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8., , STRAIGHT LINE, , 1. INTRODUCTION, Co-ordinate geometry is the branch of mathematics which includes the study of different curves and figures by, ordered pairs of real numbers called Cartesian co-ordinates, representing lines & curves by algebraic equation. This, mathematical model is used in solving real world problems., , 2. CO-ORDINATE SYSTEM, Co-ordinate system is nothing but a reference system designed to locate position of any point or geometric, element in a plane of space., , 2.1 Cartesian Co-ordinates, Let us consider two perpendicular straight lines XOX’ and YOY’ passing through the origin, , y, , O in the plane. Then,, Axis of x: The horizontal line xox’ is called axis of x., Axis of y: The vertical line yoy’ is called axis of y., Co-ordinate axis: x-axis and y-axis together are called axis of co-ordinates or axis of, reference., , =90, x’, , x, , O, y’, Figure 8.1, , Origin: The point ‘O’ is called the origin of co-ordinates or just the origin., , Oblique axis: When xox’ and yoy’ are not at right angle, i.e. if the both axes are not perpendicular, to each other,, then axis of co-ordinates are called oblique axis., , 2.2 Co-ordinate of a Point, The ordered pair of perpendicular distances of a point from X- and Y-axes are called co-ordinates of that point., If the perpendicular algebraic distance of a point p from y-axis is x and from x-axis is y, then co-ordinates of the, point P is (x, y). Here,, (a) x is called x-co-ordinate or abscissa., (b) y is called y-co-ordinate or ordinate., (c), , x-co-ordinate of every point lying upon y-axis is zero., , (d) y-co-ordinate of every point lying upon x-axis is zero., (e) Co-ordinates of origin are (0, 0)., Note: A point whose abscissa and ordinate are both integers is known as lattice point.

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M a them a tics | 8.3, , ,  π, π, Illustration 2: Find the distance between P  2, −  and Q  3,  . , 6, ,  6, , Sol: The distance between two points =, PQ = r12 + r22 − 2r1r2 cos(θ1 − θ2 ) =, , (JEE MAIN), , r12 + r22 − 2r1r2 cos(θ1 − θ2 ) Therefore,, ., ,  π π, 4 + 9 − 2.2.3cos  − −  = =,  6 6, , π, 4 + 9 − 12cos   =, 3, , 1, 13 − 12. =, 2, , 7, , Illustration 3: The point whose abscissa is equal to its ordinate and which is equidistant from the points A(1, 0),, B(0, 3) is , (JEE MAIN), Sol: Given, abscissa = ordinate. Therefore distance can be found by considering the co-ordinates of required point, be P(k, k)., Now given PA = PB ⇒, , (k − 1)2 + k 2 =, , k 2 + (k − 3)2, , 2k2 – 2k + 1 = 2k2 – 6k + 9 ⇒ 4k = 8 ⇒ k = 2, , 4. SECTION FORMULA, Let R divide the two points P(x1, y1) and Q(x2, y2) internally in the ratio m:n., Let (x, y) be the co-ordinates of R., y, , Draw PM, QN, RK perpendicular to the x-axis., , Q, , Also, draw PE and RF perpendicular to RK and QN., Here,, , PR m, = ., RQ n, , x’, , Triangles PRE and RFQ are similar., ∴, , O, , PR PE, PE m, =, ⇒, =, RQ RF, RF n, , x − x1, , x2 − x, , =, , y − y1, , y2 − y, , Figure 8.4, , =, , my 2 + ny1, m, ⇒y=, The co-ordinates of R are, n, m+n, ,  mx2 + nx1 my 2 + ny1 , ,, , , m + n ,  m+n, , PR ' m, = , triangles PER’ and QR’F are similar., QR ' n, , y, F, , PR ', PE, =, R 'Q R 'F , , mx2 − nx1, m, , i.e., x =, m−n, x − x2 n, my 2 − ny1, Similarly, y =, ., m−n,  mx − nx1 my 2 − ny1 , ,, The co-ordinates of R’ are  2, , m − n ,  m−n, x − x1, , =, , R’, Q, , P, , But PE = x – x1 and R’F = x – x2., ∴, , N x, , ER m, =, FQ n, , If R’ divides PQ externally, so that, ∴, , M K, , mx2 + nx1, m, ⇒ x=, m+n, n, , In the same way,, i.e.,, , F, , y’, , But PE = x – x1 and RF = x2 – x., ∴, , R, P E, , x’, , M N, , O, y’, , Figure 8.5, , E, K, , x

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8 . 4 | Straight Line, , Alternate Method:, , PR ', m m, =, − = By changing n into –n in the co-ordinates of R, we can obtain the co-ordinates, R 'Q, n –n, , of R’:, , mx2 − nx1 my 2 − ny1, ,, m−n, m−n,  x + x 2 y1 + y 2 , ,, Cor. The mid-point joining the two points (x1, y1) and (x2, y2) is  1, , 2  .,  2,  x + λx2 y1 + λy 2 , Cor. From the above cor., the co-ordinates of a point dividing PQ in the ratio λ:1 are  1, ,,  . Considering, 1+ λ ,  1+ λ, λ as a variable parameter, i.e. of all values positive or negative, the co-ordinates of any point on the line joining the, points (x1, y1) and (x2, y2) can be expressed in the above forms., , 5. SPECIAL POINTS OF A TRIANGLE, 5.1 Centroid, , C(x3, y3), , Let the vertices of the triangle ABC be (x1, y1), (x2, y2) and (x3, y3), respectively.,  x + x3 y 2 + y 3 , ,, The mid-point D of BC is  2,  G, the centroid, divides AD internally, 2, 2 , , in the ratio 2:1., Let G be (x, y),, then x =, y=, , 2. ( (x2 + x3 ) / 2 ) + 1.x1, , 2+1, 2. ( (y 2 + y 3 ) / 2 ) + 1.y1, 2+1, , =, , =, , x1 + x2 + x3, 3, , y1 + y 2 + y 3, 3, , F, 2, (x1, y1)A, , 1 D, G, E, , B(x2, y2), , Figure 8.6, , and, ,  x + x 2 + x3 y1 + y 2 + y 3 , ∴ G is  1, ,,  ., 3, 3, , , , 5.2 Incentre, Let A (x1, y1), B (x2, y2), C (x3, y3) be the vertices of the triangle., Let AD bisect angle BAC and cut BC at D., , A, , BD AB c, We know that = =, DC AC b, Hence the co-ordinates of D are, , cx3 + bx2 cy 3 + by 2, , ,, c+b, c+b, , Let (x, y) be the incentre of the triangle, , y, , C, D, , B, BC b + c, ca AI AB, c, b+c, =, ∴, ∴ BD =, = =, =, Figure 8.7, DB, c, b + c ID BD ( ca / (b + c) ), a, (b + c) ( (cx3 + bx2 ) / (c + b) ) + ax1 ax1 + bx2 + cx3, =, x =, ,, b+c+a, a+b+c, (b + c) ( (cy 3 + by 2 ) / (c+ b) ) + ay1 ay1 + by 2 + cy 3, (b + c) ( (cx3 + bx2 ) / (c + b) ) + ax1 ax1 + bx2 + cx3, =, =, , y =, b+c+a, a+b+c, b+c+a, a+b+c, (b + c) ( (cy 3 + by 2 ) / (c+ b) ) + ay1 ay1 + by 2 + cy 3, =, b+c+a, a+b+c, , CD b, =, BD c, , ∴ x, , I

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8 . 6 | Straight Line, , HA = distance of orthocentre from vertex A = 2R cos A, ∴, (c), , AH, AG HG, =2=, =, ⇒ G divides line joining H and O in 2:1., OD, GD GO, In an isosceles triangle centroid, orthocentre, incentre, circumcentre lie on the same line., , 5.6 Nine-Point Circle, Nine-point circle can be constructed for any given triangle, and is so named because it touches nine significant, concyclic points throughout the triangle., These nine points are as follows:, ••, , Mid-point of each side of the triangle, , ••, , Mid-point of the line segment from each vertex of the triangle to the orthocentre., , ••, , Foot of each altitude, , L, , L, Figure 8.13, , MASTERJEE CONCEPTS, ••, , The centroid, incentre, orthocentre and circumcentre coincide in an equilateral triangle., , ••, , Orthocentre, centroid and circumcentre are always collinear, and centroid divides the line joining, orthocentre and circumcentre in the ratio 2:1., , ••, , In an isosceles triangle, centroid, orthocentre, incentre and circumcentre lie on the same line., , Saurabh Gupta (JEE 2010, AIR 443), , Illustration 4: If G be the centroid of the triangle ABC, prove that AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2). ,  (JEE MAIN), Sol: Distance formula of two points can be used to prove AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2)., In triangle ABC, let B be the origin and BC the x-axis. Let A be (h, k) and, , A(h, k), , a+h k , C be (a, 0). Then centroid G is , , .,  3 3, LHS, , = AB2 + BC2 + CA2 = (h – 0)2 + (k – 0)2 + a2 + (h – a)2 + (k – 0)2 , , = 2h2 + 2k2 + 2a2 – 2ah, , C (a, 0), , B, Figure 8.14

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8 . 8 | Straight Line, , Note:, (a) If area of the triangle is zero, then the three points are collinear., (b) The area of a polygon with vertices A1(xi, yi), i = 1, …, n (vertices taken in anti-clockwise order), 1, [(x y – x2y1) + (x2y3 – x3y2) +…+(xny1 – x1yn)], 2 1 2, , 6. LOCUS, Locus is a set of points which satisfies a given geometrical data. Thus, for example, locus of a point moving at a, constant distance from a given point is a circle. Locus of a point which is equidistance from two fixed points is a, perpendicular bisector of the line joining the two points., All the points in a locus can be represented by an equation. For example,, (a) If the distance of the point (x, y) from (2, 3) is 4, then, (x – 2)2 + (y – 3)2 = 42., i.e. , , x2 + y2 – 4x – 6y – 3 = 0., , This equation will represent a circle with its centre at (2, 3) and radius 4., (b) If (x, y) be the point equidistant from the points (3, 4) and (2, 1), then, (x – 3)2 + (y – 4)2 = (x – 2)2 + (y – 1)2, i.e., , x + 3y = 10., , From the geometrical constraint, which governs the motion, we can find a relation (locus) between the coordinates of the moving point in any of its positions. Equation of locus is therefore merely on equation, relating the x and y co-ordinates of every point on the locus., Steps to find locus, (i), , Assume the co-ordinates of point for which locus is to be determined as (h, k)., , (ii) Apply the given geometrical conditions., (iii) Transform the geometrical conditions into algebraic equation and simplify., (iv) Eliminate variables (if any)., (v) Replace h → x and k → y to get the equation of locus., Note:, ••, ••, , Locus should not contain any other variables except x and y., , The algebraic relation between x and y satisfied by the co-ordinates at every point, on the curve and not off the curve is called the equation of curve., , y, Illustration 6: Find the equation of locus of a point which moves so that its distance, from the point (0, 1) is twice the distance from x-axis. , (JEE MAIN), , P(x, y), , (0, 1), N, , Sol: Here we can obtain the equation of locus of given point by using given condition, and distance formula of two points., Let the co-ordinates of such a point be (x, y). Draw PM ⊥ to x-axis., Hence, PM = y, PN = 2PM (given), , x’, , M, , 0, y’, , Figure 8.16, , x

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M a them a ti cs | 8.11, , Therefore slope of L1 is, , m1= tan q1, , Slope of L2 is m2 = tanq2, Now in ∆ABC,, q1 + π – q2 + θ = π, θ = θ2 – θ1, tanθ = tan (q2 – q1) ⇒ tanθ =, Note:, (i), , tan θ2 − tan θ1, , 1 + tan θ2 . tan θ1, , ⇒ tan θ =, , m2 − m1, , 1 + m1m2, , , this gives the acute angle between lines., , If m1 = m2, then θ = 0º, i.e. lines are parallel or coincident., , (ii) If m1m2 = −1, then θ = 90 º , i.e. lines are perpendicular to each other., π, 1, and slope of one of the lines is . Find the slope of the other., Illustration 8: If the angle between two lines is, 2, 4, , (JEE MAIN), , Sol: We know that, tan θ =, Let m1 =, , m2 – m1, , 1+m1m2, , , where m1 and m2 are the slope of lines and θ is the angle between them., , m – (1 / 2), 1, π, π, m – (1 / 2), , m2 = m and θ = So, tan =, ⇒ 1 =±, ⇒ m = 3 or –(1/3), 1+(1 / 2)m, 2, 4, 4 1+(1 / 2)m, , Illustration 9: Line through the point (–2, 6) and (4, 8) is perpendicular to the line through the point (8, 12) and, (x, 24). Find the value of x. , (JEE MAIN), Sol: Given two lines are perpendicular to each other. Therefore, product of their slope will be -1., 8–6, 2 1, Slope of the line through the points (–2, 6) and (4, 8) is m1=, = =, 4 – (–2) 6 3, 24 – 12, 12, =, Slope of the line through the points (8, 12) and (x, 24) is m2 =, x–8, x–8, Since two lines are perpendicular m1m2 = – 1, ⇒, , 1 12, ×, = –1, 3 x–8, , 4, ⇒x=, , 7.3 Collinearity, C, B, A, , Figure 8.23, , If three points A, B, C are collinear, then, Slope of AB = Slope of BC = slope of AC

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8 . 1 2 | Straight Line, , MASTERJEE CONCEPTS, Collinearity of three given points:, Three given points A, B, C are collinear if any one of the following conditions is satisfied., ••, , Area of triangle ABC is zero., , ••, , Slope of AB = Slope of BC = Slope of AC., , ••, , AC = AB + BC., , ••, , Find the equation of the line passing through two given points, if the third point satisfies the equation of, the line, then three points are collinear., 1, If any one line is parallel to y-axis, then the angle between two straight lines is given by tan θ = ± ,, m, where m is the slope of other straight line., , A line of gradient m is equally inclined with the two lines of gradient m1 and m2., Then, , m1 – m, , 1+m1m, , =-, , m2 – m, , 1+m2m, , ., Aman Gour (JEE 2012, AIR 230), , 7.4 Equation of a line, (a) Point slope form: Suppose P0(x0, y0) is a fixed point on a non-vertical line L whose, , P(x, y), , slope is m. Let P(x, y) be an arbitrary point on L. Then by definition, the slope of L, y – y0, is given by m =, ⇒ y – y 0 = m(x – x0 ), x – x0, , P0(x0, y0), , This is called point slope form of a line., , Figure 8.24, , (b) Two point form: Let line L passes through two given points P1(x1, y1) and P2(x2, y2)., Let P(x, y) be a point on the line. So slope P1P = slope P1P2, ⇒, , ⇒, , y – y1, x – x1, , =, , y – y1 =, , y 2 – y1, , P2(x2, y2), , x2 – x1, , y 2 – y1, x2 – x1, , P(x, y), , (x – x1 ), , This is called two-point form of the line., (c), , P1(x1, y1), , Figure 8.25, , Slope intercept from: Case-I: If slope of line is m and makes y-intercept c, then equation is, (y – c) = m (x – 0), , ⇒, , y = mx + c, , Case-II: If slope of line is m and makes x-intercept d, then equation is, y = m(x – d), These equations are called slope intercept form., , (0, c), (d, 0), , Figure 8.26

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8 . 1 4 | Straight Line, , Illustration 10: A straight line is drawn through the point P(2, 3) and is inclined at an angle of 30° with positive, x-axis. Find the co-ordinate of two points on it at a distance 4 from P on either side of P. , (JEE MAIN), Sol: By using formula, The equation of line, , x – x1, cos θ, , =, , y – y1, sinθ, , = ±r ⇒, , x – x1, cos θ, , =, , y – y1, sinθ, , = r , we can obtain co-ordinates of point., , y –3, x–2, =, = ± 4 ⇒ x = 2 ± 2 3, y = 3 ± 2, cos 30° sin 30°, , (, , So, co-ordinate of two points are 2± 2 3 ,3± 2, , ), , Illustration 11: If two vertices of a triangle are (–2, 3) and (5, –1). Orthocentre of the triangle lies at the origin and, centroid on the line x + y = 7, then the third vertex lies at , (JEE MAIN), (A) (7, 4) , , (B) (8, 14), , (C) (12, 21) , , (D) None of these, , Sol: (D) The line passing through the third vertex and orthocentre must be perpendicular to line through (–2, 3), and (5, –1). Therefore, product of their slope will be -1., Given the two vertices B(–2, 3) and C(5, – 1); let H(0, 0) be the orthocentre; A(h, k) the third vertex., Then, the slope of the line through A and H is k/h, while the line through B and C has the slope, (–1 –3)/(5 + 2) = – 4/7. By the property of the orthocentre, these two lines must be perpendicular,,  k  4 , k 7, So we have    −  = – 1 ⇒ = , h 4, h, 7,  , , , Also, , … (i), , 5 − 2 + h −1 + 3 + k, +, = 7 ⇒ h + k = 16, 3, 3, , … (ii), , Which is not satisfied by the points given in (A), (B) or (C)., Illustration 12: In what direction should a line be drawn passing through point (1, 2) so that its intersection point, 6, with line x + y = 4 is at a distance of, units. , (JEE ADVANCED), 3, Sol: By using x = x1 + r cos θ and y = y1 + r sin θ, we can obtain the required angle., For co-ordinates of B, Substitute r=, , 6, 3, , ∴ x = 1+, , 6, 6, cos θ & =, y 2+, sin θ , 3, 3, , Substituting in x + y =4, ⇒ 1+, ∴, , 6, 6, cos θ + 2+, sin θ = 4, 3, 3, 1, 2, , cos θ +, , 3, sin θ =, 2, 2, , 1, , sin (45° + θ) = sin 60°, ∴ θ = 15°, or, sin (45° + θ)= sin 120° ∴ θ = 75°, , B, A(1,2), , , ∴ (cos θ + sin θ) =, (Multiple by, , 1, 2, , ), , 3, 6, Figure 8.29, , 6, 3

