] = P(B/A) = 6/18 = 1/3, , Event B: number is less than 4, , 202, , 1, ∴ A = {1, 3, 5} ∴ P(A) = 3/6 = 2

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B = {1, 2, 3} ∴ A∩B = {1, 3}, ∴ P(A∩B) =, , SOLVED EXAMPLES, , 2 1, =, 6 3, , Ex. 1) Two cards are drawn from a pack of, 52 cards one after other without replacement., What is the probability that both cards are ace, cards?, , ∴ P(number is less than4 given that it is odd), 1 1 2, = P(B/A) = P(A∩B)/ P(A) =   /   =, 3  2 3, , Soln.: Let event A: first card drawn is an Ace, card., , Ex. 2) If P(A') = 0.7, P(B) = 0.7, P(B/A) =, 0.5, find P(A/B) and P(A∪B)., , Let event B: second card drawn is an Ace, card., , Solution : Since 1 – P(A') = 0.7, , ∴ required probability = P(both are Ace, cards), , P(A) = 1 – P(A') = 1 − 0.7 = 0.3, Now P(B/A) = P(A∩B)/P(A), , = P(A∩B) = P(A)P(A/B), , ∴ 0.5 = P(A∩B)/ 0.3, , Now P(A) =, , ∴ P(A∩B) = 0.15, , 4, 1, =, 52 13, , Since first ace card is not replaced in the, pack, therefore now we have 51 cards containing, 3 ace cards, , Again P(A/B) = P(A∩B)/P(B), = 0.15/0.7, , ∴ Probability of getting second are card under, the condition that first ace card is not replaced in, 3, 1, =, the pack = P(B/A) =, 51 17, , ∴ P(A/B) = 3/14, , Further, by addition theorem, P(A∪B) = P(A) + P(B) - P(A∩B), , ∴ P(both are ace cards) = P(A∩B), , = 0.3 + 0.7 – 0.15 = 0.85, 9.3.3 Multiplication theorem:, , = P(A)P(B/A) =, , Statement: Let S be sample space associated, with the given random experiment. Let A and B, be any two events defined on the sample space, S. Then the probability of occurrence of both the, events is denoted by P(A∩B) and is given by, , 1 1, 1, × =, 13 17 221, , Ex. 2) An urn contains 4 black and 6 white, balls. Two balls are drawn one after the other, without replacement, what is the probability that, both balls are black?, Solution : Let event A: first ball drawn in, black., , P(A∩B) = P(A).P(B/A), , Event B: second ball drawn is black., , B P(A ∩ B), Since P(B/A) ==, A, P(A), , ∴ Required probability = P(both are black, balls), , ∴ P(A∩B) = P(A).P(B/A), , = P(A∩B) = P(A)P(B/A), 4, Now P(A) =, 10, , Similarly P(A∩B) = P(B).P(A/B), , 203

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Since first black ball is not replaced in the, urn, therefore now we have 9 balls containing 3, black balls., , = P(A){1 – P(B)}, = P(A).P(B'), ∴ A and B' are also independent., b) P(A'∩B') = P(A∪B)', (By De Morgan’s Law), , ∴ Probability of getting second black ball, under the condition that first black is not replaced, 3, in the pack = P(B/A) = 9, , = 1 - P(A∪B), = 1 –[P(A) + P(B) – P(A∩B)], , ∴ P(both are black balls) = P(A∩B), , = P(A)P(B/A) =, , = 1 – P(A) – P(B) + P(A).P(B), , from (I), , 3, 2, 4, × 9 = 15, 10, , 9.3.4 Independent Events:, , = [1– P(A)] – P(B)[1 – P(A)], , Let S be sample space associated with the, given random experiment. Let A and B be any, two events defined on the sample space S. If, the occurrence of either event, does not affect, the probability of the occurence of the other, event, then the two events A and B are said to be, independent., , = [1– P(A)] [1 – P(B)], = P(A').P(B'), ∴A' and B' are also independent., SOLVED EXAMPLES, , Thus, if A and B are independent events, then, P(A/B) = P(A/B') = P(A) and, , Ex. 1: Two cards are drawn at random one, after the other. Given that first card drawn in nonface red card, what is the probability that second, card is face card, if the cards are drawn, , P(B/A) = P(B/A') = P(B), Remark: If A and B are independent events, then P(A∩B) = P(A).P(B), , i) without replacement? ii) with replacement?, Solution : Let event A: first card drawn is, a non-face red card and event B: second card, drawn is face card., , note that P(A∩B) = P(A).(B/A), , , = P(A).P(B), , ∴ P(A∩B) = P(A).P(B), ∴ P(A) =, , In general, if A1, A2, A3, … An are n mutually, independent events, then, , 20 5, 12 3, =, =, and P(B) =, 52 13, 52 13, , ∴ required probability = P(second card, drawn is face card given that it is a red card), , P (A1∩A2∩…. ∩An) = P(A1).P(A2)…P(An), Theorem:, , i) Without replacement: Since first non-face, red card is not replaced, therefore now we have, 51 cards containing 12 face cards., , If A and B are independent events then, a) A and B' are also independent event, , 12, ∴ P(B/A) = 51 ≠ P(B). In this case A and B, , b) A' and B' are also independent event, Proof: Since A and B are independent,, , therefore P(A∩B) = P(A).P(B) … (1), , are not independent., , ii) With replacement: Since first non-face, red card is replaced, therefore now again we have, 52 cards containing 12 face cards., , a) P(A∩B') =P(A) – P(A∩B), = P(A) – P(A).P(B) [From (1)], 204

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3, 12, ∴ P(B/A) = 52 = 13 = P(B)., In this case A and B are independent., , = P(A).P(N/A) + P, +, =, , i) P(A∩B) ii) P(A∩B') iii) P(A'∩B), iv) P(A'∩B') v) P(A∪B), , +, +, , +, +, , =, , EXERCISE 9.3, , 3, 2, Solution : P(A) = 5 ∴P(A') = 1 -P(A) = 5, 2, 1, P(B) = 3 ∴P(B') = 1 – P(B) = 3, 2, i) P(A∩B) = P(A)P(B) = 5, 1, ii) P(A∩B') =P(A)P(B') = 5, 4, iii) P(A'∩B) =P(A')P(B) = 15, 2, iv) P(A'∩B') =P(A')P(B') = 15, 13, v) P(A∪B) = P(A) + P(B) - P(A∩B) = 15, Ex. 3) Three professors A, B and C appear in, an interview for the post of Principal. Their, chances of getting selected as a principal, 2, 4, 1, are, ,, ,, . The probabilities they, 9, 9, 3, 3, introduce new course in the college are ,, 10, 1 4, ,, respectively. Find the probability that, 2 5, the new course is introduced., , 1), , A bag contains 3 red marbles and 4 blue, marbles. Two marbles are drawn at random, without replacement. If the first marble, drawn is red, what is the probability the, second marble is blue?, , 2), , A box contains 5 green pencils and 7 yellow, pencils. Two pencils are chosen at random, from the box without replacement. What is, the probability that both are yellow?, , 3), , In a sample of 40 vehicles, 18 are red, 6 are, trucks, of which 2 are red. Suppose that a, randomly selected vehicle is red. What is the, probability it is a truck?, , 4), , From a pack of well-shuffled cards, two cards, are drawn at random. Find the probability, that both the cards are diamonds when, i) first card drawn is kept aside, ii) the first card drawn is replaced in the, pack., , Solution : Let A, B, C be the events that prof. A,, B and C are selected as principal., 2, 4, 1, 3, Given P(A) = , P(B) = , P(C) =, =, 9, 9, 3, 9, Let N be the event that New Course in, introduced P(N/A) =, , P(N/B) =, ,, 4, P(N/C) = 5, N = (A∩N) ∪ (B∩, )∪(, ∩N), ∴ P(N) = P(A∩N) + P(B∩, P(, ∩N), , × P(N/C), , =, , Ex.2: If A and B are two independent events, 3, 2, and P(A) = 5 , P(B) = 3 , find, , ×, , 5), , A, B, and C try to hit a target simultaneously, but independently. Their respective, 3 1, probabilities of hitting the target are ,, 4 2, 5, and . Find the probability that the target, 8, a), b), c), d), , ) +, , 205, , is hit exactly by one of them, is not hit by any one of them, is hit, is exactly hit by two of them