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Mathematics | 8.15, , Illustration 13: If sum of the distances of the points from two perpendicular lines in a plane is 1, then find its locus., (JEE ADVANCED), Sol: If (h, k) be any point on the locus, then |h| + |k| = 1, , Y, , Let the two perpendicular lines be taken as the co-ordinate axes., ⇒ locus of (h, k) is |x| + |y| = 1, , (0, 1), , This consists of four line segments which enclose a square as, , =1, , , , y, , shown in figure., , y=, , +, , x-, , x, , -1, , O, , (-1, 0), , (1, 0), , x, +, y, =1, , y, x-, , X, , =1, , (0, -1), Figure 8.30, , Illustration 14: If the circumcentre of a triangle lies at the origin and the centroid is the mid-point of the line, (JEE ADVANCED), joining the points (a2 + 1, a2 + 1) and (2a, –2a), then the orthocentre lies on the line. , (A) y = (a2 + 1)x, , (B) y = 2ax, , (C) x + y = 0 , , (D) (a – 1)2 x – (a + 1)2 y = 0, , Sol: (D) We know from geometry that the circumcentre, centroid and orthocentre of a triangle lie on a line. So the, ,  (a + 1)2 (a − 1)2 , ,, orthocentre of the triangle lies on the line joining the circumcentre (0, 0) and the centroid , ,  2, 2 , , 2, 2, (a + 1), (a − 1), y=, x or (a – 1)2 x – (a + 1)2 y = 0., 2, 2, , MASTERJEE CONCEPTS, Equation of parallel and perpendicular lines:, • Equation of a line which is parallel to ax + by + c = 0 is ax + by + k = 0., • Equation of a line which is perpendicular to ax + by + c = 0 is bx – ay + k = 0., • If y = m1x + c1, y = m1x + c2, y = m2x + d1 and y = m2x + d2 are sides of a parallelogram then its, area is, , (c1 – c2 )(d1 – d2 ), m1 – m2, , ., , y, x, + =2., x1 y1, c2, ., • Area of the triangle made by the line ax + by +c = 0 with the co-ordinate axes is, 2 | ab |, , • The equation of a line whose mid-point is (x1, y1) in between the axes is, , • A line passing through (x1, y1) and if the intercept between the axes is divided in the ratio m:n at this, point then the equation is, , nx my, +, = m+n., x1 y1, , • The equation of a straight line which makes a triangle with the co-ordinates axes whose centroid is, y, x, +, =1., (x1, y1) is, 3x1 3y1, B Rajiv Reddy (JEE 2012, AIR 11)

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8 . 1 6 | Straight Line, , 7.5 Foot of the Perpendicular, The foot of the perpendicular (h, k) from (x1, y1) to the line ax + by + c = 0 is given by, h – x1, , k – y1, , A(x1, y1), , ax+by+c=0, , –(ax1 +by1 + c), , ., a2 +b2, Hence, the co-ordinates of the foot of perpendicular is, a, , =, , b, , =, ,  b2 x – aby – ac a2 y – abx – bc , 1, 1, 1, 1, , ., ,, 2, 2, 2, 2, , , a, +b, a, +b, , , , B(h, k), , Figure 8.31, , The image of a point with respect to the line mirror: The image of A(x1, y1) with respect to the line mirror ax +, by + c = 0, B (h, k) is given by, , h – x1, a, , =, , k – y1, b, , =, , –2(ax1 +bx1 + c), a2 +b2, , Special Cases, , ., , (a) Image of the point P(x1, y1) with respect to x-axis is (x1, – y1)., (b) Image of the point P(x1, y1) with respect to y-axis is (– x1, y1)., (c), , Image of the point P(x1, y1) with respect to the line mirror y = x is Q(y1, x1)., , (d) Image of the point P(x, y) with respect to the origin is the point (-x, -y)., Illustration 15: Find the equation of the line which is at a distance 3 from the origin and the perpendicular from, the line makes an angle of 30° with the positive direction of the x-axis. , (JEE MAIN), Sol: By using xcosα + ysinα = P, we can solve this problem. Here α = 30° and P = 3., So equation is x cos 30° + y sin 30° = 3 x, , 3 y, + =3, 2 2, , ⇒, , 3x + y = 6, , Position of a point w.r.t. a line, Let the equation of the given line be ax + by + c = 0 and let the co-ordinates of the two given points be P(x1, y1), and Q(x2, y2). Let R1 be a point on the line., The co-ordinates of R1 which divides the line joining P and Q in the ratio m:n are, , m –ax1 – by1 – c, =, n, ax2 +by 2 + c, , ., , Thus, the two points (x1, y1) and (x2, y2) are on the same (or opposite) sides of the straight line ax + by + c = 0, m, whether Point R1 divides internally or externally or sign of, ., n, Note:, , ⇒ A point (x1, y1) will lie on the side of the origin relative to a line ax + by + c = 0, if ax1 +by1 + c and c have the, same sign., ⇒ A point (x1, y1) will lie on the opposite side of the origin relative to the line ax + by + c = 0, if, ax1 + by1 + c and c have the opposite sign., Illustration 16: For what values of the parameter α does the point M (α, α + 1) lies within the triangle ABC of, vertices A(0, 3), B(– 2, 0) and C(6, 1). , (JEE ADVANCED), , Sol:Here, the point M will be inside the triangle if and only if |Area ∆MBC| + |Area ∆MCA| + |Area ∆MAB| = |Area, ∆ABC|. And each individual area must be non-zero.

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M a them a ti cs | 8.17, , α α +1 1, 1, 1, –2, 0, 1 = 7α + 6 , Area MBC =, 2, 2, 6, 1, 1, , Area MCA =, , α α +1 1, 1, 1, 6, 1, 1 = –8α +12, 2, 2, 0, 3, 1, , Area MAB =, , α α +1 1, 1, 1, 0, 3, 1 = α+4, 2, 2, –2, 0, 1, , Figure 8.32, , 0 3 1, 1, 1, Area ABC =, –2 0 1 = .22, 2, 2, 6 1 1, The above equation has critical points – 4, –, For α ≤ –4, the equation is, , 6, 3, and ., 7, 2, , –7α – 6 – 8α + 12 – α – 4 =22, ⇒ α=–, , 5, 5, which is not a solution, since – > –4, 4, 4, , , 6, 6, For a∈  –4, –  , then equation is – 7α – 6 – 8α + 12 + α + 4 = 22 ⇒ α = –, 7, 7, , 6, which is solution of equation but area MBC = 0 ⇒ M lies on BC ⇒ α = – is not the desired value., 7,  6 3, For a∈  – ,  , the equation is 7α + 68α + 12 + α + 4 = 22.,  7 2, ,  6 3, ⇒ All α in the interval  – ,  satisfy the equation.,  7 2, 3, , 3, Finally over  , ∞  , we get α = implies area MCA become zero., 2, 2, , ,  6 3, ⇒ The desired values of α lie in the interval  – ,  .,  7 2, , 7.6 Length of the Perpendicular, , P(x1,y1), , The perpendicular distance ‘p’ of a point P(x1, y1) from the line , ax + by + c = 0 is, , p=, , p, , | ax1 +by1 + c |, a2 +b2, , M, , Figure 8.33, , (a) Distance between parallel lines: The distance between the parallel lines a, c1 – c2, a2 +b2, , passing through P(x’, y’) and making an angle α with the line y = mx + c, (where m = tan θ) are, , , , P(x’,y’), (, , (b) Lines making angle α with given line: The equations of the two straight lines, , y=mx+c, (, , x+ by + c1 = 0 and ax + by + c2 = 0 is, , ax+by+c, , Figure 8.34

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M a them a ti cs | 8.19, , 8. FAMILY OF LINES, Consider two intersecting lines L1: a1x + b1y + c1 = 0 and L2: a2x + b2y + c2 = 0, then, , Type-1: The equation of the family of lines passing through the intersection of the lines, L1 + λL2 = 0, ⇒, , (a1x +b1y + c1) + λ(a2x +b2y + c2) = 0 where λ is a parameter., , Type-2: Converse, L1 + λL2 = 0 is a line which passes through a fixed point, where L1 = 0 and L2=0 are fixed lines, and the fixed point is the intersection of L1 and L2., Type-3: Equation of AC ≡ u2u3 – u1u4 = 0 and BD ≡ u3u4 – u1u2= 0, , B, , u1a1x+b1y+c1=0, , u4a2x+b2y+d1=0, , u2a2x+b2y+d2=0, A, , C, , u3a1x+b1y+c2=0, , D, , Figure 8.38, , Note that second degree terms cancel and the equation u2u3 – u1u4 = 0 is satisfied by the co-ordinate points B, and D., Illustration 18: If a, b, c are in A.P., then prove that the variable line ax + by + c = 0 passes through a fixed point., , (JEE MAIN), Sol: By using given condition we can reduce ax + by + c = 0 to as L1 + λL2 = 0. Hence we can obtain co-ordinate, of fixed point by taking L1 = 0 and L2=0., 2b =a + c , , ⇒ c = 2b – a , , ⇒ ax + by + 2b – a = 0, , ∴ a (x – 1) + b(y + 2) = 0 This is of the form L1 + λL2 = 0, where b/a = l, ∴ Co-ordinates of fixed point is (1, – 2)., , 9. ANGULAR BISECTOR, 9.1 Bisectors of the Angle Between Two Lines, (a) Equations of the bisectors of angle between the lines ax + by + c = 0 and a1x + b1y + c1 = 0 are, ax +by + c, a2 +b2, , =±, , a1 x +b1 y + c1, a12 +b12, , (ab1 ≠ a1b), , (b) To discriminate between the bisectors of the angle containing the origin and that of angle not containing, the origin, rewrite the equations, ax + by + c = 0 and a’x+ b’y + c’ = 0 such that the terms c, c’ are positive,, ax +by + c, , a'x +b'y + c', , gives the equation of the bisector of the angle containing origin and, a +b, a'2 +b'2, ax +by + c, a'x +b'y + c', =–, gives the equation of the bisector of the angle not containing origin., 2, 2, a +b, a'2 +b'2, , then, , (c), , 2, , 2, , =+, , Acute angle bisector and obtuse angle bisector can be differentiated from the following methods:, Let two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 intersect such that constant terms are positive.

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8 . 2 0 | Straight Line, , If a1a2 + b1b2 < 0, then the angle between the lines that contain the origin is acute and the equation for, ax1 +by1 + c, a x +b2 y + c2, a x +b1 y + c, a x +b2 y + c2, is the, . Therefore, the acute angle bisector is 1, =– 2, =+ 2, 2, 2, 2, 2, 2, 2, a +b, a +b, a2 +b2, a22 +b22, , equation of other bisector. If, however, a1a2 + b1b2 > 0, then the angle between the lines containing the origin, is obtuse and the equation of the bisector of the obtuse angle is, a1 x +b1 y + c1, a12 +b12, , =–, , a2 x +b2 y + c2, a22 +b22, , is acute angle bisector., , a1 x +b1 y + c1, a12, , +b12, , =+, , a2 x +b2 y + c2, a22 +b22, , ; therefore, , (d) Few more methods of identifying an acute and obtuse angle bisectors are as follows:, Let L1 = 0 and L2 = 0 are the given lines and u1 = 0 and u2 = 0 are the bisectors, between L1 = 0 and L2 = 0. Take a point P on any one of the lines L1 = 0 or L2 = 0 and, draw a perpendicular on u1 = 0 and u2 = 0 as shown. If, |p| < |q| ⇒ u1 is the acute angle bisector., , L1=0, P p, q, , L2=0, , |p| > |q| ⇒ u1 is the obtuse angle bisector., |p| = |q| ⇒ the lines L1 and L2 are perpendicular., Note: The straight lines passing through P(x1, y1) and equally inclined with the lines, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are those which are parallel to the bisectors, between lines and passing through the point P., , u1=0, , u2=0, , Figure 8.39, , MASTERJEE CONCEPTS, (a) Algorithm to find the bisector of the angle containing the origin: Let the equations of the, two lines be a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. The following methods are used to find the, bisector of the angle containing the origin:, Step I: In the equations of two lines, check if the constant terms c1 and c2 are positive. If the terms are, negative, then make them positive by multiply both the sides of the equation by –1., Step II: Obtain the bisector corresponding to the positive sign, i.e. a1 x +b1 y + c1, a12, , +b12, , =, , a2 x +b2 y + c2, a22 +b22, , L1, , Acute bisector, , L2, Obtuse bisector, Figure: 8.40, , This is the required bisector of the angle containing the origin, i.e. the bisectors of the angle between the, lines which contain the origin within it., (b) Method to find acute angle bisector and obtuse angle bisector, (i) Make the constant term positive by multiplying the equation by –1., (ii) Now determine the sign of the expression a1a2 + b1b2.

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Mathematics | 8.21, , MASTERJEE CONCEPTS, (iii)If a1a2 + b1b2 > 0, then the bisector corresponding to ‘+ve’ and ‘–ve’ signs give the obtuse and acute, angle bisectors, respectively, between the lines., (iv)If a1a2 +b1b2 < 0, then the bisector corresponding to ‘+ve’ and ‘–ve’ signs give the acute and obtuse, angle bisectors, respectively., Both the bisectors are perpendicular to each other. If a1a2 + b1b2 > 0, then the origin lies in the, obtuse angle and if a1a2 + b1b2 < 0, then the origin lies in the acute angle., T P Varun (JEE 2012, AIR 64), , MASTERJEE CONCEPTS, , Incentre divides the angle bisectors in the ratios (b + c):a, (c + a):b and (a + b):c . Angle bisector divides, the opposite sides in the ratio of remaining sides. , , Figure: 8.41, , BD AB c, =, =, DC AC b, Aishwarya Karnawat (JEE 2012, AIR 839), , 9.2 Bisectors in Case of Triangle, Two possible models are as follows:, Case-I: When vertices of a triangle are known, compute the sides of the, , A(x1,y1), , triangle and the incentre. All the internal bisectors can be known, using the, co-ordinates of incentre and vertices of triangle., , Note: If the triangle is isosceles/equilateral, then one can easily get the, incentre., Case-II: When the equations of the sides are given,compute tan A, tan B,, tan C by arranging the lines in descending order of their slope. Compute the, acute/obtuse angle bisectors as the case may be. Plot the lines approximately, and bisectors containing or not containing the origin., , c, , (x2,y2)B, , b, , a, Figure 8.42, , C(x3,y3)

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8 . 2 2 | Straight Line, , Illustration 19: The line x + y =a meets the x- and y-axes at A and B, respectively. A triangle AMN is inscribed in, the triangle OAB, O being the origin, with right angle at N. M and N lie respectively on OB and AB. If the area of, the triangle AMN is 3/8 of the area of the triangle OAB, then AN/BN is equal to., (JEE ADVANCED), (A) 3 , , (B) 1/3 , , (C) 2 , , (D) 1/2, , Sol: (A) Here simply by using the formula of area of triangle, , , Y, , 1, i.e. {x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)} , we can solve the problem., 2, , Let, , AN, = λ . Then the co-ordinates of N are, BN, , B(0, a), N, ,  a, λa , ,, , ,, 1, +, λ, 1, +λ, , , M, , where (a, 0) and (0, a) are the co-ordinates of A and B, respectively., Now equation of MN perpendicular to AB is, , y–, , λa, a, =x–, , 1+ λ, 1+ λ, , or x – y =, , O, ,  λ –1 , 1–λ, a So the co-ordinates of M are  0,, a, 1+λ,  λ +1 , , A(a, 0), , X, , Figure 8.43, , Therefore, area of the triangle AMN is, , =, , 1, 2, ,   –a , λa2, 1 – λ 2, +, =, a, a,  , , , 2, (1 + λ )2,   λ + 1  (1 + λ ), , , Also area of the triangle OAB = a2/2., So that according to the given condition:, , λa2, 2, , (1 + λ ), , =, , 3 1 2, . a ⇒ 3l2 – 10λ + 3 = 0 ;, 8 2, , ⇒, , λ = 3 or λ = 1/3., , For λ = 1/3, M lies outside the segment OB and hence the required value of λ is 3., , 10. PAIR OF STRAIGHT LINES, 10.1 Pair of Straight Lines Through Origin, (a) A homogeneous equation of degree two of the type ax2 + 2hxy + by2 = 0 always represents a pair of straight, lines passing through the origin and if, (i) h2 > ab ⇒ lines are real and distinct., (ii) h2 = ab ⇒ lines are coincident., (iii) h2 < ab ⇒ lines are imaginary with real point of intersection, i.e. (0, 0), (b) If y = m1x and y = m2x be the two equations represented by ax2 + 2hxy + by2 = 0, then, , 2h, a, m1 + m2 =, –, and m1m2 =, b, b, Angle between two straight lines:, (c), , If θ is the acute angle between the pair of straight lines represents by ax2 + 2hxy + b, then, tan θ =, , 2 h2 – ab, a+b, , The condition that these lines are:, (i) At right angles to each other if a + b = 0, i.e. sum of coefficients of x2 and y2 is zero.