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6), , The probability that his wife who is 40 years, 3, old will be alive till she becomes 65 is ., 8, What is the probability that, 25 years hence,, a) the couple will be alive, b) exactly one of them will be alive, c) none of them will be alive, d) at least one of them will be alive, , The probability that a student X solves, 2, a problem in dynamics is, and the, 5, probability that student Y solves the same, 1, problem is . What is the probability that, 4, i) the problem is not solved, ii) the problem is solved, iii) the problem is solved exactly by one of, them, , 7), , A speaks truth in 80% of the cases and B, speaks truth in 60% of the cases. Find the, probability that they contradict each other in, narrating an incident., , 8), , Two hundred patients who had either Eye, surgery or Throat surgery were asked, whether they were satisfied or unsatisfied, regarding the result of their surgery., , 11) A box contains 10 red balls and 15 green, balls. Two balls are drawn in succession, without replacement. What is the probability, that,, a) the first is red and the second is green?, b) one is red and the other is green?, 12) A bag contains 3 yellow and 5 brown balls., Another bag contains 4 yellow and 6 brown, balls. If one ball is drawn from each bag,, what is the probability that,, a) both the balls are of the same color?, b) the balls are of different color?, , The follwoing table summarizes their, response., Surgery Satisfied Unsatisfied, , 13) An urn contains 4 black, 5 white and 6 red, balls. Two balls are drawn one after the other, without replacement. What is the probability, that at least one of them is black?, , Total, , Throat, , 70, , 25, , 95, , Eye, , 90, , 15, , 105, , Total, , 160, , 40, , 200, , 14) Three fair coins are tossed. What is the, probability of getting three heads given that, at least two coins show heads?, , If one person from the 200 patients is selected, at random, determine the probability, a) that the person was satisfied given that, the person had Throat surgery, b) that person was unsatisfied given that the, person had eye surgery, c) the person had Throat surgery given that, the person was unsatisfied, 9), , 15) Two cards are drawn one after the other from, a pack of 52 cards without replacement., What is the probability that both the cards, drawn are face cards?, 16) Bag A contains 3 red and 2 white balls and, bag B contains 2 red and 5 white balls. A bag, is selected at random, a ball is drawn and put, into the other bag, and then a ball is drawn, from that bag. Find the probability that both, the balls drawn are of same color., , Two dice are thrown together. Let A be the, event 'getting 6 on the first die' and B be the, event 'getting 2 on the second die'. Are the, events A and B independent?, , 17) (Activity) : A bag contains 3 red and 5 white, balls. Two balls are drawn at random one, after the other without replacement. Find the, probability that both the balls are white., , 10) The probability that a man who is 45 years, 5, old will be alive till he becomes 70 is ., 12, 206

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Solution : Let,, , Also, P(A ∩ Ei) = P(A).P(Ei/A), , A : First ball drawn is white, , i.e. P(Ei /A) =, , B : second ball drawn in white., , P( A ∩ Ei ), P( E ).P ( A / Ei ), = n i, P( A), ∑ P( A ∩ Ei ), i =1, , P(A) =, , Three types of probabilities occur in the above, formula P(Ei), P(A/Ei), P(Ei/A), , After drawing the first ball, without replacing, it into the bag a second ball is drawn from, the remaining, balls., , i), , ), , The probabilities occur in the above formula, P(Ei), i = 1, 2, 3, ... n are such that P(E1), + P(E2) + .... + P(En) = 1 are called prior, probabilities, since they are known before, conducting experiment., , 18) A family has two children. Find the, probability that both the children are girls,, given that atleast one of them is a girl., , ii) The probabilities P(A/Ei) tell us, how likely, the event A under consideration occurs,, given each and every prior probability. They, may refer to as likelihood probabilities of, the event A, given that event Ei has already, occurred., , ∴ P(B/A) =, ∴ P(Both balls are white) = P(A∩B), , , = P(, , , , =, , , , =, , ).P(, , iii) The conditional probabilities P(Ei/A), are called posterior probabilities, as they, obtained after conducting experiment., , 9.4 Bayes' Theorem:, (This is also known as Bayes' Law and, sometimes Bayes' Rule). This is a direct, application of conditional probabilities. Bayes', theorem is useful, to determine posterior, probabilities., , Bayes' theorem for n = 3 is explained in the, following figure., , Theorem : If E1, E2, E3 ... En are mutually, exclusive and exhaustive events with P(Ei) ≠, 0, where i = 1, 2, 3 ... n then for any arbitrary, event A which is a subset of the union of, events Ei such that P(A) > 0, we have, P(Ei / A) =, , P( A ∩ Ei ), P( E ).P ( A / Ei ), = n i, P( A), ∑ P( A ∩ Ei ), i =1, , Proof : We have A = (A ∩ E1) ∪ (A ∩ E2) ∪, (A ∩ E3) ... ∪ (A ∩ En), A ∩ E1, A ∩ E2, A ∩ E3 ... A ∩ En are, mutually exclusive events, So, P(A) = P[(A ∩ E1) ∪ (A ∩ E2) ∪, (A ∩ E3) ... ∪ (A ∩ En)], P(A) =, , n, , ∑, i =1, , P(A ∩ Ei), , Fig. 9.2 (a) & (b), 207

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0.5 and 0.8 respectively. If the bonus scheme has, been introduced, what is the probability that X is, appointed as the manger?, , SOLVED EXAMPLES, Ex. 1: A bag contains 6 red, 5blue balls and, another bag contains 5 red and 8 blue balls. A, ball is drawn from first bag and without noticing, colour is put in the second bag. A ball is drawn, from the second bag. Find the probability that, ball drawn is blue in colour., , Solution: Let E1: Person X becomes, manager, E2: Person Y becomes manager, Let E3: Person Z becomes manager, ∴P(E1) =, , Solution: Let event E1: Red ball is drawn, from the first bag and event E2: Blue ball is drawn, from the first bag., 5, ∴ P(E1) = 6/11 and P(E2) =, (Note that E1, 11, and E2 are mutually exclusive and exhaustive, events), , 4, 2, 3, ; P(E2) = ; P(E3) =, 9, 9, 9, , (Note that E1, E2 and E3 are mutually, exclusive and exhaustive events), Let event A: Bonus is introduced., ∴ P(A/E1) = P(Bonus is introduced under the, condition that person X becomes manager) = 0.3, P(A/E2) = P(Bonus is introduced under the, condition that person Y becomes manager) = 0.5, , Let event A: Blue ball is drawn from the, second bag, , and P(A/E3) = P(Bonus is introduced under, the condition that person Z becomes manager) =, 0.8, ∴ P(A) = P(A∩E1)+ P(A∩E2) + P(A∩E3), , ∴P(A/E1) = P (Blue ball is drawn from, the second under the condition that red ball is, 8, transferred from first bag to second bag) =, 14, , = P(E1)P(A/E1) + P(E2)P(A/E2) + P(E3), P(A/E3), , Similarly, P(A/E2) = P (Blue ball is drawn, from the second under the condition that blue ball, 9, is transferred from first bag to second bag) =, 14, , 4, 2, 3, =   (0.3) +   (0.5) +   (0.8), 9, 9, 9, =, , ∴ required probability = P(Blue ball is drawn, from the second bag), , 23, 45, , ∴ required probability = P(Person X becomes, manager under the condition that bonus scheme, is introduced), , ∴ P(A) = P(A∩E1) + P(A∩E2), , = P(E1)P(A/E1) + P(E2)P(A/E2), , = P(E1/A) = P(A∩E1) / P(A), ,  6  8   5  9 , =    +   ,  11   14   11   14 , , =, ,  48   45  93, =, , =,  154   154  154, , (2 / 15), (23 / 45), , =, , 6, 23, , Ex. 2: The chances of X, Y, Z becoming, managers of a certain company are 4:2:3. The, probabilities that the bonus scheme will be, introduced if X, Y, Z become managers are 0.3,, , Ex. 3: The members of the consulting firm, hire cars from three rental agencies, 60% from, agency X, 30% from agency Y and 10% from, agency Z. 9% of the cars from agency X need, 208

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repairs, 20% of the cars from agency Y need, repairs and 6% of the cars from agency Z need, repairs. If a rental car delivered to the consulting, firms needs repairs, what is the probability that it, came from rental agency Y?, Solution : If A is the event that the car needs, repairs and B, C, D are the events that the car, comes from rental agencies X, Y or Z. We have, P(B) = 0.6, P(C) = 0.3, P(D) = 0.1, P(A/B) = 0.09,, P(A/C) = 0.2 and P(A/D) = 0.06, , 3), , There is a working women's hostel in a, town, where 75% are from neighbouring, town. The rest all are from the same town., 48% of women who hail from the same, town are graduates and 83% of the women, who have come from the neighboring town, are also graduates. Find the probability that, a woman selected at random is a graduate, from the same town., , 4), , If E1 and E2 are equally likely, mutually, exclusive and exhaustive events and, P(A/E1) = 0.2, P(A/E2) = 0.3. Find P(E1/A)., , 5), , Jar I contains 5 white and 7 black balls. Jar, II contains 3 white and 12 black balls. A fair, coin is flipped; if it is Head, a ball is drawn, from Jar I, and if it is Tail, a ball is drawn, from Jar II. Suppose that this experiment is, done and a white ball was drawn. What is, the probability that this ball was in fact taken, from Jar II?, , 6), , A diagnostic test has a probability 0.95 of, giving a positive result when applied to a, person suffering from a certain disease, and, a probability 0.10 of giving a (false) positive, result when applied to a non-sufferer. It, is estimated that 0.5% of the population, are sufferers. Suppose that the test is now, administered to a person about whom we, have no relevant information relating to, the disease (apart from the fact that he/she, comes from this population). Calculate the, probability that:, , P(A) = P(A∩B) + P(A∩C) + P(A∩D), = P(B).P(A/B) + P(C).P(A/C) +, P(D).P(A/D), = 0.6 × 0.09 + 0.3 × 0.2 + 0.1 × 0.06, = 0.054 + 0.06 + 0.006, ∴ P(A) = 0.12, P(C/A) =, =, , P(A∩C) P(C).P(A/C), P(A), P(A) =, , 0.06, 0.3 × 0.2, 0.12 = 0.12, , = 0.5, EXERCISE 9.4, 1), , 2), , There are three bags, each containing 100, marbles. Bag 1 has 75 red and 25 blue, marbles. Bag 2 has 60 red and 40 blue marbles, and Bag 3 has 45 red and 55 blue marbles., One of the bags is chosen at random and a, marble is picked from the chosen bag. What, is the probability that the chosen marble is, red?, , a) given a positive result, the person is a, sufferer., b) given a negative result, the person is a, non-sufferer., , A box contains 2 blue and 3 pink balls and, another box contains 4 blue and 5 pink balls., One ball is drawn at random from one of the, two boxes and it is found to be pink. Find the, probability that it was drawn from (i) first, box (ii) second box., , 7), , 209, , A doctor is called to see a sick child. The, doctor has prior information that 80% of, the sick children in that area have the flu,, while the other 20% are sick with measles., Assume that there is no other disease in that, area. A well-known symptom of measles, is rash. From the past records, it is known