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Mathematics | 8.23, , (ii) Coincident if h2 = ab and (ax2 + 2hxy + by2) is a perfect square of ( ax + by)2 ., (iii) Equally inclined to the axis of x if h = 0, i.e. coefficient of xy =0., Combined equation of angle bisectors passing through origin: The combined equation of the bisectors of, , the angles between the lines ax2 + 2hxy + by 2 = 0 (a pair of straight lines passing through origin) is given by, x2 – y 2 xy, ., =, a–b, h, , 10.2 General Equation for Pair of Straight Lines, (a), , ax2 + 2hxy + by 2 + 2gx + 2fy + c = 0 represents a pair of straight lines if:, a h g, 2, 2, 2, abc + 2fgh – af – bg – ch = 0 , if h b f = 0, g f c, , (b) The slope of the two lines represented by a general equation is the same as that between the two lines, represented by only its homogeneous part., , 10.3 Homogenisation, The equation of the two lines joining the origin to the points of intersection of the line lx + my +n = 0 and the curve, ax2 + 2hxy + by 2 + 2gx + 2fy + c = 0 is obtained by homogenising the equation of the curve using the equation of, the line., The combined equation of pair of straight lines joining origin to the points, of intersection of the line given by lx + my + n = 0 , , y, , …. (i), , The second degree curve:, ax2 + 2hxy + by 2 + 2gx + 2fy + c = 0 , Using equation (i) and (ii), , lx+my+n=0, , …. (ii), 2, ,  lx + my ,  lx + my ,  lx + my , ax2 + 2hxy + by 2 + 2gx ,  + 2fy ,  + c,  =0 , –n, –n, , , , ,  –n , , O, , …. (iii), , x, Figure 8.44, ,  lx + my , Obtained by homogenizing (ii) with the help of (i), by writing (i) in the form: ,  = 1.,  –n , , MASTERJEE CONCEPTS, Through a point A on the x-axis, a straight line is drawn parallel to y-axis so as to meet the pair of straight, lines., ax2 + 2hxy + by 2 = 0 in B and C. If AB = BC, then 8h2 = 9ab., Krishan Mittal (Jee 2012, Air 199)

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8 . 2 4 | Straight Line, , Illustration 20: The orthocentre of the triangle formed by the lines xy = 0 and x + y = 1 is , (A) (1/2, 1/2) , , (B) (1/3, 1/3) , , (C) (0, 0) , , (JEE MAIN), , (D) (1/4, 1/4), , Sol: (C) Here the three lines are x = 0, y = 0 and x + y = 1., Since the triangle formed by the line x = 0, y = 0 and x + y = 1 is right angled, the orthocentre lies at the vertex, (0, 0), the point of intersection of the perpendicular lines x = 0 and y = 0., 0 then, Illustration 21: If θ is an angle between the lines given by the equation 6x2 + 5xy – 4y 2 + 7x + 13y – 3 =, equation of the line passing through the point of intersection of these lines and making an angle θ with the positive, x-axis is , (JEE ADVANCED), (A) 2x + 11y + 13 = 0 , , (B) 11x – 2y + 13 = 0, , (C) 2x – 11y + 2 = 0 , , (D) 11x + 2y – 11 = 0, , Sol: (B) By taking the term y constant and using the formula of roots of quadratic equation, we can get the, equation of two lines represented by the given equation and then by using tan θ =, required result., , 2 h2 – ab, , we will get the, a+b, , Writing the given equation as a quadratic in x, we have, 6x2 + (5y + 7) x – (4y 2 – 13y + 3) =, 0⇒ x=, =, , –(5y + 7) ± (5y + 7)2 + 24(4y 2 – 13y + 3), 12, , 2, , –(5y + 7) ± 121y – 242y + 121, –(5y + 7) ± 11(y – 1) 6y – 18, –16y + 4, or, = =, 12, 12, 12, 12, , ⇒ 2x – y + 3 = 0 and 3x + 4y – 1 = 0 are the two lines represented by the given equation and the point of, intersection is (– 1, 1), obtained by solving these equations., Also tan θ =, , 2 h2 – ab, 2 (5 / 2)2 – 6(–4), , where a = 6, b = – 4, h = 5/2 =, =, a+b, 6–4, , So the equation of the required line is y=, –1, , 121 11, =, 4, 2, , 11, (x + 1) ⇒ 11x – 2y + 13 = 0, 2, , Illustration 22: If the equation of the pair of straight lines passing through the point (1, 1), and making an angle, θ with the positive direction of x-axis and the other making the same angle with the positive direction of y-axis is, x2 – (a + 2)xy + y 2 + a(x + y – 1) =, 0 , a ≠ – 2, then the value of sin 2θ is , (JEE ADVANCED), (A) a – 2 , , (B) a + 2 , , (C), , 2, 2, (D), a, (a + 2), , Sol: (C) As both line passes through (1, 1) and one line makes angle θ with x-axis and other line with y–axis, slopes, of line are tan θ and cot θ, Equations of the given lines are y – 1 = tan θ (x – 1) and y – 1 = cot θ (x – 1), So, their combined equation is [(y – 1) – tan θ (x – 1)] [(y – 1) – cot θ (x – 1) ] = 0, ⇒ (y – 1)2 – (tan θ + cot θ) (x – 1) (y – 1) + (x – 1)2 = 0, ⇒ x2 – (tan θ + cot θ) xy + y2 + (tan θ + cot θ – 2) (x + y – 1) = 0, Comparing with the given equation we get tan θ + cot θ = a + 2, ⇒, , 1, 2, = a + 2 ⇒ sin2θ =, sin θ cos θ, a+2

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Mathematics | 8.25, , Illustration 23: If two of the lines represented by x 4 + x3 y + cx2 y 2 – xy 3 + y 4 = 0 bisect the angle between the, other two, then the value c is, (JEE ADVANCED), (A) 0 , , (B) –1 , , (C) 1 , , (D) – 6, , Sol: (D) As the product of the slopes of the four lines represented by the given equation is 1 and a pair of line, represents the bisectors of the angles between the other two, the product of the slopes of each pair is –1., So let the equation of one pair be ax2 + 2hxy – ay2 = 0., The equation of its bisectors is, , x2 – y 2 xy, ., =, 2a, h, , By hypothesis x 4 + x3 y + cx2 y 2 – xy 3 + y 4, ≡ (ax2 + 2hxy – ay2) (hx2 – 2axy – hy2) = ah(x4 + y4) + 2(h2 – a2) (x3y – xy3) – 6ahx2y2, Comparing the respective coefficients, we get ah =1 and c = –6ah = –6, , 11. TRANSLATION AND ROTATION OF AXES, , Y, , Y’, P, , 11.1 Translation of Axes, Let OX and OY be the original axes, and let the new axes, parallel to original axes, be O’X’, and O’Y’. Let the co-ordinates of the new origin O’ referred to the original axes be (h, k). If, the point P has co-ordinates (x, y) and (x’, y’) with respect to original and new axes,, respectively, then x = x’ + h; y = y’ + k, , X’, , O’, , X, , O, , Figure 8.45, , 11.2 Rotation of Axes, Let OX and OY be the original system of axes and let OX’ and OY’ be the new system, of axes and angle XOX’ = θ (the angle through which the axes are turned). If the point, P has co-ordinates (x, y) and (x’, y’) with respect to original and new axes, respectively,, then, x = x’ cos θ – y’ sin θ and y = x’ sin θ + y’ cos θ, , Y, P, , Y’, , , O, , in matrix form it is as follows:, , X’, , X, , Figure 8.46, ,  x  cos θ − sin θ   x' ,  =,  ,  y   sin θ cos θ   y ', , MASTERJEE CONCEPTS, If origin is shifted to point (α, β) , then new equation of curve can be obtained by putting x + α in place, of x and y + β in placed of y., Vaibhav Krishnan (JEE 2009, AIR 22), , Illustration 24: The line L has intercepts a and b on the co-ordinate axes. The co-ordinate axes are rotated through, a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then, , 1, 2, , a, , –, , 1, p, , (A) –1, , 2, , +, , 1, b, , 2, , –, , 1, q2, , (JEE MAIN), , is equal to , (B) 0 , , (C) 1 , , (D) None of these

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M a them a ti cs | 8.27, , PROBLEM-SOLVING TACTICS, (a) In most of the questions involving figures like triangle or any parallelogram, taking origin as (0,0) helps a lot, in arriving at desired solution. One must ensure that conditions given are not violated., (b) One must remember that in an isosceles triangle, centroid, orthocentre, incentre and circumcentre lie on the, same line., (c), , The centroid, incentre, orthocentre and circumcentre coincide in an equilateral triangle., , (d) If area of the triangle is zero, then the three points are collinear., (e) Find the equation of the line passing through two given points, if the third point satisfies the equation of the, line, then three points are collinear, (f), , Whenever origin is shifted to a new point (α, β), then new equation can be obtained by putting x + α in place, of x and y + β in placed of y., , FORMULAE SHEET, (a) Distance Formula: The distance between two points P(x1, y1) and Q(x2, y2) is, PQ =, , (x1 – x2 )2 + (y1 – y 2 )2 =, , (x1 – x2 )2 + (y1 – y 2 )2, , And between two polar co-ordinate A(r1, q1) and B(r2, q2) is AB =, , r12 + r22 − 2r1r2 cos(θ1 − θ2 ), , (b) Section Formula: If P(x1, y1), Q(x2, y2) and the point R(x, y) divide the line PQ internally in the ratio m:n then, the co-ordinates of R will be, , x=, , mx2 + nx1, m+n, , and y =, , my 2 − ny1, m−n, ,  mx + nx1 my 2 + ny1 , ,, i.e. R  2, , m + n , ,,  m+n, ,  x + x 2 y1 + y 2 , ,, And if R is a mid-point of line PQ, then the co-ordinates of R will be  1, , 2 ,  2, (c), , Centroid of Triangle: If A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of the triangle ABC and G is,  x + x 2 + x3 y1 + y 2 + y 3 , Centroid, then co-ordinate of G will be  1, ,,  ., , 3, 3, , , , ax1 + bx2 + cx3, ay1 + by 2 + cy 3, (d) Co-ordinates, of Incentre: x =, =, , y, a+b+c, a+b+c, (e) Co-ordinates of Ex-centre: As shown in figure, ex-centres of ∆ABC, , A, I3, , I2, C, , B, , I1, , with respect to vertices A, B and C are denoted by , I1, I2 and I3, respectively,,  −ax1 + bx2 + cx3 −ay1 + by 2 + cy 3 , I1 = , ,,  ;, −a + b + c, −a + b + c, , ,  ax + bx2 − cx3 ay1 + by 2 − cy 3 , ,, I3 =  1, , a+b−c, a+b−c, , , , Figure 8.47, ,  ax1 – bx2 + cx3 ay1 − by 2 + cy 3 , ,,  ,, I2 = , a−b + c, a−b + c, , 

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M a them a ti cs | 8.29, , Solved Examples, JEE Main/Boards, , =, , Example 1: Find the ratio in which y – x + 2 = 0 divides, the line joining A (3, – 1) and B (8, 9)., , 20 2 – 10 2 + 30 2 5, =, 2, 5 2 +5 2 + 6 2, y=, , ay1 + by 2 + cy 3, , a+b+c, Sol: By considering the required ratio be λ:1, and using, section formula, we can solve above problem., –10 2 + 20 2 + 30 2 5, = =, 2, 5 2 +5 2 + 6 2, The point of division P is internal as A and B lie on, opposite sides of given line., Example 3: A rectangle PQRS has its side PQ parallel to, Let required ratio be λ:1., the line y = mx and vertices P, Q, S lie on lines y = a, x = b,  8λ + 3 9λ – 1 , and x = –b, respectively. Find the locus of the vertex R., Since, P , ,,  lies on y – x + 2 = 0,,  λ +1 λ +1 , Sol: Here sides PQ and QR must be perpendicular to, 9λ – 1 8λ + 3, 2, each other. Therefore product of their slopes will be –1., +2 =, –, 0 or λ =, ∴, λ +1, λ +1, 3, Let R(h, k) be any point on the locus and let S and Q, y-x+2=0, have co-ordinates (–b, β) and (b, α), respectively, as T is, mid-point of SQ and PR., 1, , α–a, A, =m, Thus P has co-ordinates (–h, a), B, P, b+h, ⇒ α = a + m (b + h), –, , Hence, required ratio is 2/3:1 or 2:3, Example 2: Find the incentre I of ∆ABC, if A is (4, – 2) B, is (– 2, 4) and C is (5, 5)., ax1 + bx2 + cx3, ay1 + by 2 + cy 3, =, ,, , y, a+b+c, a+b+c, we can obtain the incentre., , =, Sol: Using x, , a = BC =, , (5 + 2)2 + (5 – 4)2 =, 5 2, 2, , (5 – 4) + (5 + 2) = 5 2, , c = AB =, , (–2 – 4)2 + (4 + 2)2 = 6 2, , ⇒α=k–, , a + m(b + h) = k –, ∴ locus of R is, , x(m2 – 1) 2 my + b + am + bm2 = 0., , ax +by + c, a2 +b2, , If incentre I is (x, y) , then,, x=, , ax1 + bx2 + cx3, a+b+c, , 1, (b – h), m, , Sol: For isosceles triangle ABC, AD is perpendicular, bisector of side BC and it also bisects angle BAC. Hence, by using equation of bisector formula, i.e., , A, , B, , 1, (b – h), m, , Example 4 Two equal sides AB and AC of an isosceles, triangle ABC have equation 7x – y + 3 = 0 and x + y – 3, = 0, respectively. The third side BC of the triangle passes, through point P(1, – 10). Find the equation of BC., , 2, , b = CA =, , α –k, 1, = slope of QR =, b–h, m, , C, , =±, , a1 x +b1 y + c1, a12 +b12, , ,, , we can obtain slope of AD., Equations of AB and AC are 7x –y + 3 = 0 and, – x – y + 3 = 0, respectively., a1 = 7; b1 = – 1, c1 = 3;

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M a them a ti cs | 8.31, , ,  5k + 1, , 3k – 5, 5k + 1 , ±4, (7) +  –2 –, – 5 =,  + 7, k +1, k +1 , ,  k +1, , , 4k + 4,10k =, 70,k =, 7, ⇒ 14k – 66 =, or 14k – 66 = – 4k – 4, 18k = 62,, k = (31/9)., Therefore, value of the k = 7, 31/9, Example 9: Prove that the sum of the reciprocals of the, intercepts made on the co-ordinate axes by any line, not passing through the origin and through the point, of intersection of the lines 2x + 3y = 6 and 3x + 2y = 6, is constant., Sol: Equation of any line through the points of, intersection of the given lines is L1 + λL2=0., , JEE Advanced/Boards, Example 1: If A(–1, 5), B(3,1) and C(5, 7) are vertices of, a ∆ABC and D, E, F are the mid-points of BC, CA and AB,, respectively, then show that area ∆ABC = 4 times area, (∆DEF)., Sol: Co-ordinates of D, E and F are first obtained by, using mid-point formula, and prove the given equation, by using formula of area of triangle., Co-ordinates of D, E, F are (4, 4), (2, 6) and (1, 3),, respectively., –1 5 1, 1, ∴ Area of ∆ABC =, 3 1 1 = 16, 2, 5 7 1, , 2x + 3y – 6 + k (3x + 2y – 6) =, 0, , Area of ∆DEF =, , (2 + 3k) x + (3 + 2k)y – 6(k + 1) =, 0, ⇒, , y, x, +, =, 1, ((6(k + 1)) / (2 + 3k)) ((6(k + 1)) / (3 + 2k)), , A, , and in this case, sum of the reciprocals of the intercepts, made by this line on the co-ordinate axis is equal to, , However, for k = – 1, the line become, x = y, which passes through the origin., Example 10: Find the straight lines represented by, 6x2+ 13xy + 6y2 +8x + 7y + 2 = 0 and also find their, point of intersection., Sol: Taking term y as a constant and using quadratic, 2, , −b ± b − 4ac, , we can obtain, 2a, equations of required straight lines and after that by, , roots formula, i.e. x =, , B, , D, , C, , Hence, area of ∆ABC = 4 area (∆DEF), , 2 + 3k + 3 + 2k 5(k + 1) 5, ., = =, 6(k + 1), 6(k + 1) 6, , solving them we will get their point of intersection., Rewrite the given equation as, x=, , E, , F, , Where k ≠ – 1, , –(13y + 8) ± (13y + 8)2 – 24(6y 2 + 7y + 2), 12, , –(2y + 1) –(3y + 2), –(13y + 8) ± (5y + 4), =, ,, 3, 2, 12, Hence, 3x + 2y + 1 = 0 and 2x + 3y + 2 = 0, 1 4 , are the required lines and they intersect at  , –  ., 5 5 , =, , 4 4 1, 1, 2 6 1 =4, 2, 1 3 1, , Example 2: Point P(a2, a + 1) is a point of the angle (which, contains the origin) between the lines 3x – y + 1 = 0,, x + 2y – 5 = 0. Find interval for values of ‘a’., Sol: Given origin and P lie on same side of each line., Substituting P in the given equation of line, we can, obtain the required interval., a2 + 2a + 2 – 5< 0 and 3a2 – (a + 1) + 1 > 0, i.e., , (a + 3) (a – 1) < 0 and a (3a – 1) > 0, 1, , ∴ a ∈ (–3, 1) and a ∈ (– ∞, 0) ∪  , ∞ , 3, , 1 , ∴ a ∈ (– 3, 0) ∪  , 1 , 3 , Example 3: Find the equations of the lines passing, through P(2, 3) and making an intercept AB of length, 2 units between the lines y + 2x = 3 and y + 2x = 5., Sol: Using equation of line in parametric form, i.e., x – x1= r cos θ and y – y1 = r sin θ, we can obtain the, required equation of line., Let equation of the line, in parametric form, be, x – 2 = r cos θ; y – 3 = r sin θ.