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that, chances of having rashes given that, sick child is suffering from measles is 0.95., However occasionally children with flu also, develop rash, whose chance are 0.08. Upon, examining the child, the doctor finds a rash., What is the probability that child is suffering, from measles?, 8), , chances of him being late to the office are, 1 1 1, ,, ,, respectively by Auto, Car and, 2 4 4, train. On one particular day he was late to the, office. Find the probability that he travelled, by car., Solution : Let A, C and T be the events that Mr., X goes to office by Auto, Car and Train, respectively. Let L be event that he is late., , 2% of the population have a certain blood, disease of a serious form: 10% have it in a, mild form; and 88% don't have it at all. A, new blood test is developed; the probability, , Given that P(A) =, , ,, , P(C) =, , 9 if the subject has the, of testing positive is 10, 6 if the subject has the mild, serious form, 10, 1 if the subject doesn't have the, form, and 10, , 9), , , P(B) =, , P(L/A) = 12 , P(L/B) =, , P(L/C) = 14, P(L) = P(A∩L) + P(C∩L) + P(T∩L), = P(A).P(L/A) + P(C).P(L/C) + P(T).P(L/T), +, , =, , disease. A subject is tested positive. What is, the probability that the subject has serious, form of the disease?, , =, , A box contains three coins: two fair coins, and one fake two-headed coin is picked, randomly from the box and tossed., a) What is the probability that it lands head, up?, b) If happens to be head, what is the, probability that it is the two-headed coin?, , P(L/C) =, , +, , =, , +, +, , P(A∩C) P(C).P(L/C), P(L), P(L) =, , =, =, 9.5 ODDS (Ratio of two complementary, probabilities):, , 10) There are three social media groups on a, mobile: Group I, Group II and Group III., The probabilities that Group I, Group II and, Group III sending the messages on sports are, 2 , 1 , and 2 respectively. The probability, 3, 5 2, of opening the messages by Group I,, Group II and Group III are 12 , 14 and 14, respectively. Randomly one of the messages, is opened and found a message on sports., What is the probability that the message was, from Group III., , Let n be number of distinct sample points, in the sample space S. Out of n sample points, m, sample points are favourable for the occurrence, of event A. Therefore remaining (n-m) sample, points are favourable for the occurrence of its, complementary event A'., n−m, ∴ P(A) = m, n and P(A') = n, , Ratio of number of favourable cases to, number of unfavourable cases is called as odds, m i.e., in favour of event A which is given by n−m, P(A):P(A'), , Ratio of number of unfavourable cases to, number of favourable cases is called as odds, against event A which is given by n−m, m i.e., P(A'):P(A), , 11) (Activity) : Mr. X goes to office byAuto, Car and, train. The probabilities him travelling by these, 2 3 2, modes are, ,, ,, respectively. The, 7 7 7, 210

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(1–p2) (1+p2) (1–p )3, 2, =, p2, 1, , SOLVED EXAMPLES, Ex. 1: A fair die is thrown. What are the odds, in favour of getting a number which is a perfect, square in uppermost face of die?, , ∴ p2 (1+p2) = 1–2p2 + p22, p2+p22 = 1–2p2 +p22, 3p2 = 1, , Soln.: Random experiment: A fair die is, thrown., ∴ Sample space S = {1, 2, 3, 4, 5, 6}, , ∴ p2 =, , 2, 1, p1 = (p2)2 =  1  =, 9, 3, 1, 1, ∴ P(A) =, and P(B) =, 9, 3, , ∴ n(S) = 6, , Let event A: die shows number which is a, perfect square., ∴A = {1, 4} ∴m = n(A) = 2, ∴A' = {2, 3, 5, 6} ∴ (n – m) = 4., , EXERCISE 9.5, , m, 2, 1, ∴P(A) =, =, =, n, 6, 3, , 4, 2, P(A') = n−m, n = 6 = 3, , 1), , If odds in favour of X solving a problem, are 4:3 and odds against Y solving the same, problem are 2:3. Find probability of:, i) X solving the problem, ii) Y solving the problem, , 2), , The odds against John solving a problem are, 4 to 3 and the odds in favor of Rafi solving, the same problem are 7 to 5. What is the, chance that the problem is solved when both, them try it?, , 3), , The odds against student X solving a, statistics problem are 8:6 and odds in favour, of student y solving the same problem are, 14:16. Find is the chance that, , ∴ Odds in favour of event, A = P(A): P(A') =, , 1/3, 1, =, 2/3, 2, , Ex. 2: The probability of one event A, happening is the square of the probability of, second event B, but the odds against the event A, are the cube of the odds against the event B. Find, the probability of each event., Solution : Let P(A) = p1 and P(B) = p2., ∴ probability on non-occurrence of the, events A and B are (1–p1) and (1–p2) respectively., We are given that p1 = (p2)2 .... (I), Odds against the event A =, , 1–p1, p1, , Odds against the event B =, , 1–p2, p2, , i) the problem will be solved if they try it, independently, ii) neither of them solves the problem, 4), , Since odds against the event A are the cube, of the odds against the event B., 1–p1  1− p 2 , =, , p1,  p2 , 1–p22 (1–p2)3, =, p22, p23, , 1, 3, , 3, , The odds against a husband who is 60 years, old, living till he is 85 are 7:5. The odds, against his wife who is now 56, living till, she is 81 are 5:3. Find the probability that, a) at least one of them will be alive 25 years, hence, b) exactly one of them will be alive 25 years, hence., , [By (I)], 211

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5), , 6), , 7), , There are three events A, B and C, one of, which must, and only one can happen. The, odds against the event A are 7:4 and odds, against event B are 5:3. Find the odds against, event C., , Multiplication theorem:, If A and B are any two events defined, on the same sample space S, then probability, of simultaneous occurrence of both events is, denoted by P(A∩B) and is given by P(A∩B) =, P(A)P(B/A), , In a single toss of a fair die, what are the, odds against the event that number 3 or 4, turns up?, , Independent events:, If the occurrence of any one event does not, depend on occurrence of other event, then two, events A and B are said to be independent., , The odds in favour of A winning a game of, chess against B are 3:2. If three games are to, be played, what are the odds in favour of A's, winning at least two games out of the three?, , i.e. if P(A/B) = P(A/B') = P(A), or P(B/A) = P(B/A') = P(B), then A and B are independent events., , Let's Remember, , ∴ P(A∩B) = P(A)P(B), , If A and B are independent events then, , A.N. Kolmogorov, a Russian mathematician, outlined an axiomatic definition of probability, that formed the basis of the modern theory. For, every event A of sample space S, we assign a, non-negative real number denoted by P(A) and is, called probability of A, which satisfied following, three axioms, , a) A and B' are also independent events, b) A' and B' are also independent events., Bayes' Theorem :, If E1, E2, E3 ... En are mutually exclusive and, exhaustive events with P(E1) ≠ 0, where i = 1,2,3, ... n. then for any arbitrary event A which is a, subset of the union of events E such that P(A) >, 0, we have, , 1) 0 ≤ P(A)≤ 1, 2) P(S) = 1, If A and B are mutually exclusive events,, then P(A∪B) = P(A) + P(B), , P(Ei/A) =, , Addition theorem:, , i =1, , If A and B are any two events defined on, the same sample space S, then probability of, occurrence of at least one event is denoted by, P(A∪B)and is given by P(A∪B) = P(A) + P(B), – P(A∩B), , MISCELLANEOUS EXERCISE - 9, , Conditional probability:, If A and B are any two events defined on the, same sample space S, then conditional probability, of event A given that event B has already occurred, is denoted by P(A/B), ∴ P(A/B) = P(A∩B) / P(B), P(B) ≠ 0, , Similarly, P(B/A) = P(A∩B) / P(A), P(A) ≠0, , P( A ∩ Ei ), P( E ).P ( A / Ei ), = n i, P( A), ∑ P( A ∩ Ei ), , 212, , I), , Select the correct answer from the given, four alternatives., , 1), , There are 5 girls and 2 boys, then the, probability that no two boys are sitting, together for a photograph is, 4, 2, 5, 1, B), C), D), A), 7, 7, 7, 21, , 2), , In a jar there are 5 black marbles and 3 green, marbles. Two marbles are picked randomly, one after the other without replacement.