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8 . 3 2 | Straight Line, , Then, A(2 + r1cos θ, 3 + r1 sin θ) and B(2 + r2 cos θ,, 3 + r2 sin θ) lie on y+2x=3 and y+2x=5, respectively., ∴ (3 + r1 sin θ) + 2 (2 + r1 cos θ) = 3, , …. (i), , and (3 + r2 sin θ) + 2 (2 + r2 cos θ) = 5, , …. (ii), , ∴ (r2 – r1) (sin θ + 2 cos θ) = 2, , y+2x=3, , P, (as |r2 – r1| = 2), , 3 cos2θ + 4 sin θ cos θ = 0, π, 3, or tan θ = – ∴ Required lines are x = 2 and, 2, 4, 4y + 3x = 18, , θ=, , Example 4: In a triangle ABC, AB = AC. If D is mid-point, of BC, E is the foot of perpendicular from D on AC, and, F is the mid-point of DE, show that AF is perpendicular, to BE., Sol: As the geometrical fact to be established does, not depend on position of ABC, we may assume that, ‘’D is the origin; BC and AD are along x and y axes, respectively (as shown)”. Therefore by using intercept, form of equation of line, we can obtain required result., Let BD = DC = a, and A and E have co-ordinates (0, b), and (h, k), respectively., x y, 1, Line AC has the equation + =, a b, , y, , 2, , ., , , π, origin makes an angle α  0 < α <  with the positive, 4, , direction of x-axis. Prove that the equations of its, diagonals are y (cos α – sin α) = x (sin α + cos α) and, y(sin α + cos α) + x (cos α – sin α) = 0, where a is the, length of a side of the square., , Sol: Using slope point form of equation of line, i.e., y – y1 = m(x – x1), we can obtain the result. Here m = tan, θ and x1 , y1 is 0., , π, y tan  α +  x, Equation of diagonal OB is=, 4, , , , π, π, or y cos  α +=,  x sin  α + , 4, 4, , , , or y (cos α – sin α) = x (cos α + sin α), , ….(i), , y, B, C, /4, , O, , A, , x, , As diagonal AC is perpendicular to diagonal OB,, equation of AC is, cos α – sin α, (x – acos α ), Y – a sin α =, cos α + sin α, , E, F, , or x (cos α – sin α) + y (cos α + sin α) = a, , C(a,0), , x, ….(i), , Example 6: Two sides of a rhombus lying in the first, quadrant are given by 3x – 4y = 0 and 12x – 5y = 0. If, the length of the longer diagonal is 12 units, find the, equations of the other two sides of the rhombus., , ….(ii), , Sol: Using formula of equation of bisector of the angle,, we can obtain equation of AC. Given the length AC, we, , Also, (k/h) (–b/a) = – 1 , ( AC ⊥ DE) , , a2 + 2b2, , h k , F is  , , a +b, 2 2, 2, , From the figure, point A is (a cos α, a sin α)., , A(0,b), , h k, 1 , ∴ + =, a b, , a +b, ab, , 2, , a2b, , Example 5: A square lying above the x-axis and has, one vertex at the origin. A side passing through the, , B, , D, , 2, , ,k=, , a2 + 2b2, –ab, Product of slopes of BE and AF is equal to (–1). Hence, AF ⊥ BE., , y+2x=5, , B(-a,0), , Slope of BE =, , ab2, , Slope of AF =, , A, , ∴ sin θ + 2 cos θ = ± 1, , By (i) and (ii) h =

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M a them a ti cs | 8.35, , x(sin α + sin β + sin γ) – y (cos α + cos β + cos γ) = 0, ,  x + 3 y2 – 1 , , the mid-point of AC lies on the, Also D  2,  2 , 2 , , , , , α, β, γ, α, β, γ, ⇒ x  4 cos cos cos  – y  1 + 4 sin sin sin  =, 0, median through B,, 2, 2, 2, 2, 2, 2, , , i.e. 6x + 10y – 59 = 0, And hence the result is α + β + γ = π.,  x +3,  y2 – 1 , ⇒ 6  2, 0,  + 10 , ,  – 59 =, Example 11: ABC is a variable triangle with the fixed,  2 ,  2 , vertex C(1, 2) and vertices A and B with co-ordinates, ….(ii), ⇒ 3x2 + 5y2 – 55 = 0 , (cos t, sin t) and (sin t, – cos t),, Solving (i) and (ii), we get x2 =10, y2 = 5,, Respectively, where t is a parameter. Find the, i.e. the co-ordinates of C are (10, 5) and thus, , locus of the centroid of the ∆ABC., Sol: We can obtain co-ordinates of centroid G(α, β),  x + x 2 + x3 y1 + y 2 + y 3 , ,, using the formula  1,  and, 3, 3, , , we will get required equation of locus of centroid by, solving them simultaneously., , the equation of AC is 6x – 7y = 25, A(3, -1), Angle bisector, Median, , Let G(α, β) be the centroid in any position .Then G(α, β),  1 + cos t + sint 2 + sint – cos t , =, –,  or, 3, 3, , , , =, ∴ α, , B(x1, y1), , ∴ or 3α – 1 = cos t + sin t , , …. (i), , 3β =2 – sin t – cos t , , …. (ii), , Squaring and adding equations (i) and (ii), we get, = (cos t + sin t)2 + (sin t – cos t)2, , 1 – 4m 17, (1 / 4) – m, (6 / 7) – (1 / 4), =, ⇒, =, 4 + m 34, 1 + (1 / 4)m 1 + (6 / 7) × (1 / 4), 2, Equation of BC is, 9, , 2, y – 5 = – (x – 10) and 6x1 + 10y1 = 59, 9, , = 2 (cos t + sin t) = 2, 2, , ∴ the equation of the locus of the centroid is (3x – 1)2+, (3y – 2)2 = 2, 9(x2 + y2) –6 x – 12y + 3 = 0, 3(x2 + y2) – 2x – 4y + 1 = 0, , Example 12: Find equations of the sides of the, triangle having (3, –1) as a vertex, x – 4y + 10 = 0 and, 6x + 10y – 59 = 0 being the equations of an angle, bisector and a median, respectively, drawn from, different vertices., Sol: Consider the vertices of the triangle to be A(3, – 1),, B(x1, y1) and C(x2, y2). Here the mid-point of AC lies on, the median through B., Equation of the median through B be 6x + 10y – 59 = 0, and the equation of the angle bisector from C be, x – 4y + 10 = 0; x2 – 4y2 + 10 = 0, , x – 4y + 10 = 0,, , ⇒m= –, , (3α – 1)2 + (3β – 2)2, , ∴, , C(x2, y2), , Let the slope of BC be m1. Since BC and AC are equally, inclined to the angle bisector, , 1 + cos t + sint, 2 – sint – cos t, =, ,β, 3, 3, , 2, , 6x + 80y - 59 = 0, x - 4y + 10= 0, , ….(i), , Solving these equations, we get X1 = – 7/2, y1 = 8, , 8 +1, (x – 3), Hence, equation of AB is y + 1 =, –7 / 2 – 3, Example 13: A triangle has the lines y = m1x and y, = m2x for two of its sides, with m1 and m2 being roots, of the equation bx2 + 2hx + a = 0. If H(a, b) is the, orthocentre of the triangle, show that the equation of, the third side is (a + b) (ax + by) =ab(a + b – 2h)., Sol: Line OD passes from orthocentre. Therefore it, must be perpendicular to the side AB. By considering, equation of AB as y = mx +c, we will get co-ordinates, of A and B. Using slope point form of equation of line,, we can solve the problem., The given lines y = m1x and y = m2x intersect at the, origin O (0, 0). Thus one vertex of the triangle is at the, origin O. Therefore, let OAB be the triangle and OA and

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8 . 3 6 | Straight Line, , since m1 and m2 are the roots of the equation, , OB be the lines, y = m1x , , , , ....(i), , and y = m2x , , .…(ii), , Y, , bx2 + 2hx + a = 0, , m1+ m2 = – 2h/b and m1m2 = a/b, From (vii), we have, c=, , B, D, , y=m2x, , From (iii), the equation of third side AB is, , y=mx+c, , y = mx –, , A, , H(a,b), y=m1x, , Let the equation of the third side AB be, y = mx +c , , or (ax + by)(a + b)= ab(a + b – 2h), …..(iii), , Given that H(a, b) is the orthocentre of the OAB,, ∴ OH ⊥ AB, ⇒ (b/a) × m = – 1 ⇒ m = –a/b , , (a + 2hm + bm2 )a, m1 (a + b), , (a – 2ha / b + ba2 / b2 )a, a, or y = – x –, b, (–a / b)(a + b), , X, , O, , –[a + 2hm + bm2 ]a, –[a / b + 2hm / b + m2 ]a, =, m(a + b), m(a / b + 1), , ….(iv), , Solving (iii) with (i) and (ii), the co-ordinates of, , Example 14: Find the co-ordinates of the centroid,, circumcentre and orthocentres of the triangle, formed by the lines 3x – 2y = 6, 3x + 4y + 12 = 0 and, 3x – 8y + 12 = 0., Sol: Solving the given equations, we can obtain the, co-ordinates of vertices of triangle. Using appropriate, formula for finding the co-ordinates of centroid,, circumcentre and orthocentre, the problem can be, solved., ,  c, cm1 , ,, A = ,  and, m, –, m, m,  1, 1 –m,  c, cm2 , ,, B = , ,  m2 – m m2 – m , , Let sides AB, BC and CA have the, , Now equation of line through A, , equations 3x – 2y – 6 = 0 , , ….(i), , perpendicular to OB is, , 3x – 8y + 12 = 0 , , ….(ii), , 3x + 4y + 12 = 0 , , ….(iii), , y–, , cm1, , m1 – m, , =–, , 1 , c ,  x –,  or, m2 , m1 – m , , Solving (ii), (iii) we get y = 0, x = – 4,, , A, , c(m1m2 + 1), x, +, y= –, , m2 m2 (m1 – m), , ….(v), , Similarly, equation of line through B, , 3x - 2y- 6 =0, , 3x - 4y+ 12=0, , B, , C, , perpendicular to OA is, y= −, , x c (m1 m2 + 1), , +, m1 m1 (m2 − m), , ….(vi), , The point of intersection of (v) and (vi) is the orthocentre, H (a, b)., ∴ Subtracting (vi) from (v), we get, x=a=, or c =, , (m1 – m)(m2 – m), m(m1m2 + 1), , C = (– 4, 0), Solving (i), (ii) we get y = 3, x = 4, B = (4, 3), , –cm(m1m2 + 1), , –[m1m2 – m(m1 + m2 ) + m2 ]a, , 3x - 8y+12=0, , Solving (i,) (iii) we get y = – 3, x = 0;, , , …..(vii), , A = (0 – 3)

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M a them a ti cs | 8.37, , Example 15: One diagonal of a square is the position, b, m+, a =±1 which is intercepted, of the line tan( ±45°) =, b, 1–m, a, between the axes. Find the co-ordinates of other two, vertices of the square. Also prove that if two opposite, vertices of a square move on two perpendicular lines,, the other two vertices also move on two perpendicular, lines., ,  x + x 2 + x3 y1 + y 2 + y 3 , ,, Centroid G =  1, , 3, 3, , , where vertices are (x, y), etc., G, ,  0 + 4 – 4 –3 + 3 + 0 , ,, =,  (0,0), 3, 3, , , , To find the circumcentre:, Let M(α, β) be the circumcentre., , Sol: Using tan θ =, , MA = MB = MC, (α – 0)2 + (β+ 3)2 = (α – 4)2+ (β – 3)2 = (α + 4)2 + (β – 0)2, α2 + β2 + 6β + 9, 2, 2, 2, 2, = α + β – 8α – 6β + 25 = α + β – 8α + 16, , , we can obtain the slopes, , 1 + m1m2, , of AB and AD. As the slope of the given lines is –b/a,, the two vertices are clearly on the diagonal BD of the, square ABCD., If m be the slope of the line inclined at an angle of 45°, to BD,, , = 6β + 9 = –8α – 6β + 25 = 8α + 16, 6β + 9= –8α – 6β + 25 and 6β + 9 = 8α + 16, , tan( ±45°) =, , 8 α + 12 β – 16 = 0, 2 α + 3 β – 4 =0 , , …. (i), , 8 α – 6 β + 7 =0 , , …. (ii), , α, Solving (i) and (ii), we get =, , m2 − m1, , m + (b / a), =±1, 1 – m(b / a), C, , 1, 23, ,=, β, 12, 18, , (0,b)D, ,  1 23 , Circumcentre =  , ,  12 18 , , E, 45, , 45, , B, (a, 0), , O, , A(0, -3), A, , E, , m=, , M, , B(4, 3), , Use =, , D, , AD is y– b = –, , C(-4, 0), , =, 0, , 23, β+2,  –1 –23 , 23, 18, Then H  ,, =, =, ⇒β –, 0, ., 3, 9,  6 9 , , a–b, (x – a), a+b, , a+b, (x – 0), a–b, , a–b b – a, The point A is , ,,  . C is obtained by using the, 2 ,  2, fact that mid-point of AC and BD is same., , Let H(α, β) be the orthocentre, 1, 1, 12, =, ⇒α – , 3, 6, , AB is y – 0 =, , By solving these equation we get, , MG 1, =, GH 2, , α+2, , a–b, (a + b), or –, a+b, a–b, , …. (i), , a+b a+b, C= , ,, , 2 ,  2, , The opposite vertices B, D move on two perpendicular, lines x-axis and y-axis. Now the point, a–b b – a, a+b a+b, A, ,, ,,  lies on y = –x and point C , , 2 , 2 ,  2,  2, lies on y = x.

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8 . 3 8 | Straight Line, , Example 16: If the image of the point (x1, y1) with, respect to the mirror ax + by + c = 0 be (x2, y2), show, that, , x2 – x1 y 2 – y1 –2(ax1 + by1 + c), = =, a, b, a2 + b2, , and a., , x1 + x2, 2, , + b., , y1 + y 2, 2, , +c =, 0 , , ….(ii), , From (ii), a(x1 + x2) + b(y1 + y2) + 2c = 0, or (ax1 +by1 + c) + (ax1 +by1 + c) = 0 , , Sol: As the line PQ joining the points P(x1, y1) and Q(x2,, y2) is perpendicular to the line ax + by +c = 0. Also the, mid-point M of PQ is on the lines ax + by +c= 0. Hence, product of their slopes will be –1 and co-ordinates of M, lies on ax + by +c= 0., , From (i),, , y1 – y 2  a ,  –  = –1 , x1 – x2  b , , , =, , (ax1 + by1 + c) – (ax2 + by 2 + c), , =, , 2(ax1 + by1 + c), , …..(i), , x1 – x2, a, , =, , y1 – y 2, b, , =, , ….(iii), , a(x1 – x2 ) + b(y1 – y 2 ), a2 + b2, , By ratio and proportion, , P(x1,y1), , a2 + b2, a2 + b2, , , using (iii), , x2 – x1 y 2 – y1 – 2(ax1 + by1 + c), = =, a, b, a2 + b2, , ax+by+c=0, M, , Q(x2,y2), , JEE Main/Boards, Exercise 1, Q.1 Find the slope of the line joining (4, – 6) and (– 2, – 5)., , Q.7 Find the coordinates of the vertices of a square, inscribed in the triangle with vertices A(0, 0), B(2, 1) and, C(3, 0); Given the two vertices are on the side AC., , Q.2 Show that the line joining (2, – 3) and (– 5, 1), is (i) Parallel to the line joining (7, – 1) and (0, 3),, (ii) Perpendicular to the line joining (4, 5) and (0, – 2)., , Q.8 Find the equation of the straight line which passes, through the origin and trisects the intercept of line, 3x + 4y = 12 between the axes., , Q.3 A quadrilateral has the vertices at the points, (– 4, 2), (2, 6), (8, 5) and (9, – 7). Show that the midpoints of the sides of this quadrilateral are vertices of a, parallelogram., , Q.9 A straight line passes through the point (3, – 2)., Find the locus of the middle point of the portion of the, line intercepted between the axes., , Q.4 Find the values of x and y for which A(2, 0),, B(0, 2), C(0, 7) and D(x, y) are the vertices of an isosceles, trapezium in which AB ||DC., , Q.10 Find the equation of the straight line which passes, through the point (3, 2) and whose gradient is 3/4. Find, the coordinates of the point on the line that are 5 units, away from the point (3, 2)., , Q.5 Find the equations of the diagonals of the rectangle,, whose sides are x = a, x =, a’, y = b and y= b’., , Q.11 Find the distance of the point (2, 5) the lines 3x, +y + 4 = 0 measured parallel to line having slope 3/4., , Q.6 In what ratio is the line joining the points (2, 3) and, (4, –5) divided by the line joining the points (6, 8) and, (– 3, – 2) ?, , Q.12 The extremities of a diagonal of a square are, (1, 1), (– 2, – 1). Obtain the other two vertices and the, equation of the other diagonal.