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balls. One ball is drawn at random from one, of the bags and it is found to be red. The, probability that it was drawn from Bag II., 35, 34, 35, 33, B), C), D), A), 69, 67, 68, 68, , What is the possibility that both the marbles, are black?, 5, 5, 5, 5, A), B), C), D), 14, 8, 8, 16, 3), , Two dice are thrown simultaneously. Then, the probability of getting two numbers, whose product is even is, 1, 5, 1, 3, B), C), D), A), 4, 7, 2, 4, , 9), , 4), , In a set of 30 shirts, 17 are white and rest are, black. 4 white and 5 black shirts are tagged, as ‘PARTY WEAR’. If a shirt is chosen, at random from this set, the possibility of, choosing a black shirt or a ‘PARTY WEAR’, shirt is, 11, 13, 9, 17, A), B), C), D), 15, 30, 13, 30, , 5), , There are 2 shelves. One shelf has 5 Physics, and 3 Biology books and the other has 4, Physics and 2 Biology books. The probability, of drawing a Physics book is, 31, 9, 1, 9, B), C), D), A), 48, 38, 2, 14, , 6), , Two friends A and B apply for a job in the, same company. The chances of A getting, selected is 2/5 and that of B is 4/7. The, probability that both of them get selected is, 1, 8, 27, 34, B), C), D), A), 35, 35, 35, 35, , 7), , 8), , The probability that a student knows, the correct answer to a multiple choice, 2, question is, . If the student does not, 3, know the answer, then the student guesses, the answer. The probability of the guessed, 1, answer being correct is . Given that the, 4, student has answered the question correctly,, the probability that the student knows the, correct answer is, 6, 7, 8, 5, B), C), D), A), 7, 8, 9, 6, , A fair is tossed twice. What are the odds in, favour of getting 4, 5 or 6 on the first toss, and 1, 2, 3 or 4 on the second die?, A) 1 : 3, , B) 3 : 1, , C) 1 : 2, , D) 2 : 1, , 10) The odds against an event are 5:3 and the, odds in favour of another independent event, are 7:5. The probability that at least one of, the two events will occur is, 71, 69, 13, 52, B), C), D), A), 96, 96, 96, 96, II) Solve the following., 1), , The letters of the word 'EQUATION' are, arranged in a row. Find the probability, that a) All the vowels are together b), Arrangement starts wiht a vowel and ends, wiith a consonant., , 2), , There are 6 positive and 8 negative numbers., Four numbers are chosen at random, without, replacement, and multiplied. Find the, probability that the product is a positive, numbers., , 3), , Ten cards numbered 1 to 10 are placed in, a box, mixed up thoroughly and then one, card is drawn randomly. If it is known that, the number on the drawn card is more than, 3, what is the probability that it is and even, number?, 1, 1, If P(A∩B) = , P (B∩C) = , P(C∩A), 2, 3, 1, = then find P(A), P(B) and P(C), If A,B,C, 6, , 4), , are independent events., 5), , Bag I contains 3 red and 4 black balls while, another Bag II contains 5 red and 6 black, 213, , If the letters of the word 'REGULATIONS' be, arranged at random, what is the probability, that there will be exactly 4 letters between R, and E?

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6), , In how many ways can the letters of the, word ARRANGEMENTS be arranged?, , 13) Find the probability that a year selected will, have 53 Wednesdays., , a) Find the chance that an arrangement, chosen at random begins with the letters, EE., , 14) The chances of P, Q and R, getting selected, 2 2 1, as principal of a college are, ,, ,, 5 5 5, respectively. Their chances of introducing, 1 1 1, IT in the college are , ,, respectively., 2 3 4, Find the probability that, , b) Find the probability that the consonants, are together., 7), , 8), , 9), , A letter is taken at random from the letters, of the word 'ASSISTANT' and another letter, is taken at random from the letters of the, word 'STATISTICS'. Find probability that, the selected letters are the same., , a) IT is introduced in the college after one, of them is selected as a principal ., b) IT is introduced by Q., , A die is loaded in such a way that the, probability of the face with j dots turning up, is proportional to j for j = 1, 2, .......6. What, is the probability, in one roll of the die, that, an odd number of dots will turn up?, , 15) Suppose that five good fuses and two, defective ones have been mixed up. To find, the defective fuses, we test them one-byone, at random and without replacement., What is the probability that we are lucky and, fine both of the defective fuses in the first, two tests?, , An urn contains 5 red balls and 2 green, balls. A ball is drawn. If it's green a red ball, is added to the urn and if it's red a green ball, is added to the urn. (The original ball is not, returned to the urn). Then a second ball is, drawn. What is the probability the second, ball is red?, , 16) Fot three events A, B and C, we know, that A and C are independent, B and C are, independent, A and B are disjoint, P(A∪C), = 2/3, P(B∪C) = 3/4, P(A∪B∪C) = 11/12., Find P(A), P(B) and P(C)., , 10) The odds against A solving a certain problem, are 4 to 3 and the odds in favor of solving the, same problem are 7 to 5 find the probability, that the problem will be solved.,  A 1, 11) If P(A) = P   = , P, B 5, , 17) The ratio of Boys to Girls in a college is 3:2, and 3 girls out of 500 and 2 boys out of 50 of, that college are good singers. A good singer, is chosen what is the probability that the, chosen singer is a girl?, , B 1,   = 3 then find,  A, , 18) A and B throw a die alternatively till one, of them gets a 3 and wins the game. Find, the respective probabilities of winning., (Assuming A begins the game)., ,  B' ,  A' , (i) P   (ii) P  ' , A, B, 12) Let A and B be independent events with, 1, P(A) = , and P (A∪B) = 2P (B) − P(A)., 4, , 19) Consider independent trails consisting of, rolling a pair of fair dice, over and over What, is the probability that a sum of 5 appears, before sum of 7?, , Find a) P(B); b) P(A/B); and c) P(B'/A)., , 214

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20) A machine produces parts that are either, good (90%), slightly defective (2%), or, obviously defective (8%). Produced parts, get passed through an automatic inspection, machine, which is able to detect any part that, is obviously defective and discard it. What is, the quality of the parts that make it throught, the inspection machine and get shipped?, , 22) In a factory which manufactures bulbs,, machines A, B and C manufacture, respectively 25%, 35% and 40% of the, bulbs. Of their outputs, 5, 4 and 2 percent, are respectively defective bulbs. A bulbs is, drawn at random from the product and is, found to be defective. What is the probability, that it is manufactured by the machine B?, , 21) Given three identical boxes, I, II and III,, each containing two coins. In box I, both, coins are gold coins, in box II, both are silver, coins and in the box III, there is one gold and, one silver soin. A person chooses a box at, random and takes out a coin. If the coin is, of gold, what is the probability that the other, coin in the box is also of gold?, , 23) A family has two children. One of them is, chosen at random and found that the child is, a girl. Find the probability that, a) both the children are girls., b) both the children are girls given that at, least one of them is a girl., , 215

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ANSWERS, , 1. ANGLE AND IT'S MEASUREMENT, , 13) (i) 72° or, , Exercise : 1.1, 1), , (ii) 60° or, , (A) (i), (iii), (iv), (vi) are co-terminal., (B) (i) III (ii) III (iii) I (iv) I (v) III, , , , (vi) I (vii) IV (viii) I (ix) III (x) III, (i), , 17p, , (v), 3), , 36, , (ii), , 151p, 360, , 25p, 18, , (vi), , −11p, , (iii), , 15, , (iv), , 131p, 360, , (i) 183°42', , 5), , 25°,, , 6), , 30°,, , 7), , 40°, 50° and 90° that is, , 8), , 420° and 480°, , 9), , 30°, 70° and 80° that is, , 4, , (iv) 97°30', , 0, , (ii) 245°19'48", , p, , 14) (i) 85°, ,  900 , (i) 105° (ii) −300° (iii) , ,  π , 0,  −45 , (iv) 110° (v) ,  or 14°19'approx",  π , , 4), , 3, , 2p, 7, , 5p, , (5) 4 : 5, , 36, p, , and 135° or, , 9, , p, , 18, , and, , p, , ), , (9) 25 sq cm, , 2, , (iii) 162°30', , (v) 50°, , (vi) 115°, , 0, , or (34.40°), , 225  π, ,  − 1 sqcm, 4 3 , , (10) 160 sq cm, , MISCELLANEOUS EXERCISE - 1, ,, , 7p, , and, , 4p, , 6 18, 9, 5p, p p, 10) 20°, 60° and 100° that is, ,, and, 9, 9 3, , (I) (i) B (ii) B (iii) A (iv) D (v)D, , (vi) C, , (vii) B (viii) B (ix) A (x) C., π, , (II) (1) 8 (2) 49  − 1 sqcm, 2 , , 11) 40°, 60°, 140° and 120°, 12) 64°, 96°, and 128° that is, , 4, , (6) 4π cm and 10π sqcm, , (, , 5p, , 3p, , (ii) 100°, , (7) 18 π − 2 2 sqcm (8), ,, , 7, ,  108 , (1) 9π cm (2) 3π cm (3) , ,  π , approx (4) 4.4cm, , 5p, , 2p, , 3, , Exercise : 1.2, , (iii) 11°27'33, , 6, , 5, 2p, , and 120° or, , and (128.57)° or, , (iv) 45° or, , 51p, 225, , 5, p, , 3p, , and 108° or, , (iii) (51.43)° or, , (ii), (v) are non co-terminal., , 2), , 2p, , 32p, 16p 8p, ,, and, 45, 45 15, , 0, , (4) 35.7 cm, 216, , (3) 3π cm, ,  450 , 49 , (5),  (6) 13:22,  π 