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M a them a ti cs | 8.39, , Q.13 In the given figure, PQR is an equilateral, triangle and OSPT is a square. If OT =2 2 units, find, the equation of the lines OT, OS, SP, OR, PR and PQ., y, , Q.24 Find the image of the point (– 8, 12) with respect, to the line mirror 4x + 7y + 13 = 0., , K, 15, , L, , Q.25 The equations of two sides of a triangle are, 3x – 2y + 6 = 0 and 4x + 5y = 20 and the orthocentre is, (1, 1). Find the equation of the third side., , 45, S, 45 45, 45, 60 45, O, M, , 45, Q, , Q.23 A ray of light is sent along the line x – 2y – 3 = 0., Upon reaching the line 3x – 2y – 5 = 0, the ray is reflected, from it. Find the equation of the line containing the, reflected ray., , x, , Q.14 Find the equation of the medians of a triangle, formed by the lines x + y – 6= 0, x – 3y – 2 =0 and, 5x – 3y + 2 =0., Q.15 Find the coordinates of the orthocentre of the, triangle whose vertices are (0, 0), 2, – 1) and (–1, 3)., Q.16 Two vertices of a triangle are (3, – 1) and (–2, 3), and its orthocentre is origin. Find the coordinates of the, third vertix., Q.17 If the lines 3x + y – 2 =0, px + 2y – 3 = 0 and, 2x – y – 3 =0 are concurrent, find the value of p., Q.18 Find the angle between the lines y – 3x – 5 = 0 and, 3y – x + 6 =., 0, Q.19 Prove that the points (2, – 1), (0, 2), (3, 3) and (5, 0), are the vertices of a parallelogram. Also find the angle, between its diagonals., Q.20 A and B are the points (–2, 0) and (0, 5). Find the, Coordinates of two points C and D such that ABCD is, a square., Q.21 Find the equations of the lines through the point, (3, 2), which make an angle of 45° with the line x – 2y = 3., Q.22 A line 4x + y = 1 through the point A(2, – 7), meets the line BC whose equation is 3x – 4y + 1 = 0 at, the point B. Find the equation of the line AC, so that, AB = AC., , Q.26 Find the equations of the straight lines passing, through the point of intersection of the lines x + 3y +, 4 = 0 and 3x + y + 4 = 0 and equally inclined to the, axis., Q.27 Show that the straight lines x (a + 2b) + y(a + 3b), = a + b, for different values of a and b pass through a, fixed point., Q.28 The equation of the base of an equilateral triangle, is x + y = 2 and the vertex is (2, – 1). Find the length of the, side of the triangle., , Exercise 2, Single Correct Choice Type, Q.1 The pair of points which lie on the same side of the, straight line, 3x – 8y – 7 = 0 is, (A) (0, – 1), (0, 0), , (B) (0, 1), (3, 0), , (C) (– 1, – 1), (3, 7), , (D) (– 4, –3), (1, 1), , Q.2 Equation of the bisector of the acute angle between, the lines, 3x – 4y + 7 = 0 and 12x + 5y – 2= 0 is, (A) 11x – 3y + 9 = 0, , (B) 11x + 3y – 9 = 0, , (C) 3x – 11y + 9 = 0, , (D) None, , Q.3 A ray of light passing through the point A (1, 2), is reflected at a point B on the x-axis and then passes, through (5, 3). Then the equation of AB is, (A) 5x + 4y = 13, , (B) 5x – 4y = – 3, , (C) 4x + 5y = 14, , (D) 4x – 5y = – 6

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8 . 4 0 | Straight Line, , Q.4 The line x + 3y – 2 = 0 bisects the angle between a, pair of straight lines of which one has equation x – 7y +, 5 = 0. The equation of the other line is , , Q.11 The straight line, ax + by = 1 , makes with the, curve px2 + 2axy + qy2 = r a chord which subtends a, right angle at the origin. Then:, , (A) 3x + 3y – 1 = 0, , (B) x – 3y + 2 = 0, , (A) r(a2 + b2) = p + q, , (B) r(a2 + p2) = q + b, , (C) 5x + 5y – 3 = 0, , (D) None, , (C) r(b2 + q2) = p + a, , (D) None, , Q.5 A is point (3, – 5) with respect to a given system of, axes. If the origin is moved to (4, – 3) by a translation, of axes, then the new co-ordinates of the point A are, given by, (A) (1, – 2), , (B) (– 1, 2), , (C) (– 1, – 2), , (D) None of these, , x y, 3 4, + =, 1 , where + =, 5, a b, a b, are concurrent at a fixed point, then point is -, , Q.6 The set of lines given by, 3 4, (A)  , , 5 5 , (C) (a, 0), , (B) (0, b), (D) None, , Q.7 If P = (1, 0 ); Q = (– 1, 0) & R= (2, 0) are three, given points, then the locus of the points S satisfying, the relation, SQ2 +SR2 = 2 SP2 is, (A) A straight line parallel to x-axis., (B) A circle passing through the origin., (C) A circle with the centre at the origin., (D) A straight line parallel to y-axis., Q.8 Area of the rhombus bonded by the four lines,, ax ± by ± c =, 0 is:, (A), (C), , c2, 2ab, 2, , 4c, ab, , (B), (D), , 2c2, ab, ab, 4c2, , Q.9 If the sum of the distances of a point from two, perpendicular lines in a plane is 1, then its locus is, (A) A square, , (B) Circle, , (C) A straight line, , (D) Two intersecting lines, , Q.10 If the straight line x + 2y = 9, 3x – 5y = 5 & ax +, by = 1 are concurrent, then the straight line 5x + 2y = 1, passes through the point, (A) (a, – b), , (B) (– a, b), , (C) (a. b), , (D) (–a, –b), , Q.12 The lines y – y1 = m(x – x1) ± a 1 + m2 are tangents, to the same circle. The radius of the circle is:, (A) a/2, , (B) a, , (C) 2a, , (D) None, , Q.13 The equation of the pair of bisectors of the angles, between two straight lines is, 12x2 – 7xy – 12y2 = 0. If, the equation of one line is 2y – x = 0, then the equation, of the other line is, (A) 41x – 38y = 0, , (B) 38x – 41y = 0, , (C) 38x + 41y = 0, , (D) 41x + 38y = 0, , Q.14 If the point B is symmetric to the point A(4, – 1), with respect to the bisector of the first quadrant, then, the length AB is, (A) 3 2, , (B) 4 2, , (C) 5 2, , (D) None, , Q.15 The co-ordinates of the points A, B, C are (– 4, 0),, (0, 2) & (– 3, 2) respectively. The point of intersection of, the line which bisects the angle CAB internally and the, line joining C to the middle point of AB is,  7 4, (A)  – , ,  3 3, ,  5 13 , (B)  – , ,  2 2 , ,  7 10 , (C)  – , ,  3 3 , ,  5 3, (D)  – , ,  2 2, , Q.16 The sides of ∆ABC are 2x – y + 5 = 0, x + y – 5 =, 0 and x – 2y – 5 = 0. Sum of the tangents of its interior, angle is, (A) 6, , (B) 27/4, , (C) 9, , (D) None, , Q.17 Equation of a straight line passing through the, origin and making with x-axis an angle twice the size, of the angle made by the line y = 0.2x with the x-axis,, is:, (A) y = 0.4x, , (B) y = (5/12)x, , (C) 6y – 5x = 0, , (D) None of these

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M a them a ti cs | 8.41, , Q.18 The shortest distance from the point M(– 7, 2) to, the circle x2 + y2 – 10x – 14y – 151 = 0 is, (A) 1, , (B) 2, , (C) 3, , (D) None, , Q.19 The image of the pair of lines represented by, ax2+2hxy + by2 = 0 by the line mirror y=0 is, (A) ax2 – 2hxy – by2 = 0, , (B) bx2 – 2hxy + ay2 = 0, , (C) bx2 + 2hxy + ay2 = 0 (D) ax2 – 2hxy +by2 = 0, Q.20 The pair of straight lines x2 – 4xy + y2 = 0 together, with the line x + y + 4 6 = 0 form a triangle which is, (A) Right angle but not isosceles, (B) Right isosceles, (C) Scalene, (D) Equilateral, Q.21 Points, A & B are in the first quadrant; point ‘O’ is, the origin. If the slope of OA is 1, slope of OB is 7 and, OA = OB, then the slope of AB is, (A) – 1/5, , (B) – 1/4, , (C) –1/3, , (D) – 1/2, , π, about the origin in, 4, , the counter clockwise direction. Then, the final position, of the point is given by the coordinates , , (, ,  1 7 , (A) , ,, ,  2 2, , (B) – 2, 7 2, ,  1 7 , (C)  –, ,, , 2 2, , , (D), , (, , 2,7 2, , (1980), , ), , ), , Q.4 The straight lines x + y = 0, 3x + y – 4 = 0,, x + 3y – 4 = 0 form a triangle which is , (A) Isosceles, , (B) Equilateral, , (C) Right angled, , (D) None of these, , (1983), , Q.5 If the sum of the distance of a point from two, perpendicular lines in a plane is 1, then its locus is, , (1992), (A) Square, , (B) Circle, , (C) Straight line, , (D) Two intersecting lines, , Q.6 The orthocentre of the triangle formed by the lines, xy = 0 and x + y = 1, is , (1995), , Previous Years’ Questions, Q.1 The points (–a, – b), (0, 0), (a, b) and are , , (III) Rotation through an angle, , (1979), , (A) Collinear, (B) Vertices of a rectangle, (C) Vertices of a parallelogram, (D) None of the above, Q.2 Given the four lines with the equations, x + 2y – 3 = 0,, 3x + 4y – 7 = 0, 2x +3y – 4 = 0, 4x + 5y – 6 = 0, then, , (1980), (A) They are all concurrent, , (B) They are the sides of a quadrilateral, , 1 1, (A)  , , 2 2, , 1 1, (B)  , , 3 3, , (C) (0, 0), , 1 1, (D)  , , 4 4, , Q.7 The graph of the function cos x cos (x + 2) – cos2, (x + 1) is , (1997), (A) A straight line passing through (0, – sin21) with slope 2., (B) A straight line passing through (0, 0)., (C) A parabola with vertex (1, – sin21)., , π, , (D) A straight line passing through the point  , – sin2 1 , 2, , , and parallel to the x-axis., , (C) Only three lines are concurrent, (D) None of these, Q.3 The point (4, 1) undergoes the following three, transformations successively, (I) Reflection about the line y = x., (II) Translation through a distance 2 unit along the, positive direction of x-axis., , Q.8 The diagonals of a parallelogram PQRS are along, the lines x + 3y = 4 and 6x –2y = 7. Then PQRS must, be a , (1998), (A) Rectangle (B) Square, (C) Cyclic quadrilateral (D) Rhombus

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8 . 4 2 | Straight Line, , Q.9 Let PS be the median of the triangle with vertices, P (2, 2), Q(6, – 1) and R(7, 3). The equation of the line, passing through (1, – 1) and parallel to PS is  (2000), , Q.16 Let PS be the median of the triangle with vertices, P(2, 2), Q(6, -1) and R(7, 3). The equation of the line, passing through (1, -1) and parallel to PS is, (2014), , (A) 2x – 9y – 7 =0, , (B) 2x – 9y – 11 = 0, , (A) 4x - 7y - 11 = 0 , , (B) 2x+ 9y+ 7 = 0, , (C) 2x + 9y – 11 = 0, , (D) 2x + 9y + 7 = 0, , (C) 4x+ 7y+ 3 = 0 , , (D) 2x - 9y - 11 = 0, , Q.10 The incentre of the triangle with vertices, (1, 3),(0, 0) and (2, 0) is , , (2000), , , 3, (A)  1,, ,  2 , , , , 2 1 , (B)  ,, , 3 3, , 2 3, (C)  ,, , 3 2 , , , ,  1 , (D)  1,, , 3, , , Q.11 The number of integer values of m, for which the, x-coordinate of the point of intersection of the lines, 3x +4y = 9 and y = mx + 1 is also an integer, is (2001), (A) 2, , (B) 0, , (C) 4, , (D) 1, , Q.12 Area of the parallelogram formed by the lines, y = mx, y = mx + 1, y = nx and y = nx + 1 equals (2001), (A), , 2, (B), |m+n|, , |m+n|, , (m – n)2, 1, (C), |m+n|, , (D), , Q.13 Let P = (– 1, 0), Q = (0, 0) and R = (3, 3 3) be, three points. Then, the equations of the bisector of the, angle PQR is , (2002), (B) x + 3y =, 0, , (C), , (D) x +, , 3x + y =, 0, , 3, y=, 0, 2, , Q.14 If the line 2x + y = k passes through the point, which divides the line segment joining the points (1, 1), and (2, 4) in the ratio 3 : 2, then k equals , (2012), (A), , 29, 5, , (B) 5, , (C) 6, , (D), , 11, 5, , Q.15 The x-coordinate of the incentre of the triangle, that has the coordinates of mid points of its sides as, (0, 1) (1, 1) and (1, 0) is , (2013), (A) 2 − 2, , (B) 1 + 2, , (C) 1 − 2, , (D) 2 + 2, , (A) 2bc - 3ad= 0 , , (B) 2bc+ 3ad= 0, , (C) 3bc - 2ad= 0 , , (D) 3bc+ 2ad= 0, , Q.18 Locus of the image of the point (2, 3) in the line, ( 2x − 3y + 4 ) + k ( x − 2y + 3) = 0, k ∈ R , is a:  (2015), (A) Straight line parallel to y-axis, (B) Circle of radius, , 2, , (C) Circle of radius, , 3, , (D) Straight line parallel to x-axis., Q.19 The number of points, having both co-ordinates, as integers, that lie in the interior of the triangle with, vertices (0, 0), (0, 41) and (41, 0), is: , (2015), (A) 861, , 1, |m–n|, , 3, x+y =, 0, (A), 2, , Q.17 Let a, b, c and d be non-zero numbers. If the point, of intersection of the lines 4ax + 2ay + c = 0 and 5bx +, 2by + d= 0 lies in the fourth quadrant and is equidistant, from the two axes then , (2014), , (B) 820, , (C) 780, , (D) 901, , Q.20 Two sides of a rhombus are along the lines,, x – y + 1 = 0 and 7x – y - 5 = 0. If its diagonals intersect, at (–1, –2), then which one of the following is a vertex, of this rhombus? , (2016), (A) (-3, -8), , 1, 8, (B)  , − , 3, 3, , ,  10, 7, (C)  − , − , 3,  3, , (D) (-3, -9)

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M a them a ti cs | 8.43, , JEE Advanced/Boards, Exercise 1, Q.1 Points O, A, B, C………are shown in figure where, OA = 2AB = 4BC =………so on. Let A is the centroid, of a triangle whose orthocentre and circumcentre are, 7 5, (2, 4) and  ,  respectively. If an insect starts moving, 2 2, from the point O(0, 0) along the straight line is zig-zag, fashions and terminates ultimately at point P(α, β), then, find the value of (α + β), y, , C, , A, , 45, , 45, 45, 45, O (0, 0), , B, , x, , Q.2 Let ABC be a triangle such that the coordinates of, A are (– 3, 1). Equation of the median through B is 2x, + y – 3 = 0 and equation of the angular bisector of C, is 7x – 4y – 1 = 0. Then match the entries of column-1, with their corresponding correct entries of column-II., Column I, , Column II, , (A) Equation of the line AB is (p) 2x + y – 3 = 0, (B) Equation of the line BC is (q) 2x – 3y + 9 = 0, (C) Equation of CA is, , (r) 4x + 7y + 5 = 0, (s) 18x – y – 49 = 0, , Q.3 The equations of the perpendicular of sides AB and, AC of triangle ABC are x – y – 4 = 0 and 2x – y – 5 = 0, respectively. If the vertex A is (–2, 3) and point of intersection, 3 5, of perpendicular bisector is  ,  , find the equation of, 2 2, medians to the sides AB and AC respectively., , Q.4 The interior angle bisector of angle A for the, triangle ABC whose coordinates of the vertices are, A(– 8, 5); B(– 15, – 19) and C(1, – 7) has the equation ax, + 2y + c = 0. Find ‘a’ and ‘c’., , Q.5 Find the equation of the straight lines passing, through (– 2, – 7) & having an intercept of length 3, between the straight lines 4x + 3y = 12, 4x + 3y = 3., Q.6 Two sides of a rhombus ABCD are parallel to the, lines y = x + 2 & y = 7x + 3. If the diagonals of the, rhombus intersect at the point (1, 2) & the vertex A is, on the y-axis, find the possible coordinates of A., Q.7 Let O(0, 0), A(6, 0) and B(3, 3 ) be the vertices, of ∆OAB. Let R be the region consisting of all those, points P inside OAB which satisfy d(P, OA) = minimum, {d(P, OB)), d(P, AB)}, where d(P, OA), d(P, OB) and, d(P, AB) represent the distance of P from the sides, OA, OB and AB respectively. If the area of region R is, 9(a – b), where a and b are coprime. Then, find the value of, (a + b)., Q.8 Find the equations of the sides of a triangle having, (4, – 1) as a vertex. If the lines x – 1 = 0 and x – y – 1 = 0, are the equations of two internal bisectors of its angles., Q.9 P is the point (– 1, 2), a variable line through P cuts, the x & y axes at A & B respectively. Q is the point on, AB such that PA, PQ, PB are in HP. Find the locus of Q., Q.10 The equations of the altitudes AD, BE, CF of a, triangle ABC are x + y = 0, x + 4y = 0 and 2x– y =, 0 respectively. The coordinates of A are (t, –t). Find, coordinates of B & C. Prove that if t varies the locus of, the centroid of the triangle ABC is x + 5y = 0., Q.11 The distance of a point (x1, y1) from each of two, straight lines which passes through the origin of coordinates is δ; find the combined equation of these, straight lines., Q.12 Consider a, ∆ABC whose sides AB, BC and CA are, represented by the straight lines 2x + y = 0. x + py = q, and x – y = 3 respectively. The point P is (2, 3), (i) If P is the centroid, then find the value of (p + q), (ii) If P is the orthocentre, then find the value of (p + q), (iii) If P is the circumcentre, then find the values of, (p + q)

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8 . 4 4 | Straight Line, , Q.13 The sides of a triangle have the combined equation, x2 – 3y2 – 2xy + 8y – 4 = 0. The third side, which is variable, always passes through the point (– 5, – 1). If the range of, values of the slope of the third line so that the origin is, an interior point of the triangle, lies in the interval (a, b),, , 1 , then find  a +, ., b2 , , Q.14 Consider a line pair 2x2 + 3xy – 2y2 – 10x + 15y, – 28 = 0 and another line L passing through origin, with gradient 3. The line pair and line L form a triangle, whose vertices are A, B and C., (i) Find the sum of the cotangents of the inter ior angles, of the triangle ABC., (ii) Find the area of triangle ABC., (iii) Find the radius of the circle touching all the 3 sides, of the triangle., Q.15 Show that all the chords of the curve, 3x2 – y2 – 2x + 4y = 0 which subtend a right angle at the, origin are concurrent. Does this result also hold for the, curve, 3x2 + 3y2 + 2x + 4y = 0 ? If yes, what is the point, of concurrency & if not, give reasons., Q.16 A straight line is drawn from the point (1, 0) to the, curve x2 + y2 + 6x – 10y + 1 = 0, such that the intercept, made on it by the curve subtends a right angle at the, origin. Find the equations of the line., Q.17 The two line pairs y2 – 4y + 3 = 0 andx2 + 4xy +, 4y2 – 5x –10y + 4 = 0 enclose a 4 sided convex polygon, find., (i) Area of the polygon., , (ii) Length of the diagonals., Q.18 Find the equations of the two straight lines which, together with those given by the equation 6x2 – xy – y2, + x + 12y – 35 = 0 will make a parallelogram whose, diagonals intersect in the origin., Q.19 A straight line passing through O(0, 0) cuts the, lines x = α, y = β and x + y = 8 at A, B and C respectively, such that OA . OB . OC = 482 and f(α, β) = 0 where, y 3, f(x,=, y), – + (3π – 2y)6 + ex + 2y – 2e – 6, x 2, (i) Find the point of intersection of lines x = α and y = β., (ii) Find the value of (OA + OB + OC), (iii) Find the equation of line OA., , Q.20 The triangle ABC, right angled at C, has median, AD, BE and CF. AD lies along the line y = x + 3, BE lies, along the line y = 2x + 4. If the length of the hypotenuse, is 60, find the area of the triangle ABC., Q.21 A triangle has side lengths 18, 24 and 30. Find, the area of the triangle whose vertices are the incentre,, circumcentre and centroid of the triangle., Q.22 The points (1, 3) & (5, 1) are two opposite vertices, of a rectangle. The other two vertices lie on the lines, y = 2x + c. Find c & the remaining vertices., Q.23 A straight line L is perpendicular to the line, 5x – y = 1. The area of the triangle formed by the line L, & the coordinate axes is 5. Find the equation of the line., Q.24 Two equal sides of an isosceles triangle are given, by the equations 7x – y + 3 = 0 and x + y –3 = 0 & its, third side passes through the point (1, – 10). Determine, the equation of the third side., Q.25 The equations of the perpendicular bisectors of, the sides AB & AC of a triangle ABC are x – y +5 = 0 &, x + 2y = 0, respectively. If the point A is (1, –2). Find the, equation of the line BC., Q.26 Let P be the point (3, 2). Let Q be the reflection, of P about the x-axis. Let R be the reflection of Q about, the lines y = – x and Let S be the reflection of R through, the origin. PQRS in a convex quadrilateral. Find the area, of PQRS., Q.27 Two parallel lines 1 and  2 having non-zero, slope, are passing through the points (0, 1) and (– 1,0), respectively. Two other lines 1 and  2 are drawn, through (0, 0) and (1, 0) which are perpendicular to 1, and  2 respectively. The two sets of lines intersect in, four points which are vertices of a square. If the area of, p, this square can be expressed is the form q where p ∈, N, then the least value of (p + q)?, Q.28 In an acute triangle ABC, the base BC has the, equation 4x – 3y + 3 = 0. If the coordinates of the, orthocentre (H) and circumcentre (P) of the triangle are, (1, 2) and (2, 3) respectively, then the radius of the circle, m, , where m and n are, circumscribing the triangle is, a, relatively prime. Find the value of (m+ n).

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M a them a ti cs | 8.45, , (You may use the fact that the distance between, orthocentre and circumcentre of the triangle is given, R 1 – 8 cos A cosB cosC ), , Q.29 The points (– 6, 1), (6, 10), (9, 6) and (– 3, – 3) are, the vertices of a rectangle. If the area of the portion of, this rectangle that lies above the x axis is a/b, find the, value of (a +b), given a and b are coprime., Q.30 Consider the triangle ABC with sides AB and AC, having the equation L1 = 0 and L2 = 0. Let the centroid., Orthocentre and circumcentre of the ∆ABC and G, H and, S respectively. L = 0 denotes the equation of sides BC., (i) If L1: 2x – y = 0 and L2: x + y = 3 and G(2, 3) then find, the slope of the line L= 0., (ii) If L1: 2x + y = 0 and L2: x – y + 2 = 0 and H(2, 3) then, find the y-intercept of L = 0., (iii) If L1: x + y – 1 = 0 and L2: 2x – y + 4 = 0 and S(2, 1), then find the x-intercept of the line L= 0., , Exercise 2, Single Correct Choice Type, Q.1 Given the family of lines, a (3x + 4y + 6) + b (x + y, + 2) = 0. The line of the family situated at the greatest, distance from the point P(2, 3) has equation:, (A) 4x + 3y + 8 = 0 , , (B) 5x + 3y + 10 = 0, , (C) 15x + 8y + 30 = 0, , (D) None, , Q.2 On the portion of the straight line, x + 2y = 4, intercepted between the axes, a square is constructed, on the side of the line away from the origin. Then the, point of intersection of its diagonals has co-ordinates:, (A) (2, 3), , (B) (3, 2), , (C) (3, 3), , (D) None, , Q.3 The base BC of a triangle ABC is bisected at the, point (p, q) and the equation to the side AB & AC are, px + qy = 1 and qx + py = 1. The equation of the, median through A is:, (A) (p – 2q) x + (q – 2p)y + 1 = 0, (B) (p + q) (x + y) – 2 = 0, (C) (2pq – 1)(px+qy –1) =(p2 + q2 – 1)(qx+py – 1), (D) None, , Q.4 The lines 3x + 4y = 9 & 4x – 3y + 12 = 0 intersect, at P. The first line intersects x-axis at A and the second, line intersects y-axis at B. Then the circum radius of the, triangle PAB is, (A) 3/2, , (B) 5/2, , (C) 10, , (D) None, , Q.5 If the lines ax + y + 1 = 0, x + by + 1 = 0 & x + y +, c = 0, where a, b & c are distinct real numbers different, from 1 are concurrent, then the value of, 1, 1, 1, +, +, =, 1–a 1–b 1–c, (A) 4, , (B) 3, , (C) 2, , (D) 1, , Q.6 The points A(a, 0), B(0, b), C(c, 0) & D(0, d) are such, that ac = bd & a, b, c, d are all non zero. The points, thus:, (A) Form a parallelogram (B) Do not lie on a circle, (C) Form a trapezium, , (D) Are concyclic, , Q.7 The angles between the straight lines joining the, origin to the points common to 7x2 + 8y2 – 4xy + 2x –, 4y – 8 = 0 and 3x – y = 2 is, π, π, π, (C), (D), (A) tan–1 2 (B), 3, 4, 2, Q.8 Distance between two lines represented by the line, pair, x2 – 4xy + 4y2 + x – 2y – 6 = 0 is, 1, (A), (B) 5, (C) 2 5, (D) None, 5, Q.9 If the straight lines joining the origin and the points, of intersection of the curve, 5x2 + 12xy – 6y2 + 4x – 2y + 3 = 0., , And x + ky – 1= 0 are equally inclined to the co-ordinate, axes then the value of k:, (A) Is equal to 1, (B) Is equal to – 1, (C) Is equal to 2, (D) Does not exist in the set of real numbers, Q.10 If the vertices P and Q of a triangle PQR are given, by (2,5) and (4,–11) respectively, and the point R moves, along the line N: 9x + 7y + 4 = 0, then the locus of the, centroid of the triangle PQR is a straight line parallel to, (A) PQ, , (B) QR, , (C) RP, , (D) N

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8 . 4 6 | Straight Line, , Q.11 Let the co-ordinates of the two points A & B, be (1, 2) and (7, 5) respectively. The line AB is rotated, through 45° in anticlockwise direction about the point, of trisection of AB which is nearer to B. The equation of, the line in new position is, , 2 at, Q.17 A is a point on either of two lines y + 3 | x |=, 4, a distance of, units from their point of intersection., 3, The co-ordinates of the foot of perpendicular from A, , (A) 2x – y – 6 = 0, , (B) x – y – 1= 0, , (C) 3x – y – 11 = 0, , (D) None of these, , on the bisector of the angle between them are, , Q.12 If the line y =mx bisects the angle between the, lines ax2 + 2hxy + by2 = 0, then m is a root of the, quadratic equation: , (A) hx2 + (a – b) x – h = 0 , ,  2 , ,2 , (A)  –, 3 , , , (B) (0, 0), ,  2 , ,2 , (C) ,  3 , , (D) (0, 4), , Q.18 All the points lying inside the triangle formed by, the points (1, 3), (5, 6) & (– 1, 2) satisfy, (A) 3x + 2y ≥ 0, , (B) 2x + y + 1 ≥ 0, , (C) (a – b)x2 + hx – (a – b) = 0 , , (C) 2x + 3y – 12 ≥ 0, , (D) – 2x + 11 ≥ 0, , (D) (a – b)x2 – hx – (a – b) = 0, , x y, + =, 1 cuts the co-ordinate axes at, a b, x y, + =, –1 at A’ (–a’, 0) &, A(a, 0) & B(0, b) & the line, a' b', , (B) x2 + h(a – b) x – 1 = 0, , Q.19 Line, , Q.13 A Triangle is formed by the lines 2x – 3y – 6 = 0;, 3x – y + 3 = 0 and 3x + 4y – 12 = 0. If the points P(α, 0), and Q(0, β) always lie on or inside the ∆ABC, then, (A) α ∈ [– 1, 2] & β ∈ [– 2, 3], (B) α ∈ [– 1, 3] & β ∈ [– 2, 4], , (A) (0, 0), , (C) α ∈ [– 2, 4] & β ∈ [– 3, 4] , (D) α ∈ [– 1, 3] & β ∈ [– 2, 3], , (B) 5x – 2y = 13, , (C) x + y = 11, , (D) 3x – 4y = – 9, , Q.15 The vertex of a right angle o triangle lies on the, straight line 2x + y – 10 = 0 and the two other vertices,, at point (2, – 3) and (4, 1) then the area of triangle in, sq. units is 33, (B) 3, (C), (D) 11, (A) 10, 5, Multiple Correct Choice Type, Q.16 The area of triangle ABC is 20 cm2. The coordinates of vertex A are (– 5, 0) and B are (3, 0). The, vertex C lies upon the line, x – y = 2. The co-ordinates, of C are, (A) (5, 3), , (B) (– 3, – 5), , (C) (– 5, – 7), , (B) (0, b’), ,  aa' , (C)  0,, ,  b , ,  bb' , (D)  0,, , a , , , Q.20 If one vertex of an equilateral triangle of side, , Q.14 In a triangle ABC, side AB has the equation 2x + 3y, = 29 and the side AC has the equation, x + 2y = 16. If the, mid-point of BC is (5, 6), then the equation of BC is, (A) x – y = – 1, , B’(0, – b’ ). If the points A, B, A’, B’ are concyclic then the, orthocentre of the triangle ABA’ is, , (D) (7, 5), , ‘a’ lies at the origin and the other lies on the line, x – 3y = 0 then the co-ordinates of the third vertex, are:,  3a a , , 3a a , (A) (0,a), (B) , , –  (C) (0, – a) (D)  –, , ,  2,  2 2, 2 , , , , Q.21 Three vertices of a triangle are A(4, 3); B(1, –1) and, C(7, k). Value(s) of k for which centroid, orthocentre,, incentre and circumcentre of the ABC lie on the same, straight line is/are:, 19, (A) 7, (B) – 1, (C) –, (D) None, 8, Q.22 Equation of a line through (7, 4) and touching the, circle x2 + y2 – 6x + 4y – 3 = 0 is, (A) 5x – 12y + 13 = 0, , (B) 12x – 5y – 64 = 0, , (C) x – 7 = 0 , , (D) y = 4, , Q.23 The circumcentre of the triangle formed by the, lines, xy + 2x + 2y + 4 = 0 and x + y +2 = 0 is, (A) (– 2, – 2), , (B) (–1 , – 1), , (C) (0, 0) , , (D) (– 1, – 2)

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M a them a ti cs | 8.47, , Q.24 The sides of a triangle are x + y = 1, 7y = x and, – 3y + x = 0. Then the following is an interior point of, the triangle., , Match the conditions expressions in column I with, statement in column II, , (A) Circumcentre , , (B) Centroid, , (C) Incentre , , (D) Orthocentre, , L1: x + 3y – 5 = 0 ;, , Q.25 Equation of a straight line passing through the, 1, point (2, 3) and inclined at an angle of tan−1   with, 2, the line y + 2x = 5 is, , (A) y =3, , (B) x = 2, , (C) 3x + 4y – 18 = 0 , , (D) 4x + 3y – 17 = 0, , Q.26 A ray of light travelling along the line x + y = 1, is incident on the x-axis and after refraction it enters, the other side of the x-axis by turning π/6 away from, the x-axis. The equation of the line along which the, refracted ray travels is, , Q.2 Consider the lines given by , , (2008), , L2: 3x – ky – 1 = 0, , L3: 5x + 2y – 12 = 0, Column I, , Column II, , (A) L1, L2, L3 are concurrent, if, , (p) k = – 9, , (B) One of L1, L2, L3 is parallel to at least, one of the other two, if, , (q) k = –, , (C) L1, L2, L3 form a triangle, if, (D) L1, L2, L3 do not form a triangle, if, , (r) k =, , 6, 5, , 5, 6, , (s) k = 5, , (A) x + (2 – 3)y =, 1 , (B) (2 – 3)x + y =, 1, , Q.3 Three lines px + qy + r = 0, qx + ry + p = 0 and, rx + py + q = 0 are concurrent, if , (1985), , (C) y + (2 + 3)x =2 + 3, , (A) p + q + r = 0 , , (B) p2 + q2 + r2 = pr + rq, , (C) p3 +q3 + r3 = 3 pqr, , (D) None of these, , (D) None of these, , Q.27 Consider the equation y – y1 = m(x – x1). If m &, x1 are fixed and different lines are drawn for different, values of y1, then, (A) The lines will pass through a fixed point, (B) There will be a set of parallel lines, (C) All the lines interest the line x = x1, , (D) All the lines will be parallel to the line, , Previous Years’ Questions, The codes (A), (B), (C) and (D) deformed as follows., , Q.4 All points lying inside the triangle formed by the, points (1, 3), (5, 0) and (– 1, 2) satisfy , (1986), (A) 3x + 2y ≥ 0 (B) 2x +y – 13 ≥ 0, (C) 2x – 3y – 12 ≤ 0 (D) – 2x + y ≥ 0, Q.5 Let L1 be a straight line passing through the origin and, L2 be the straight line x+y=1. If the intercepts made by the, circle x2 + y2 – x + 3y = 0 on L1 and L2 are equal, then which, of the following equation can represent L3?, (1999), (A) x + y = 0 (B) x – y = 0, , (A) Statement-I is true, statement-II is also true;, statement-II is the correct explanation of statement-I., (B) Statement-I is true, statement-II is also true;, statement-II is not the correct explanation of, statement-I., (C) Statement-I is true; statement-II is false., (D) Statement-I is false; statement-II is true., , (C) x + 7y = 0 (D) x – 7y = 0, , Q.1 Lines L1: y – x = 0 and L2: 2x + y =0 intersect the line L3:, y + 2 = 0 at P and Q, respectively. The bisector of the acute, angle between L1 and L2 intersects L3 at R. , (2007), , Q.7 The straight line 2x – 3y = 1 divides the circular, region x2 + y2 ≤ 6 into two parts. If, , Statement-I: The ratio PR: RQ equals 2 2 : 5 ., Because, , (2011), , Statement-II: In any triangle, bisector of an angle, divides the triangle into two similar triangles., , Q.6 If the distance between the plane Ax – 2y + z = d and, , x –1 y –2 z –3, the plane containing the lines = =, and, 2, 3, 4, x–2 y –3 z– 4, = =, is 6 , then |d| is……….., (2010), 3, 4, 5, ,  3   5 3   1 1   1 1  , S =  2,  ,  ,  ,  , –  ,  ,   , ,  4   2 4   4 4   8 4  , , Then the number of point(s) lying inside the smaller, part is……………