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135p, , (7) 15π cm and, , 2, , 20p, , (10), , sq cm, , (11) 60°, 80°, 100°, 120° that is, , (9) 17°11'20", , p 4p 5p 2p, ,, ,, ,, 3 9 9 3, , 3, 2. TRIGONOMETRY - I, Exercise : 2.1, , (1), 0°, , 30°, , 45°, , 60°, , 150°, , 180°, , 210°, , sinθ, , 0, , 1, 2, , 1, 2, , 3, 2, , 1, 2, , 0, , 1, −2, , cosθ, , 1, , 3, 2, , 1, 2, , 1, 2, , tanθ, , 0, , 1, 3, , 1, , 3, , N.D., , 2, , 2, , 2, 3, , secθ, , 1, , 2, 3, , 2, , 2, , −, , cotθ, , N.D., , 3, , 1, , 1, 3, , -60°, , -90°, , θ, , cosecθ, , θ, sinθ, , -30°, −, , 3, 2, , cosθ, tanθ, , 1, 2, , −, , 1, 3, , -45°, −, , 1, 2, , −, , 3, 2, , −1, , 1, 2, , 1, 2, , 0, , −1, , − 3, , N.D., , cosec, , −2, , − 2, , sec, , 2, 3, , 2, , cot, , − 3, , −1, , −, , 2, 3, 2, , −, , 1, 3, , −1, , 300°, −, , 3, 2, , 3, 2, , −1, , −, , 1, 3, , 0, , 1, 3, , − 3, , N.D., , −2, , −, , 2, 2, 3, , 1, −2, , 1, 2, , −, , 3, 2, , −, , 330°, , 2, 3, , 2, 3, , 3, 2, −, , 1, 3, , −2, 2, 3, , −1, , −, , − 3, , N.D., , − 3, , −, , 1, 3, , − 3, , -120°, , -225°, , -240°, , -270°, , -315°, , 1, 2, , 3, 2, , 1, , 1, 2, , 1, 2, , 0, , 1, 2, , 3, 2, , −, −, , −, , 1, 2, , −, , 1, 2, , 3, , −1, , − 3, , N.D., , 1, , 2, 3, , 2, , 2, 3, , 1, , 2, , −2, , N.D., , 2, , 0, , 1, , N.D., , −2, , − 2, , 0, , 1, 3, , −1, , 217, , −, , 2, , −, , 1, 3

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(2) (i) Positive, , (ii) Positive, , (iii) Negative, , Exercise : 2.2, , (3) cos 4° ˃ cos 4c , cos 4° ˃ 0, cos 4° < 0, (4) (i) III, , (ii) III, (1), , 1+ 2, (5) (i), (ii) 1 + 2, (iii) 0, 2, 4, 4, 3, (6) sinθ = − , cos θ = , tan θ = − ,, 3, 5, 5, 5, 5, 3, cos ecθ = − , s ecθ = , cot θ = − ,, 3, 4, 4, (7) −, , 1, 2, , 2, , 2, , (5) cos θ = ±1, , 4, 3, , cot θ =, 3, 4, , 7, , (9) 1 or, , 24, 24, 7, , sin A =, , tan A = −, 25, 7, 25, , (6), , 25, , (, , (14) (i), , 4, 5, , tan x =, 3, 3, , (, , (7) 30°, , (8)60°, , (11) −8, , (ii) (−1, 0), , 0, (13) (i) 5 2 , 45, , (iii), , 1, 2, , 13, 12, , (10), , (12) (i) (0, 3), , 25, 7, cosec A =, , cot A = −, 24, 24, , (iv) sin x = −, , 2, , (v)  3 y − 5  −  2 x − 3  = 1,  3   4 , , (ii) 2, , sec x = −, , 8, 11, , x −5  y −3, (iv) ,  −,  =1,  6   8 , , 4, 3, 5, (iii) sin x = − , cos x = − , cosec x = −, 5, 5, 4, , , ), , 2, , (ii) cos A = −, , , 3+ 2, , (2) −5 (3), , (iii) x2 + y2 = 41, , 5, 5, 4, (9) (i) sin θ = − , cosecθ = − , secθ = −, 4, 3, 5, tan θ =, , ), , (4) (i) 16x2 − 9y2 = 144 (ii) 16x2 − 9y2 = 576, , 119 144, ,, 120 25, , (8) (i), , (, 3(, , 2 1+ 3, , ), , 2 , 2250, , 3, 2, , (ii) (2, 600), , ), , (iv) (2, 1500), , (ii), , 1, 2, , (iii), , 1, 3, , MISCELLANEOUS EXERCISE - 2, , 12, 5, , cos x =, ,, 13, 13, (I), , 12, 13, , cosec x = −, ,, 5, 5, 13, sce x =, 12, cot x = −, , 218, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , B, , A, , A, , B, , A, , B, , D, , C, , B, , B

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(II), 90°, , 120°, , sin, , 1, , 3, 2, , −, , 1, 2, , cos, , 0, , 1, 2, , −, , 1, 2, , tan, , N.D., , cosec, , 1, , 2, 3, , − 2, , sec, , N.D., , −2, , − 2, , −2, , N.D., , 2, , −2, , cot, , 0, , − 3, , 1, , 1, 3, , 0, , 1, , 1, 3, , −210°, , −300°, , −330°, , 1, 2, , 3, 2, , 1, 2, , 1, 2, , 3, 2, , 3. TRIGONOMETRY - II, , 1, 3, , Exercise : 3.1, , −, , 3, 2, , −, , 1, 3, , −, −, , 3, , 2, , 2, 3, , 2, , 2, 3, , − 3, , c, , 3, , −, , (2) (i) Positive, , 240°, −, −, , 1, , 2, 3, , 2, , (3) (i) IV, , 1, 3, , 225°, , −, , 270°, , 1, 2, , 3, 2, , −1, , 1, 2, , 0, , 1, 2, , 3, , N.D., , −1, , 2, 3, , −1, , − 2, , −, , -120°, 3, 2, , −, −, , −, , 1, 2, , -150°, −, −, , -180°, , 1, 2, , 0, , 3, 2, , −1, , 3, , 1, 3, , 0, , 2, 3, , −2, , N.D., , −, , 2, 3, , −1, , 3, , N.D., , 2 xy, x2 − y 2, (8) cos θ = 2, , tan θ =, x + y2, 2 xy, (9) −1, , 3 +1, 3 +1, 3 −1, (ii), (iii), (iv) 1, 1− 3, 2 2, 2 2, , Q.1 (i), Q.3 (i), , 33, 65, , (ii), , −16, 65, , (iii), , −33, 56, , Exercise : 3.2, , (ii) Negative (iii) Negative, , (ii) III, , 315°, , Q.1 (i) −, , (iii) II, , (4) sin1856 ˃ sin 2006, , 1, 2, , (iv) −, , (5) sin (−310°), , 219, , 1, 2, , (ii), (v) 1, , 1, 2, , (iii), (vi), , 1, 2, 1, 3

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(vii) −2, , (viii) − 2, , (ix), , Q.4 M11 =11, C11 = 11, M12 = 7, C12 = −7,, M13= −3, C13 = −3, , 2, 3, , M21 =−23, C21 = 23, M22 = −11, C22 = −11,, M23 = 19, C23 = −19, , (x) − 3, Exercise : 3.3, , Q.1 (i), , (ii), , M31 =−5, C31 = −5, M32 = −5, C32 = 5,, M33 = 5, C33 = 5, , 2 −1, 2− 2, OR, 2, 2 2, 2 +1, OR, 2 2, , Q.5 −28, Q.6 −2, Exercise : 4.2, , 2+ 2, 2, , Q.1 (i) 0, , −120 −119 120, ,, ,, Q.2, 169 169 119, , Q.5 (i) x = −, , 7π, π, + sin, 12, 12, (iii) cos 6θ + cos 2θ (iv) cos110° + cos40°, (ii) sin, , (2) C, , (6) B, , (7) C, , (3) D, , Q.1 (i) 1, 2, 3, , (4) C, , (5) C, , (8) B (9) A, , (10) A, , Q.3 (1) Consistent (ii) Not Consistent, (iii) Consistent, Q.4 (i) 16, , Q.3 x = 11, , , (ii) 2, , Q.5 (i) 16 sq. unit, (iii) 10 sq. unit, , (ii) 25 sq. unit, 8, , Q.6 21 sq. unit, , (iii) 46, , Q.7 1 or −5, , (iv) abc + 2fgh − af2 − bg2 − ch2, Q.2 (i) x = 0, x = −1, x = 2, , (iii) 2,2,−1, , 1, 1, ,, , 1., 2, 4, Q.2 3, 5, 7, , Exercise : 4.1, (ii) −10, , (ii) −5, 3, 4, , (iv) −, , 4. DETERMINANTS AND MARTICES, , Q.1 (i) −2, , (ii) x = 1 or 2 or 3., , Exercise : 4.3, , MISCELLANEOUS EXERCISE - 3, Q.1 (1) B, , 7, 3, , (iii) 0, , Q.6 x = 0 or 12, , Exercise : 3.4, Q.1 (i) sin6x + sin 2x, , (ii) 0, , Q.8 (i) Collinear, , (ii) x = −2, , (iii) Collinear, , y = 52 ., , 220, , (ii) Non - Collinear