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8 . 4 8 | Straight Line, , Q.8 For points P = (x1, y1) and Q = (x2, y2) of the, coordinate plane, a new distance d(P,Q) is defined by, d(P, Q) = | x1 – x2 | + | y1 – y2 | . Let Q = (0, 0) and A =, (3, 2). Prove that the set of points in the first quadrant, which are equidistant (with respect to the new distance), from O and A consists of the union of a line segment, of finite length and an infinite ray. Sketch this set in a, labelled diagram. , (2000), Q.9 A straight line L through the origin meets the line x, + y = 1 and x + y = 3 at P and Q respectively. Through, P and Q two straight lines L1 and L2 are drawn, parallel, to 2x – y = 5 and 3x + y = 5 respectively. Lines L1 and L2, intersect at R, show that the locus of R as L varies, is a, straight line. , , (2002), Q.10 Let S be a square of unit area. Consider any, quadrilateral which has one vertex on each side of S. If a, b,, c and d denote the length of the sides of the quadrilateral,, , (1997), prove that 2 ≤ a2 + b2 + c2 + d2 ≤ 4. , , Q.11 Using coordinate geometry, prove that the three, altitudes of any triangle are concurrent. (1998), Q.12 A Straight line L through the point (3, -2) is, inclined at an angle 60° to the line 3x + y =., 1 If L, also intersects the x-axis, then the equation of L is , , (2011), Q.13 For a > b > c > 0, the distance between (1, 1) and, the point of intersection of the lines ax + by + c = 0 and, (2013), bx + ay + c = 0 is less than 2 2 . Then , (A) a + b – c > 0, , (B) a – b + c < 0, , (C) a – b + c > 0, , (D) a + b – c < 0, , Q.14 For a point P in the plane, let d1 (P ) and d2 (P ) be, the distance of the point P from the lines x – y = 0 and x, + y = 0 respectively. The area of the region R consisting, of all points P lying in the first quadrant of the plane, and satisfying 2 ≤ d1 (P ) + d2 (P ) ≤ 4 , (2014), , MASTERJEE Essential Questions, JEE Main/Boards, , JEE Advanced/Boards, , Exercise 1, , Exercise 1, , Q.10, , Q.13, , Q.23, , Q.2, , Q.6, , Q.10, , Q.24, , Q.26, , Q.27, , Q.15, , Q.19, , Q.24, , Q.2, , Q.6, , Q.8, , Q.12, , Q.18, , Q.21, , Q.23, , Q.26, , Q.27, , Exercise 2, Q.11, , Q.12, , Q.16, , Q.18, , Previous Years’ Questions, Q.2, , Q.7, , Q.11, , Q.13, , Q.10, , Exercise 2, , Previous Years’ Questions, Q.2, , Q.5, , Q.9, , Q.10, , Q.7

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M a them a ti cs | 8.49, , Answer Key, JEE Main/Boards, Q.14 x = 2, x + 9y – 14 = 0, 7x – 9y – 2 = 0, , Exercise 1, , Q.15 (– 4, – 3), , 1, 6, , Q.1 –, , Q.16 –, , 36 45, ,–, 7, 7, , Q.4 (7, 0) or (2, 5), , Q.17 5, , Q.5 (b – b')x + (a'– a)y + ab'– a'b, , Q.18 30° or 150°, , = 0,(b – b')x + (a – a')y + a'b'– ab= 0, Q.6 5 : 97 externally, , =, θ tan–1 ( −, Q.19, , 22, ), 3, , Q.20 D = (– 7, 2), C = (– 5, 7), , 3   9 , 9 3, 3 3 , Q.7 E  ,0  , F  ,0  , G  ,  , H  , , 2, 4, 4, 4, ,  , , , , 2 4 , , Q.21 3x – y – 7 = 0 and x + 3y – 9 = 0, , Q.8 3x – 8y = 0, 3x – 2y = 0, , Q.23 29x – 2y – 31 = 0 , , Q.9 3y – 2x = 2xy, , Q.24 (– 16, – 2), , Q.10 (7, 5) and (– 1, – 1), , Q.25 26x – 122y – 1675 = 0, , Q.11 5, , Q.26 x + y + 2 = 0, ,  3 3  3 13  3 1 3 , , ,=–,   , – , Q.12 C =  – ,  – ,, A,  2 2  2 22  2 2 2 , , Q.27 P (2, – 1), , Q.13 OT → y = x, OS → y = –x, SP → y = x + 4,, PQ → y = –x + 4, PR → y = (2 –, , Q.22 52x + 89y + 519 = 0 or 4x + y = 1, , Q.28, , 2, 3, , 3 )x + 4, , Exercise 2, Single Correct Choice Type, Q.1 C, , Q.2 A, , Q.3 A, , Q.4 C, , Q.5 C, , Q.6 A, , Q.7 D, , Q.8 B, , Q.9 A, , Q.10 C, , Q.11 A, , Q.12 B, , Q.13 A, , Q.14 C, , Q.15 D, , Q.16B, , Q.17 B, , Q.18 B, , Q.19 D, , Q.20 D, , Q.21 D, , Previous Years Questions, Q.1 A, , Q.2 C, , Q.3 C, , Q.4 A, , Q.5 A, , Q.6 C, , Q.7 D, , Q.8 D, , Q.9 D, , Q.10 D, , Q.11 A, , Q.12 D, , Q.13C, , Q.14 C, , Q.15 A, , Q.16 B, , Q.17 C, , Q.18 B, , Q.19 C, , Q.20 B

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8 . 5 0 | Straight Line, , JEE Advanced/Boards, , Q.15 (1, – 2), yes ( −1, −2 ), , Exercise 1, , Q.16 x +y = 1; x + 9y = 1, Q.17 (i) area – 6 sq. units, (ii) diagonals ar, , Q.1 8, , Q.18 6x2 – xy – y 2 – x – 12y – 35 = 0, , Q.2 (A) R; (B) S; (C) Q, , Q.19 (i) α =2 and β =3; (ii) 9 2 ; (iii) x – y = 0, , Q.3 x + 4y = 4 ; 5x + 2y = 8, , Q.20 400 sq. units, , Q.4 a = 11, c = 78, , Q.21 3 units, , Q.5 7x + 24y - 182 = 0 or x = – 2, , Q.22 c = – 4; B (2, 0) ; D (4, 4), ,  5, Q.6 (0, 0) or  0, ,  2, Q.7 5, , Q.24 x – 3y – 31 = 0 or 3x+ y + 7 = 0, , Q.8 2x – y +3 =0, 2x + y – 7 =0, x – 2y – 6 =0, , Q.25 14x + 23y = 40, , Q.9 y = 2x , , Q.26 15, ,  2t –t   t , Q.10 B  – .  .C  .t ,  3 6  2 , , Q.27 6, , Q.23 x + 5y + 5 2 = 0 or x + 5y – 5 2 = 0, , Q.28 63, , Q.11 (y12 – δ2 )x2 – 2x1 y1 xy + (x12 – δ2 )y 2 =, 0, , Q.29 533, , Q.12 (a) 74 ; (b) 50 ; (c) 47, , Q.30 (a) 5 ; (b) 2 ; (c), , Q.13 24, Q.14 (a), , 5 &, , 3, 2, , 50, 63, 3, (8 5 – 5 10 ), ; (b), ; (c), 7, 10, 10, , Exercise 2, Single Correct Choice Type, Q.1 A, , Q.2 C, , Q.3 C, , Q.4 B, , Q.5 D, , Q.6 D, , Q.7 D, , Q.8 B, , Q.9 B, , Q.10 D, , Q.11 C, , Q.12 A, , Q.13 D, , Q.14 C, , Q.15 B , , Multiple Correct Choice Type, Q.16 B, D, , Q.17 B, C, , Q.18 A, B, D, , Q.19 B, C, , Q.20 A, B, C, D, , Q.21 B, C, , Q.22 A, C, , Q.23 B, C, , Q.24 B, C, , Q.25 B, C, , Q.26 A, C, , Q.27 B, C, , Q.4 A, C, , Previous Years Questions, Q.1 C, , Q.2 A → s; B → p, q; C → r; D → p, q, s, , Q.3 A , C, , Q.5 B , C, , Q.6 6, , Q.14 6, , Q.7 2, , Q.13 A, , 53

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M a them a ti cs | 8.51, , Solutions, JEE Main/Boards, Exercise 1, Sol 1: Slope of line joining [4, –6] & [–2, –5], =, , 1, −5 + 6, =, −6, −2 − 4, , Sol 2: Line 1 joining (2, –3) and (–5, 1), ⇒ Slope = −, , 4, 7, ,  2 → line joining (7, –1) and (0, 3), −4, ⇒ Slope =, 7, ,  3 → line joining (4, 5) and (0, –2), ⇒ Slope =, 1 ||  2, , 7, 4, , and , , 5, 3, 1, 2, Slope of GH =, =, =, 17 5, 12, 4, −, 2 2, −1 +, , 5, 2 = −13, Slope of EH =, 5, 7, −1 −, 2, 4+, , EF || GH and FG || EH, hence midpoints E, F, G, H form a, parallelogram., Sol 4: AB || DC, , 2−0, = –1, 0−2, 7−y, y −7, =, Slope of DC =, 0−x, x, , Slope of AB =, , y – 7 = –x ⇒ x + y = 7 – x,  m2 ⋅ m3 =, −1, , A, , …(i), , B, , i.e. 1 and  2 are perpendicular to  3, Sol 3: A(–4, 2) ; B(2, 6) ; C(8, 5); D(9, –7) mid points of,  −4 + 2 2 + 6 , AB = E , ,,  = (–1, 4), 2 ,  2, , Trapezium is isosceles i. e. AD = BC, (x – 2)2 + y2 = 25, x2 + y2 – 4x = 21, , 2 + 8 6 + 5,  11 , BC = F , ,,  = 5, , 2 ,  2,  2, , x2 + y2 + 2xy = 49, ,  8 + 9 5 − 7   17, , CD = G , ,,  =  , − 1, 2  2,  2, , , y= 7-x, ,  9 − 4 −7 + 2 , AD = H , ,, =, 2 ,  2, ,  5 −5 ,  , , 2 2 , ,  11, ,  2 − 4, 3, 1, Slope of EF = , =,  =, 12, 4, 5, +, 1, , , , , , 11, +1, −13, 13, Slope of FG = 2, =, =, 7, −7, 17, 5−, 2, , C, , D, , …(ii), , 4x + 2xy = 28 ⇒ 2x(2 + y) = 28 from equation (i), ⇒ x(9 – x) = 14 ⇒ x2 – 9x + 14 = 0, ⇒x = 7, 2, If x = 7, y = 0 and if x = 2, y = 5, Ans is (7, 0) and (2, 5), Sol 5:, , D, A, , x=a, , C y = b’, B, , x = a’, , y=b

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8 . 5 2 | Straight Line, , Coordinates are A(a, b), B(a′, b), C(a′, b′) and D(a, b′), , H lies of x = 2y ⇒ a = 2c ⇒ b = 3c, , b′ − b, a′ − a, , Slope of line AC =, , It is a square hence b – a = c, G lies on x + y = 3 ⇒ b + c = 3, , b′ − b, Slope of line BD =, a − a′, , 4c = 3, 3, 9, ; b = ; , 4, 4, , a=, , Equation of line AC, , ⇒c=, ,  b′ − b , ⇒y–b= ,  (x – a),  a′ − a , ⇒ (b′ – b)x + (a – a′)y – ab′ + a⇒b = 0, , Coordinates of square are, , Equation of line BD, ⇒y–b=, , b′ − b, (x – a′), a − a′, , 6, 4, , 3 , 9 , 9 3, 3 3 , E  ,0  , F  ,0  , G  ,  , H  , , 2, 4, 4, 4, , , , , , , 2 4 , , Sol 8:, , ⇒ (b′ – b)x – (a – a′)y – a'b' + ab = 0, , A, , (0,3), , Sol 6: 1 (line joining (2, 3) and (4, –5)), , B, (4,0), , 3x+4y=12, ,  2 (line joining (6, 8) and (–3, –2)), (6, 8), m, , 1, F, , (2, 3), , (4, -5), , (-3, -2), , Lets assume it divides 1 in ratio m : 1,  4m + 2 3 − 5m , Coordinates of F are , ,, ,  m+1 1+m , , y −8, 10, =, x−6, 9, , Equation of  2 is, ,  1(4,0) + 2(0,3) , 4 6, A= , ⇒  , , 3, , , 3 3,  2(4,0) + 1(0,3) , 8 3, B= , ⇒  , , 3, , , 3 3, , Line OA → y =, , 3x, ⇒ 3x – 2y = 0, 2, , Line OB → y =, , 3x, ⇒ 3x – 8y = 0, 8, , Sol 9:, , 3 − 5m, −8, 10, 10, −5 − 13m, ⇒ 1+m, =, ⇒, =, 4m + 2, 9, 9, −2m − 4, −6, m+1, −5, ⇒ 45 + 117m = 20m + 40 ⇒ m =, 97, i. e. in ratio 5 : 97 (externally)., Sol 7:, , B, A, (3,-2), , y+2, =m, x −3, y + 2 = m(x – 3), , 2 , A ⇒  3 + ,0  [x intercept], m , , , B, H, A, , E, , G, F, , C, , E = (a, 0), F = (b, 0), G = (b, c), H = (a, c), , B ⇒ (0, –2 – 3m) [y intercept], , , 2,  3 + m −2 − 3m , Mid point of AB , ,, , 2,  2, , , 

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M a them a ti cs | 8.53, , y=, , −3m − 2, 3m + 2, ,x=, 2, 2m, , Slope of AC =, , −3, 2, ,  2y + 2 , m = –, ,  3 ,  −2y − 2 ,  −2y − 2 , ⇒ 2x ,  = 3,  +2,  3 ,  3 , , BD =, , ⇒ –4xy – 4x = –6y – 6 + 6, , 2, 3 =1, tan 45º =, 2m, 1+, 3, , y −2, 3, =, and y = 3x – 1, x −3, 4, , Points which are 5 units away from (3, 2) are, , (3 ± 5, , cos θ, 2 ± 5 sin θ ), , tan θ =, , Length of side =, , 3, ⇒(3 ± 4, 2 ± 3)⇒ [7, 5] [–1, –1], 4, , Sol 11: Coordinates of A at r units from B, A(2 + r cosθ, 5 + r sin θ), , Let slope of AB be m, , m, 5, 5m, −1, −1, =, ⇒ m = 5 or, =, ⇒m=, 3, 3, 3, 5, 3, , , 13, 13, A = 1 +, cos θ, 1 +, sin θ , 2, 2, , , −1, −5, Now cos θ =, , sin θ =, 26, 26,  1 −3 , ⇒A  , , 2 2 , , m=, , B, 2,5, 3/4, r, 3x + y + 4 = 0, , A, , 13, 2, , m−, , ⇒ 2xy + 2x – 3y = 0, Sol 10:, , 13, , −1, , sin θ =, 5, , 1, 26, , −5, , , cosθ =, , 26, , , , 13, 13, C = 1 +, cos θ,1 +, sin θ , 2, 2, , ,  3 3, ⇒ C = − , ,  2 2, , 3, tan θ =, 4, , , 4r, 3r , A 2 + ,5 + , 5, 5, , , Sol 13: OT → y = x [O(0, 0), T (2, 2)], , A lies on given line, , SP → y = x + 4 [as OP = 4], , , 3r, 4r , +4=0, ⇒ 3 2 +  + 5 +, 5, 5 , , 15 + 3r = 0 ⇒ r = –5, , OS → y = –x [O(0, 0), S (–2, 2)], PQ → y = –x + 4 [as OQ = 4], PR → y = mx + 4, , A[2 – 4, 5 – 3] = [–2, 2], , m = tan 150 →, , r = 5 units, , ⇒ y = (2 –, OR, , Sol 12:, D, , (-2,-1), , =2–, , 3, , 2 cos75, 4, , 2 sin 75], , , ( 3 − 1), ( 3 + 1) , R 4 + 4 2, ,4 2, , 2 2, 2 2 , , , C, B (1,1), , Slope of BD =, , 3 +1, 3 )x + 4, , → y = mx, , R[4 + 4, , m, A, , 3 −1, , 2, 3, , R[4 + 2( 3 –1), 2( 3 +1)], R[2( 3 +1), 2( 3 +1)], ⇒y=x

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8 . 5 4 | Straight Line, , Sol 14 :, , Sol 16:, , A, x-3y=2, C, , G E, F, , A (3, -1), , x+y-6=0, , O, (0, 0), , 5x-3y+2=0, , B, , B, (-2, 3), , Solving, , Orthocentre is (0, 0), , x + y – 6 = 0, x – 3y – 2 = 0 gives A[5,1], , (slope)OA = −, , x + y – 6 = 0, 5x – 3y + 2 = 0 gives B[2,4], x – 3y – 2 = 0, 5x – 3y + 2 = 0 gives C[–1,–1], , Equation of median, 1, y −1, AF ⇒, = 2 ⇒ x + 9y = 14, 9, x −5, −, 2, BG, CE, , 7, y +1, ⇒, =, ⇒ 7x – 9y = 2, 9, x +1, , 1, 3, b−3, , a+2, , …. (i), , (slope)OB =, , −3, 2, , (slope)AC =, , 2, b +1, =, , 3, a−3, , …. (ii), , (slope)BC = 3 =, ,  7 5  1 3 , Midpoints are E  ,  , F  ,  , G[2, 0], 2 2 2 2, , 1, y−4, ⇒, =, ⇒x=2, 0, x−2, , C, (a, b), , Solving (i) and (ii), we get, b = 3a + 9 and 3b = 2a – 9, 9a + 27 = 2a – 9, a= −, , 36, −108 + 63, 45, ; b=, = −, 7, 7, 7, ,  36, 45 , ⇒ C − , −, , 7 ,  7, , Sol 15:, C(-1,3), , Sol 17: Lines are concurrent intersection of 1 and M  3, gives x = 1 y = –1, It lies on  2 ⇒ p – 2 – 3 = 0 ⇒ p = 5, , A(0,0), B(2,-1), , Slope of line BC =, , Sol 18: 1 ⇒ y =, , −4, 3, , Slope of line perpendicular to BC =, , m1 =, , 3, 4, , 3−, , ⊥ bisector through A, y=, , 3x, 4 , , …. (i), , Slope of line AC = –3, , x = –4;, , y = –3, , Ans is (–4, –3), , 1+, , 1, 3, , 1, 3, =, 3, , 1, 3, , 3, , Sol 19:, , ⊥ bisector through B, Solving (i) & (ii), we get, , tan θ =, , m2 =, , ⇒ θ = 30º, 150º, , 1, Slope of line perpendicular to AC =, 3, 3y = x – 5, , 3 , , 3x + 5, , …. (ii), , 3, 2, , A(2, –1) , , mAB = −, , B(0, 2) , , mBC =, , 1, 3, , C(3, 3) , , mCD =, , −3, 2, , 2 ⇒ y =, , x−6, 3