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Exercise : 4.4, , MISCELLANEOUS EXERCISE - 4 (A), (I), 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , B, , B, , B, , B, , B, , C, , C, , D, , D, , C, , (II) Q.1, Q.2, , (i) −113, , 1, , 0 4,  8 27 ,  −2 −5, , , 1, , , 1, Q.1 (i) , 0  (ii)  −1 −4  (iii)  27 64 , , , 3, , 5,  64 125,  0 −3, , , 1, 2, , 2, , , (ii) −76, , −2, , Q.3 (i) 0 , , Q.2 (i) Upper triangular matrix, , (ii) 0, , (ii) Skew - symmetric matrix, , Q.4 (i) M11 =14, C11 = 14, M12 = −4, C12 = 4,, M13= 8, C13 = 8, , (iii) Column matrix, , M21 =16, C21 = −16, M22 = −2, C22 = −2,, M23 = 4, C23 = −4, , (v) scalar matrix, , (iv) row matrix, (vi) Lower triangular matrix, , M31 =−4, C31 = −4, M32 = 5, C32 = −5,, M33 = −1, C33 = −1, , (vii) diagonal matrix, (viii) symmetric matrix, , (ii) M11 =0, C11 = 0, M12 = 11, C12 = −11,, M13= 0, C13 = 0, , (ix) Identity matrix, (x) symmetric matrix, , M21 =−3, C21 = 3, M22 = −1, C22 = 1, M23, = 1, C23 = −1, , Q.3 (i) Singular, (iii) Non-Singular, , M31 =2, C31 = 2, M32 = −8, C32 = 8,M33 =, 3, C33 = 3, Q.5 (i) −, , 1, or 2, 3, , Q.9 (i) 1, 2, 1, (iv), Q.10 (i), , (ii), , Q.11 (i) 4 , , 2, 3, , (ii) 5, 25, (ii), 2, , −6, 7, , (iv) Non-Singular, , (ii) 6, , (iii), , 5 1 −1, Q.5 , , 3 2 0 , , (ii) 1, 2, 3 (iii) 1, 2, −1, , Q.12 (i) 0 or 8 , Q.13, , Q.4 (i), , 3 1, 7, , Q.6  −2 −4 1,  5 9 1, , 9, 3 1, , − ,, 2, 2 2, 1, , 3, , (ii) Singular, , (iii) 5, , Q.7 a = −4,, , 13, (iii), 2, , Q.8 x = −, , (ii) 1 or 34, , 32 sq. unit, , Q.14 `1750, `1500, `1750, 221, , b=, , 3, ,, 5, , 3, y = 5 i,, 2, , c = −7, z=, , 2, , 49, 8

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Q.9 (i) Symmetric, (ii) Neither Symmetric nor Skew Symmetric, , Q.10 a = 1,, , b = 0,, ,  0 −1 −2 , Q.10 A = 1 0 −1 Skew Symmetric matrix, , ,  2 1 0 , , Q.4 , , Exercise : 4.6, , Q.5, ,  −10 −1 1 , , , C =  7 −9 3 ,  −4 6 2 , , 8, − , 3,  , B=, 3 , , ,  1,  8, , − 1,  8, , 1, 4, , 1, 2 , , 10, , 0 − 3, , 0, 0, , , 16 , − , 3, , 5 , , , 1 0 , 0 1 , , , , Q.8 A − B is singular, Q.9 x = −, , 1, ,, 4, , y=, , (ii) [8], , Q.3 AB ≠ BA, , 2 , ,  −1 5 , , , 6 19 , , X=, 5, 5, , , 19 26 ,  5, 5 , , 14, , 3 − 3, Q.6 A = ,  −2, 1, , , Q.7, ,  6 −12 9 , Q.1 (i)  4 −8 6 ,  2 −4 3, , 5 4,  −3 23, , , , 1,  3,  8 − 4, X=  3 1  , Y=, −, ,  8 2 , , d=, , 9, 5, , (ii) Profit of suresh book shop on P, C, M, is ` 665, ` 705.50, `890.50 respectively. That, of Ganesh `700, `750, `1020 respectively., , Exercise : 4.5, , Q.3 , , 2, ,, 5, , Q.11(i) 1760, 2090,, , (iii) Skew Symmetric, , Q.2 , , c=, , 9, 2, , 222, , Q.8 , ,  −5 −15,  33 35 , , , , Q.10, , 10 10 4 , 25 39 2 , , , 35 7 22 , , Q.11, , α=1, , Q.13, , k = −7, , Q.17, , a = 2, b = −1, , Q.18, , 5, 3, X=  , 7,  3 , , Q.19, , K=1, , Q.20, , x = −5/3, , Q.21, , x = 19, y = 12, , Q.22, , x = −3, , , Q.24, , Jay `104 and Ram `150., , y = 1,, , z = −1

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 2 cos α, (ii)  0, ,  0, , Exercise : 4.7, 2, , (ii)  6,  1, , 1 −4 , Q.1 (i) , , 3 5 , , −4 , 0 , 5 , , 1  1 −2 2, 1  −5 10 6 , B= , , 0 25, 16  −4 0 − 5, , Q.5 α = 60° or, ,  −16 14 , Q.7 CT = , ,  −6 −10 , , 1 , ,  0, 4, 2, Q.12 (i) , +, 1 − 5 5,   2,  2, , p, 3, , Q.16 x = 2, y = 2, ,  35 −10 , , , (ii)  25 15 ,  −15 10 , , 8 , 8 , −18, ,  2 −1, Q.18 , ,  3 −2 , Q.19 (i) x = 7, y = −44, , −5 , 2 , , 0, , , (ii) x = 5, y = −1, , Q.20 (i) x = −9, y = −3, z = 0., (ii) x = 31, y = 53, z = 19.,  8 −7 , T, Q.21 AB = ,  AB=, −, 12, 22, , , ,  2 0 −4 ,  7 −2 6 , , , 15 −6 30 , , Q.25 (i) , , Kantaram, , T, , 1 − 5, 6,  0 5, 1, 1, , (ii) 1 − 4 − 4  +  −5 0, 2, 2,  −5 − 4 4 ,  −3 − 6, , 3, 6 , 0 , , Shantaram, , Rice  ` 33000, , ` 39000 , , Wheat  ` 28000, Groundnut  ` 24000, , ` 31500, , (ii) , , Kantaram, , , , MISCELLANEOUS EXERCISE - 4 (B), (I), 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , B, , C, , A, , D, , A, , C, , B, , A, , A, , C, , sin α, cos α, 0, , Shantaram, , Rice  ` 3000, , , Wheat  ` 2000, Groundnut  ` 0 , , (II) Q.1 (i) diag [−1 1 3] (ii) diag [23 −32 −18],  cos α, , Q.2 (i)  − sin α,  0, , 1, 7, , 1, 7 , , (ii) A = 16  4, , , both are skew symmetric., ,  7, , Q.8 (i)  −5,  12, , 0, 0 , 2 , ,  1,  7, B= ,  0, , , 1 4 −4 , Q.3 (i) A= , 7 0 4 , ,  0 2 4, , , AT =  −2 0 2 ,  −4 −2 0 , ,  0 −2 −4 , Q.2 A =  2 0 −2 ,  4 2 0 , , 0, 2 cos α, 0, , 0, 0 , 1 , 223, , , , ` 24000 , ` 3000 , , , , ` 1500 , ` 8000 

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5. STRAIGHT LINE, Exercise : 5.1, , 6., , m = 1, c = −1, , 7., , x+y–7=0, , 8., , a) 2x + y – 4 = 0, c) 2x + 4y – 13 = 0, , 1. 2x – 4y + 5 = 0, 9., , 2. 9x – y + 6 = 0, 3. 3x2 + 3y2 + 4x – 24y + 32 = 0, , a) 3,2, , b), , 2 3, ,, 3 2, , c) −6,4, , 10. x−y+2 = 0 , 3x − y = 0, , 4. x + y – 11x – 11y + 53 = 0, 2, , b) 2x – 5y + 14 = 0, , 2, , 11. x+y = 7 , 4x−3y = 0, , 5. 3x + 4y – 41 = 0, 6. x2 + y2 – 4x – 11y + 33 = 0, , 12. A : 5x+y−15=0 , B : 3x+4y−14=0,, C : 2x−3y−1 = 0, , 7. (a) (–1, 0), , (b) (0, 2), , 13. 9x+y+7 = 0, 8x + 22y−31 = 0 , 2x-4y+9 = 0, , 8. (a) (6,7), , (b) (4,6), , 5 4, 14.  , , 7 7, 15. 3x−4y = 25, , 9. (–3, 11), 10. (a) 3X – Y + 6 = 0, (b) X 2 + Y 2 + X + 4Y – 5 = 0, , Exercise : 5.4, , (c) XY = 0, (d) Y 2 – 4X = 0, , 1., , Exercise : 5.2, 1., , a) 2, , 2., , −3, 2, , 4, c) not defined. d) 0., 7, 1, 3., 4. 1, 5. 135°, 3, , 7., , –1, , 8. k = 1, , c) Slope −, , b), , 2., , 9. 45°, , Exercise : 5.3, 1., , a) y = 5 b) x = –5 c) y = –1 or y = 7, , 2., , a) y = 3 b) x = 4, , 3., , a) x = 2 b) y = –3, , 4., , a) 4x – y – 8 = 0 b) x = 2, , 5., , a) y =− 3 x, , 1, , intercepts 0, 2, , a) 2x – y – 4 = 0, , b) 0x + 1y – 4 = 0, , c) 2x + y – 4 = 0, , d) 2x – 3y + 0 = 0, , 4., , (1, −3), , 5. ±24, , 6. (1,2), , 7., , (1, –1), , 5 2, 8.  , , 3 3, , 9. (5,5), , 10. x + 3y = 3, 13., , 2, 5, , 14., , 11. 2, 25, 117, , 16. y + 2 = 0, , 12. 4, 15. (3, 1) and (–7, 11), , 17. 8x + 13y – 24 = 0, , 18. x – 3y + 5 = 0, , b) y = −3x, , c) x – 2y – 7 = 0, , 2, , X-intercept 3, Y-Intercept 2, 3, b) Slope 3, X-intercept 3, Y-Intercept –9, a) Slope −, , 19. 2x + y + 13 = 0,, , d) 2x – 3y + 9 = 0, , x – 9y + 73 = 0,, ,  −1 −10 , 11x – 4y – 52 = 0 ,  ,, ,  19 19 , 20. (2, 2), , e)− 3 x + y − −4 3 −3 = 0, f) 3x – y = 0, 224

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(1) (i) (1, −2); 3, , (ii) (3, 4); 7, , 8 6, (ii)  ,  ,, 5 5, , (iii) (3, 1), 4, , Exercise : 6.3, (1) (i) x = 3 cos q, y = 3 sin q, , (15) k = 8, , (16) 3x + 2y − 26 = 0, , (17) x − 2y = 5, , (18) x +, , (19) (−3, 0), , 3 y = 10, , (21) 2x + y ± 4 5 = 0, , (20) −61, , (ii) x = −1 + 3 cos q, y = 2 + 3 sin q, , (22) 3x + 2y ± 2 13 = 0, , (iii) x = 3 + 5 cos q, y = −4 + 5 sin q,, 5, 3, , 3x − 4y = 0, , (14) 7 , , (3) x2 + y2 − 4x − 6y − 12 = 0, , 2, 3, , y+4 = 0, , (13) (i) (2, −4),, , Exercise : 6.2, , (23) x − 5y ± 6 26 = 0, , 5, 3, , (2) x = + cosθ , y = −1 + sin θ , , (24) 3x − y − 27 = 0 and 3x − y + 13 = 0, , (3) 3x − 2y = 0 , , (25) x2 + y2 = 18, , (5) 4x − y − 18 = 0, , (26) (i) xy = 0, MISCELLANEOUS EXERCISE - 6, , (ii) 5y2 − 2xy = 5a2, , (iii) x2 − a2 = c(x2 − a2), , (I), 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , C, , C, , A, , C, , A, , C, , D, , C, , B, , A, , (II) (1)  1 , −1 , 17 , 2, , , , 2, , 7. CONIC SECTIONs, Exercise : 7.1, , (2) (3,2), 4, 1), , (3) x2 + y2 + 4x − 2y = 0, (4) x2 + y2 − 4x − 6y = 0, , 8  4 2, iii.  0, 2  , 3y + 2 = 0 , ,  ± , , 3  3 3,  3, , (6) 5x2 + 5y2 +34x + 8y −3 = 0, (8) x -, , 3, , y + 16 = 0, , iv. (0, –2), y – 2 = 0 , 8 , (± 4, –2), , (9) x2 + y2 = 50, (10) x + y − 4x + 6y −3 = 0, 2, , 2, , 2), 3), 4), 5), , (11) (i) x-intercept = 12 , r - intercept = 9, (ii) x-intercept = 9 , r - intercept = 15,  1 −13 , (12) (i)  ,,  , 3x − 4y − 11 = 0, 5 5 , (ii) (1, 2),, , 24  6 12 , 6, i.  , 0  , 5x + 6 = 0 ,, ,, ,± , 5  5, 5, 5 , ii. (–5, 0) , x – 5 = 0 , 20 , (–5, ± 10), , 6), , x + 3y − 7 = 0, 226, , v.  − 4 , 0  , 3x – 4 = 0,,  3 , 2, x = –20y, 3y2 = 16x, y2 = –28x, i) y2 = 36x, ii) y2 =, 9, 3, i) –, ii) –, 2, 2, , 16, ,, 3, , 9, x, 2, ,  4 8, − ,± ,  3 3

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7), 8), , 4 or 8,  1  10, i)  , 2  ,, 3  3, , 7  35, 7, ii)  ,   ,, 2 8, 2, , (vii) 3x2 + 5y2 = 32, , (16, 8), (16,–8), 18 units, 18 sq. units, (5, 0),  9, 13) (1, 2) , 1,  ,,  4, 4y – 7 = 0 ,, x=1, 14) i) x – y + 3 = 0 , 3x – 2y + 4 = 0, ii) 3x – y + 3 = 0 , 3x – 2y + 12 = 0, , (4) e =, , 1, 3, , (11) (i) y + 2 = 0, 8x − y − 18 = 0, (ii) y + 2 = 0, 6x + y = 16, (iii) 5x − y = 9, x + y = 3, (iv) 4x + 6y = ± 15 , (vi) 2x − y = ± 9 , (vii) 3x − 4y = ± 2 65, , 3, 2, ,, ,, 2 , 0), x = ±, 2, 3, , 2, 1, 3, , , ,, ,, ,0 x = ±, , 3, 2,  2 3 , , (12) x2 + y2 = 8 , , (13) x2 − xy −5 = 0, , (15) bx − ay = 0 , , (17) x + y = ± 5, , (18) 4 sq. units, , 1, 4, ,, 3, 3, , (v), , 29, , (v) x + y = ±, , 2 2 , 3 2., , (iii), , (8) (1, 2), , (10) k = ± 12 2, , 25 18, 25, (1) (a) 10, 6, (± 4, 0), x = ±, ;, , 8,, ., 4, 5, 2, (b) 4, 2 3 , (±10), x = ±4, 3, 2, 8., , 2, 2, (2) (i) x  y  1, 64 55, ,  16 9 , (7)  , ,  5 5 , , , 4 2,  1 ,, , 3  ., , , Exercise : 7.2, , 2, , 1,, 3, , 2 2, 3, , (9) The line is a tangent and point of contact, , 29, = 7.25cm, 4, , (d), , (3) e =, , (6) 4 sq. unit, , 15) k = 24, 17) x + 2y + 4 = 0, 18) y = –3x, , (c) 2 3 , 2, (±, , x2 y 2, , 1, 15 6, , x2 y 2, , 1, (ix), 9, 5, , 9), 10), 11), 12), , 19), , (viii), , Exercise : 7.4, x2 y 2, , 1, (ii), 25 9, , x2 y 2, , 1, 9, 8, , (iv), , x2 y 2, , 1, 25 16, , (vi), , (1) (i) 10, 8,, , x2 y 2, , 1, 72 64, , (ii) 8, 10,, , x2 y 2, , 1, 16 12, 227, , 41, , (±, 5, 41, , (0, ±, 4, , 41 , 0), x  , 41 ) y  , , 25 32, ,, 41 5, , 16, 25, ,, 41, 2

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(iii) 6, 8,, , 9 32, 5, , (± 5, 0). x = ± ,, 5 3, 3, , (iv) 4, 2 21 ,, , MISCELLANEOUS EXERCISE - 7, , 5, 4, , (± 5, 0). x = ± , 21., 2, 5, , 4, (v), , 4, 2 ,, 3, , (I), ,  4  x=± 1, , 4 3, ,0 ,, ±, 3, 3 , , , (, , ), , 2 , ±4 2 , 0 , x = ±2 2 , 8, , (vi) 8, 8,, , 34 , (0,± 34 ), y = ± 25 , 18, 34 5, 5, 25 288, 13, (viii) 10, 24,, , (0, ±13), y = ± 13 ,, ., 5, 5, (vii) 10,6,, , (ix) 20, 10,, , (3) e = 2, x 2 y32, −, =1, (ii), 16 9, , 10 x, y, x2 y 2, −, = 1 (iv) 9 − 36 = 1, 4 5, 2, , (v), , 2, , 2, , 2, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , A, , C, , A, , C, , A, , B, , C, , C, , B, , B, , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , 19, , 20, , C, , C, , B, , B, , B, , C, , B, , A, , C, , A, , 2), , i) (12,12) ii) (27,−18), , 4), , 3x + 4y + 12 = 0, , 5), , x−y+2=0, , 6), , 9x − 4y + 4 = 0, x − 4y + 36 = 0, , 8), , x + y + 2 = 0, (2,−4), , b), , v)6 vi), c), , 2 y=2, , ii) (±3,0), , d), , (iii) 5x − 6 3y = 30, (iv) 3 2 x − 4y = 12, , i) 24,10 ii)(±13, 0) iii) x = ±, , (8) ± 5, , i) 8,8, v) 8 2, , (v) 5x − 4y = 16, , 288, 13, , vi), , 228, , 144, 13, , 2, x2 y 2, +, ii), =1, 25 9, , iii) 3x2 + 5y2 = 32, , (10) 3x + 2y = ± 4, , 25, 32, iv), 3, 5, , iv), , 25, 6, , ii) ( ±4 2 ,0) iii) x = ±2 2 iv) 8, , x2 y 2, 14) i), =1, +, 64 55, , (9) x + y = ± 4, , iii) y = ±, , 50, 3, , v) 26 vi), , (ii) x − y = 1, , (7) (−6, −2), , 25, 2, i) 10,8, , 23, 18, iv), v) 8, 4, 5, , vi), , 2, , x2 y 2, 9x2 9 y 2, −, =1, −, = 1 (ix), (vii), 16 9, 16, 20, (6) (i) 3x −, , 3) (8,8) and (8,−8), , 13) a) i) 10,6 ii)(±4,0) iii) x =, , x, y, x, y, −, =1, −, = 1 (vi), 49, 9, 9 27, 2, , 3, ,  6, 24 ±12 6 , , , i)  0,  5y + 6 = 0,, ,, 5  5 5,  5, , (x) 4, 4 3 , 2, (±4, 0), x = ±1, 12., , (iii), , 2, , 17  17 17 ,  17 , (II) 1) i)  , 0  , 8x + 17 = 0,, , , , 2  8 4,  8 , , 20, 5, , (± 5 , 0) x = ±, ,5, 5, 2, , x2 y 2, −, = 1 , (2), 24 25, x2 y 2, (5) (i), −, =1, 4 21, , 1

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15) e = ±, , 1, 17) y+2=0 or 8x−y−18=0, 3, , 18) 2x + 3y = 25, , 5), , S.D. = 3.76, , 6), , (C.V.)P = 27.27;, , i) Worker P is more consistent., , 19) (1,2), , ii) Worker Q seems to be faster in completing, the job., , 20) x2 − xy − 5 = 0, x2 4 y 2, −, =1, 36 25, , 22) i), , iii), , ii), , 7), , x2 y 2, −, =1, 16 20, , ii) 10x − 3 3y = 15, , iii) 8x − 5y = 20 3, 25) y = 2x ± 4, , 8), , (C.V)A = 18.6; (C.V)B = 18.7, Series B is more variable, , 9), , (C.V)A = 80; (C.V)B = 74.5, Team B is more consistent., , 10) (C.V)M = 10; (C.V)S = 12, The subject Statistic shows higher vairablility, in marks., , 26) k(x2 − a2) = 2xy, 8. MEASURES OF DISPERSION, , MISCELLANEOUS EXERCISE - 8, , Exercise : 8.1, 38, , (C.V.)2 = 2.5, , ii) Second department has larger variability, in wages., , 23) i) 7x − 2y + 17 = 0, , 1), , (C.V.)1 = 1.07, , i) First department has larger bill, , x2 4 y 2, −, =1, 4, 9, , 24) (3,2), , (C.V.)Q = 33.33;, , 2) 717 3) 11 4) 5, , (I), 5) 10, , Exercise : 8.2, 1), , σ2 = 8; σ = 2.82, , 2), , σ = 380; σ = 19.49, , 3), , σ = 32.39; σ = 5.69, , 4), , σ2 = 4.026; σ = 2.006, , 5), , σ = 3.0275; σ = 1.74, , 6), , x = 58.2; σ = 653.76; σ = 25.56, , 7), , σ2x = 41.25; σx = 6.42, , 8), , 5 and 7, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , C, , A, , B, , D, , A, , C, , B, , B, , C, , B, , (II), , 2, 2, , 2, , 2, , Exercise : 8.3, , 1), , Range = 48, , 2), , Range = 89, , 3), , Range = Rs. 30, , 4), , Range = 60, , 5), , Variance = 7.44 , σ = 2.72, , 6), , Variance = 2000 , S. D. = 44.72, , 7), , S. D. = 1.35, , 8), , S. D. = 13.42, , 9), , S. D. = 16.85, , 10) A. M. = 72; S. D. = 12.2, , 1), , σc = 5.15, , 11) Mean = 19.15; S. D. = 4.66, , 2), , σc = 3.14, , 12) Mean = 41; S. D. = 7.1, , 3), , C.V. = 6.32, , 4), , C.V. = 20, , 13) Number of boys = 75, combined S. D. = 10.07, 229

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14) combined S. D. = 2.65, , (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),, (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),, (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}, , 15) C.V. = 26.65, 16) (C.V.)B = 6.67 (C.V.)G = 6.38, Series of boys is more variable, , a), , A : {(1, 2), (2, 1), (1, 3), (2, 2), (3, 1), (1, 5),, (2, 4), (3, 3), (4, 2), (5, 1), (2, 6), (3, 5),, (4, 4), (5, 3), (6, 2), (3, 6), (4, 5), (5, 4),, (6, 3), (6, 6), , 19) C.V. = 31.35, , b), , B : {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), , 20) (C.V.)x = 9.21; (C.V.)Y = 5.91, The variation is greater in the area of the, field., , c), , C : {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),, (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),, (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}, , 21) (C.V.)U = 37.67; (C.V.)V = 55.5, i) Company U gives higher average life, , d), , D : {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),, (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),, (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}, , e), , A and B are mutually exclusive but not, exhaustive., , f), , C and Dare mutually exclusive and, exhaustive., , 5), , a) S = {(5, 5), (5, 6), (5, 7), (5, 8), (6, 5),, (6, 6), (6, 7), (6, 8), (7, 5), (7, 6), (7, 7),, (7, 8), (8, 5), (8, 6), (8, 7), (8, 8)}, , b), , S = {(5, 6), (5, 7), (5, 8), (6, 5), (6, 7), (6, 8),, (7, 5), (7, 6), (7, 8), (8, 5), (8, 6),, (8, 7),}, a) 1, b) 5/12, c) 1/6, d) 1/9, 9, , 17) (C.V.)I = 22.22 (C.V.)II = 20.83, Brand-I is more variable, 18) C.V. = 29.76, , ii) Company U shows greater consistency in, performance., 22) (C.V.)1 = 15.50 (C.V.)2 = 19.96, Height shows more variability, , 9. PROBABILITY, Exercise : 9.1, 1), , 2), , S ={RR, GR, BR, PR, RG, GG, BG, PG,, RB, GB, BB, PB, RP, GP, BP, PP}, a) A = {RR, GR, RB, RP, GR, BR, PR}, , 6), , b) B = {RG, RB, RP, GR, GB, GP, BR, BG,, BP, PR, PG, PB}, , 7), , a), , 8, 221, , 8), , a), , 6, b) 997, 5525, 1700, , 9), , a) 1/2, , S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5),, (H, 6), (T, 1), (T, 2), (T, 3), (T, 4),, (T, 5), (T, 6), a) A = {(T, 1), (T, 3), (T, 5)}, b) B = (H, 2), (H, 3),(H, 5),(T, 2), (T, 3), (T, 5),, c) C = (H, 1), (H, 4),, , 3), , i) 56, , ii) 120, , iii) 720, , iv) 1140, , 4), , S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),, (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),, (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),, 230, , b), , 13, 102, , 12, 51, , c), , d), , c) 22, 425, , 25, 13, e), 102, 34, d) 16, 5525, , b) 1/2, , c)7/10, , 10) a) 4/25, , b) 8/75, , c) 7/25 d) 1/15, , 11) a) 2/7, , 12) i) 25/81, , 13) i) 1/6, , ii), , 5, 6, , ii) 5/18

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14) i) 1/3, 15), , ii) 2/3, , iii) 1/30, , iv) 4/15, , 6), , 4! 3, =, 16) 1/105 17) i) 7/33 ii) 14/55, 44 32, , 2/3, , 2) i) 1, , 3), , i) 0.85, , 4), , a) 22/75 b) 47/75, , 5), , 0.69, , 6) 5/18, , 7), , a) 1/4, , b) 3/8, , 8), , 1/2, , 9) m = 6, , 10) i) 7/33, , ii), , 21, 55, , 0.018, = 0.108, 0.166, 1, 2, 9) (a) Total Probability =, b), 2, 3, 20, 10), 59, 7), , iii) 0.15, , c) 3/4, 33, 50, , 11), , 1), , 2/7, , 2) 7/22, , 4), , i) 1/17, , ii) 1/16, , 5), , a) 17/64, , b) 3/64, , 6), , i) 9/20, , ii) 11/20 iii) 9/20, , 8), , a) 14/19 (0.733) b) 1/7 (0.143) c) 5/8(0.625), , 9), , Independent, b) 23/48, , 11) a) 1/4, , b) 1/2, , 12) a) 21/40, , b) 19/40, , 15) 11/221, , 95, = 0.748, 127, , 8), , Exercise : 9.5, , Exercise : 9.3, , 10) a) 5/32, , p (T / S ) P( S ) 0.8955, =, 1 − P(T ), 0.8958, , b) P(S'/T') =, , ii) 8/13, , ii) 0.74, , 0.00475, 0.10425, , a), , Exercise : 9.2, 1), , T = Test positive, S = Sufferer, P(T) = Total, probability = 0.10425, , 3) 1/9, , 1), , i), , 3, 5, , 4), , a), , 61, 96, , 7), , 81 : 44, , c) 61/64 d) 29/64, 7) 11/25, , ii) 25/52, , 3), , 16/99, , 4) 4/5, , 5) 12/37, , 6) 2:1, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , D, , A, , A, , D, , B, , C, , D, , D, , C, , B, , 2), , 505, 1001, , II) 1) a), , Exercise : 9.4, 2) i) 27/52, , 5) 65:23, , , , 1, 18), 3, , 0.60, , 23, 48, , b), , (I), , 13) 10/21 14) 1/4, , 1), , 73, 16, 32, 3) a), b), 105, 21, 105, , 2), , MISCELLANEOUS EXERCISE - 9, , 61, c) 35/96 d), 96, , 16) 901/1680, , 3, 5, , ii), , 4), , 1, 1, 2 , 1, 3, , a), , 1, , 66, , 9), , 32, 16, 10), 49, 21, , 12) a), 231, , 15, 1, b), 56, 14, , 2, 5, , 5), b), , b), , 6, 55, , 1, 99, , 1, 4, , 6) n(s) =, 7), , 19, 90, , 11) i), c), , 3, 5, , 4, 5, , 3), , 4, 7, , 12 !, , ( 2!), , 4, , 8), , 3, 7, , ii), , 2, 3, , 13), , 5, 28

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14) a), , 23, 60, , 16) P(A) =, , b), , 8, 23, , 15), , 1, 21, , 1, 1, 1, , P(B) = 2 , P(C) = 2, 3, , 232, , 17), , 1, 11, , 18) P(A win) =, , 19), , 2, 5, , 20), , 23), , 1, 2, , 90, 92, , 21), , 6, 5, , P(B win) =, 11, 11, 2, 3, , 22), , 28, 69