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M a them a ti cs | 8.55, , D(5, 0) , , mAD =, , 1, 3, , A (2, -7), , BC || AD and AB || CD, , B, 4x+y=1, , ⇒ ABCD is a parallelogram, , C, 3x-4y+1=0, , ⇒ Diagonals are AC & BD, Slope of AC →, , −1 − 3, =4, 2−3, , Solving equations we get B, 3x + 1, = 1 – 4x ⇒ 19x = 3, 4, 3, 7, ⇒x=, ,y=, 19, 19, , 2 − 0 −2, =, Slope of BD →, 0 −5 5, m1 = 4; , , m2 =, , −2, 5, , 3 7, ∴ B , ,  19 19 , , 2, 22, −22, 5, tan θ =, =, =, 8, 3, −3, 1−, 5, 4+, , AB =, ,  −22 , φ = tan–1 , ,  3 , , =, , Sol 20: A(–2, 0) ; B(0, 5), 5, r =, mAB =, 2 AB, , 35 + 3r cosθ – 4r sinθ = 0, 1+, , D =  −2 + 29 cos θ,0 + 29 sin θ , , , , Sol 21:, , 35, (35)2 + (140)2, 17, =, 19, 19, , 3(2 + r cosθ) – 4(–7 + r sinθ) + 1 = 0, , C = 0 + 29 cos θ,5 + 29 sin θ , , , , C = [–5, 7] , , 2, 29, , ; cos θ =, , D = [–7, 2], , y −2, =m, x −3, , −5, 29, , 3 17, 4 17, cosθ –, sinθ = 0, 19, 19, , −52, tan θ = –4 or, 89, Equation of AC ⇒, , 17, , −52, y+7, = –4 or =, 89, x−2, , 52x + 89y + 519 = 0 or 4x + y = 1, , m=1/2, , m=–2/3, , m=m, , x–2y–3=0, , 2m − 1, = ±1, m+2, , 1, 3, , ⇒y – 2 = 3x – 9 ⇒ y = 3x – 7, , 1, & y – 2 = − (x – 3) ⇒ x + 3y = 9, 3, Sol 22:, , 19, , –3 cosθ + 4 sinθ =, , Sol 23:, , 1, 2, tan 45 =, m, 1+, 2, m−, , m = +3, m = −, , 2, , C = (2 + r cosθ, –7 + r sinθ), , 29, , −2, tan θ =, ; sin θ =, 5, , 2, , , , 3   7, + 7, 2 −  + , 19, 19, ,  , , , m=3/2, , 3x–2y–5=0, , Points of intersection is (1, –1) and both the lines, x – 2y – 3 = 0 and reflected Line are equally inclined to, normal on 3x – 2y 5 = 0, , 2, −2 1, −, 3 = 3 2, 2m, 1, 1−, 1−, 3, 3, m+

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8 . 5 6 | Straight Line, , ⇒, , 3m + 2, −7, =, 3 − 2m, 4, , ⇒ Slope of BC =, , 33, y+, y −b, 13, 2 = 13, ⇒, =, ⇒, x−a, 61, 61, x + 13, , ⇒ 2(6m + 4) = –21 + 14m, ⇒ 2m = +29, ⇒ m=+, , 29, 2, , ⇒ (y + 1) =, , 61 × 33, 2, , ⇒ 61y = 13x + 13×13 –, 29, (x – 1), 2, , ⇒ 26x – 122y – 1675 = 0, , ⇒ 2y + 31 = 29x, , Sol 26: The equation of line passes through point of, intersection of x+3y+y=0 and 3x+y+4=0 is, , Sol 24:, , 3x+y+4 + λ (x+3y+4) = 0, (-8, 12), , (λ+3)x + (1+3 λ)y + 4 + 4 λ =0, 4x+7y+13, , a+8, b − 12, =, = –2, 4, 7, , &, , Slope of line = ± 1, λ+3, =, 0, and, 1 + 3λ, , −, , λ+3, =, −1, 1 + 3λ, , ⇒ λ + 3 = - 1 -3 λ, or, ⇒ λ = -1 or, , 65, a+8, b − 12, =, = –2 ., 65, 7, 4, , ⇒ a = –16, , The obtained line is equally inclined to axes, then, , −, , (a, b), , ⇒, , 13, 61, , λ + 3 = 1 +3 λ, ⇒λ=1, , Eqn of line is (-1 + 3)x + (1 - 3)y + 4 - 4=0, ⇒ 2x − 2y= 0, , b = –2, , (1+3)x + (1 + 3)y + 4 + 4=0, , Image ≡ (–16, –2), Sol 25:, , ⇒ x= y and, , 4x + 4y + 8 =, 0, x+y+2 =, 0, , ⇒, ⇒, , A, , Sol 27: x(a + 2b) + y(a + 3b) – a – b = 0, 3x-2y+6, , 4x+5y=20, (1,1), , B, ,  10 84 , A=  , ,  23 23 , , a(x + y – 1) + b(2x + 3y – 1) = 0, Fixed point is the intersection of given 2 lines, , C, , x + y = 1 & 2x + 3y = 1, ⇒ y = –1; x = 2; P (2, – 1), , B = (a, b) ⇒ 3a – 2b + 6 = 0, , Sol 28:, , b −1, 5, =, ⇒ 4b = 5a – 1, a−1, 4, Solving these, we get, 6a + 12 = 5a – 1, a = –13 , , −33, b=, 2, , 84, −1, −61, AB slope = 23, =, 10, 13, −1, 23, , (2, -1), , 60, x+y=2, , Distance of vertex from line is, Length of base = 2 a, =2, , 2, 3, , −1, 2, , =, , 1, 2

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M a them a ti cs | 8.57, , sin 60º =, , 1, 2a, , 2, , ⇒a=, , 3, , 1, 1 1, − −, 3, = 3 7, m, 1, 1−, 1−, 3, 21, , 2, 3, , =, , m+, , Exercise 2, , ⇒, , Single Correct Choice Type, , 3m + 1, −10 3m + 1, −1, =, ⇒, =, 20, 3−m, 3−m, 2, , ⇒ 6m + 2 = m – 3 ⇒ m = –1, , Sol 1: (C) 3x – 8y – 7 = 0, 3(0) − 8( −1) − 7 =, 1, (A),  different sides, 3(0) − 8(0) − 7 =−7 , (B), , 3(0) − 8(1) − 7 =−15,  different sides, 3(3) − 8(0) − 7 =, 2 , , (C), , 3( −1) − 8(1) − 7 =−2 ,  same sides, 3(3) − 8(7) − 7 =−54 , , Sol 2: (A) 3x – 4y + 7 = 0;, ,  −1 7 , It passes through  , ,  10 10 , 7, y−, 10 = –1⇒ x + y = 7 – 1 ⇒ 5x + 5y = 3, 1, 10, 10, x+, 10, , Sol 5:(C) X = x – h = 3 – 4 = –1, Y = y – k = –5 + 3 = –2, , 12x + 5y – 2 = 0, , x y, + =1, a b, , p1p2 + q1q2 = 36 – 20 = 16 > 0, , Sol 6:(A), , 3x − 4y + 7, −(12x + 5y − 2), =, 5, 13, , 3, 4, +, = 1 are concurrent at a fixed point., 5a 5b, , 39x – 52y + 91 = –60x – 25y + 10, , Point is x =, , 99x – 27y + 81 = 0 ⇒ 11x – 3y + 9 = 0, Sol 3: (A), (5,3), , (1,2), A, , C, , B, , Sol 7: (D) P (1, 0) ; Q (–1, 0) ; R (2, 0), (x+1)2+y2+(x–2)2+y2 = 2(x–1)2+2y2, , (5,-3), , C’, , C’ is the reflection of C w.r.t. x-axis., ∴ Eq. of AB = Eq. of AC’, AC’, , 3 4 , ⇒  , , 5 5 , , 2SP2 = SR2 + SQ2, , , , , , ⇒ x2+1+2x+x2+4–4x=2(x2+1–2x), ⇒ 5 – 2x = 2 – 4x, 3, 2x = –3 ⇒ x = −, 2, Which is a straight line parallel to y-axis., , y +3 2+3, 5, =, = −, x −5 1 −5, 4, , Sol 8: (B) ax ± by ± c = 0, , ⇒ 5x + 4y = 13, , ⇒ ±, , Sol 4: (C), , x-7y+5=0 (m = 1/7), , , , , 3, 4, and y =, 5, 5, , x+3y=2 (m=-1/3), , y, x, ±, =, 1, ca cb, , 2c2, 1 c c , ∴ Area = 4  ⋅ ⋅  =, ab, 2 a b, Sol 9: (A) Lets assume 2 lines are x and y axis so the, distances from 2 lines are …. .

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M a them a ti cs | 8.59, , Eq. AD , , Sol 20: (D) x2 + y2 – 4xy = 0, , y=x+4, , Intersection point –1 – x = x + 4 ⇒ 2x = –5; 2y = 3, , m1 . m2 = 1, , Sol 16: (B) 2x – y + 5 = 0 , , m 1 + m2 =, , x + y –5 = 0 , , m = –1, , x – 2y – 5 = 0 , , m=, , m=2, , 1, 2, , 1, 1, −2, −1 −, 3, 2, ∑ tan θ =, +, + 2, 1, 1−2, 1+1, 1−, 2, , =3+3+, , ⇒ m1 = 2 –, m2 = 2 +, , +4, = +4, 1, 3, 3 & m3 = −1, , It forms an equilateral triangle., , 3, 3, 27, = 6+ =, 4, 4, 4, , Sol 17: (B) y = 0.2x, m = 0.2, tan θ = 0.2, tan 2θ =, ⇒y=, , 2 tan θ, 2, , 1 − tan θ, , =, , 0.4, 0.4, 5, =, =, 1 − 0.04, 0.96, 12, , 5x, 12, , Sol 18: (B) x2 + y2 – 10x – 14y – 151 = 0, M(–7, 2), , Sol 21: (D) y = x (A lies on line 1), , A(x, x), , y = 7x (B lies on this line  2 ), (y, 7y), OA = OB ⇒ x 2 = y 50, , 49 + 4 + 70 – 28 – 151 < 0, , x = 5y, , Point is inside, , Slope of line AB =, , A(–7, +2) and Centre (5, 7), r=, AC =, , 7y − x, 2y, −1, =, =, 2, y−x, −4y, , 25 + 49 + 151 = 15, 144 + 25 = 13, , (r – AC) = 2 (minimum distance), Sol 19: (D) ax2 + 2hxy + by2 = 0, Image by line mirror y = 0, , Previous Years’ Questions, Sol 1: (A) The point O(0,0) is the mid point of A(-a,-b), and B(a,b)., Therefore, A,O,B are collinear and equation of line AOB, b, is y = x, a, Since, the fourth point D(a2, ab) satisfies the above, equation. Hence, the four points are collinear., , Replace y by (–y), ax2 – 2hxy + by2 = 0, , Sol 2: (C) Given lines, x+2y–3=0 and 3x+4y–7=0 intersect, at (1,1), which does not satisfy 2x+3y–a=0 and 4x+5y–, 6=0. Also, 3x+4y–7=0 and 2x+3y–4=0 intersect at, (5,-2) which does not satisfy x+2y–3=0, 4x+5y–6=0. Lastly, intersection point of x+2y-3=0 and, 2x+3y-4=0 is (-1,2) which satisfy 4x+5y-6=0. Hence,, only three lines are concurrent.

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8 . 6 0 | Straight Line, , Sol 3: (C) Let B, C, D be the position of the point A(4, 1), after the three operations I, II and III respectively. Then,, B(1, 4), C(1+2, 4) ie, (3, 4). The point D is obtained from, C by rotating the coordinate axes through an angle π/4, in anticlockwise direction., Therefore, the coordinates of D are given by, , π, 1, = −, 4, 2, π, π, 7, and Y =3sin + 4 cos =, 4, 4, 2, x = 3cos, , 1+9 =, , 10 ,| BC |=, , and, , | CA |=, , 9 +1 =, , 10, , 7 + 6 3 − 1 13, ,, = ,1, 2, 2, 2, , Now, slope of PS = m =, , 2 −1, 2, = −, 2 − 13 / 2, 9, , Now, equation of the line passing through (1, -1) and, parallel to PS is, , Sol 10: (D) Let the vertices of triangle be A (1, 3) , B(0,, 0 ) and C (2, 0)., , Sol 4: (A) The point of intersection of three lines are, A(1, 1), B(2, -2), C(-2, 2)., , | AB |=, , ∴ S≡, , 2, y + 1 =− (x − 1), 9, ⇒ 2x + 9y + 7 =, 0, ,  1 7 , ∴ Coordinates of D are  −, ,, , 2 2, , , Now,, , Sol 9: (D) Since, S is the mid point of Q and R., , 16 + 16 = 4 2,, , ∴ Triangle is an isosceles, Sol 5: (A) By the given condition, we can take two, perpendicular lines as x and y axes. If (h, k) is any point, on the locus, then |h|+|k|=Sol 1 Therefore, the locus is, |x|+|y|=1 This consist of a square of side 1., Hence, the required locus is a square., , Here AB = BC = CA = 2., Therefore, it is an equilateral triangle. So the incentre, coincides with centroid., ∴, , 0 +1+ 2 0 + 0 + 3 ,  1 , I=, ,,  =  1,, , , , 3, 3, 3, , , , , Sol 11: (A) On solving equations 3x + 4y = 9 and y =, mx + 1, we get, , x=, , 5, 3 + 4m, , Now, for x to be an integer, 3 + 4m = ± 5 or ± 1, , Sol 6: (C) Orthocentre of right angled triangle is at, the vertex of right angle. Therefore, orthocentre of the, triangle is at (0, 0)., , The integral values of m satisfying these conditions are – 2, and – 1, , Let y cos x cos(x + 2) − cos2 (x + 1), Sol 7: (D)=, , Sol 12: (D) Let lines OB : y = mx, , = cos(x + 1 − 1)cos(x + 1 + 1) − cos2 (x + 1), 2, , 2, , 2, , = cos (x + 1) − sin 1 − cos (x + 1), , CA : y = mx + 1, BA : y = nx + 1 , , y = − sin2 1., , and OC : y = nx, , This is a straight line which is parallel to x-axis It passes, through ( π / 2, − sin2 1)., , The point of intersection B of OB and AB has x-coordinate, , Sol 8: (D) Slope of line x+3y=4 is -1/3, , 1, ., m–n, , y, , And slope of line 6x-2y=7 is 3.,  −1 , −1, Here, 3 ×   =,  3 , Therefore, these two lines are perpendicular which, show that both diagonals are perpendicular., , A, C, x’, , B, , D, O, , Hence, PQRS must be a rhombus., y’, , x

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M a them a ti cs | 8.61, , Now, area of parallelogram, , 2×, , OBAC = 2 × area of DOBA, 1, 1, 1, = 2 × × OA × DB = 2 × ×, 2, 2 m–n, , =, , 30, 8 14, =, k, +, =, k ⇒, 5, 5 5, , ⇒k=, 6, Sol 15: (A) Sides OA = OB = 2 and AB = 2 2, , 1, 1, =, m–n |m–n|, , X-coordinates of incentre of ∆ OAB, , Sol 13: (C) The equation of the line passing through, points P(-1, 0) and Q(0, 0) is;, Y = 0, , … (i), R(3, 3 3), , Q(0, 0), , P(-1, 0), , ≡, , Equation of the line passing through points Q(0, 0) and, R(3,, , 3 ) is;, , =, , y −0 3 3, =, x−0 3−0, , y, 3 3, ⇒=, x, 3, , ⇒ y =3x , , … (ii), , Therefore, the equations of the bisector of the angle, PQR is, y, 2, , 1, , =, ±, , 3x − y, 3+1, , 3x − y, ⇒ y =, 2, ⇒, , ⇒ y=, ±, , or, , y=, −, , 2 2 ×0 + 2 × 0 + 2 × 2, 2 2 +2+2, 4, , 2, =, = 2− 2, 4 + 2 2 2+ 2 2, ,  6 + 7 −1 + 3 , ,, Sol 16: (B) S ≡ , ;, 2 ,  2, , Slope of PS =, , 3x − y, 2, ,  13 , ≡  , 1,  2, , , 2 −1, 2, = −, 13, 9, 2−, 2, , 3x + y, 2, , 3x, =, − 3y 0 or =, 3x + y 0 ., , Sol 14: (C) Point T is given by, , Equation of line passes through having slope −, ,  3 × 2 + 2 × 1 3 × 4 + 2 × 1   8 14 , T ≡, ,,  ≡ ,, , 3+2, 3+2, ,  5 5 , , y + 1 =−, , 2, 9, , 2, ( x − 1) ⇒ 9y + 9 =−2x + 2, 9, , ⇒ 2x + 9y − 7 =, 0, Sol 17: (C) The given equation of lines are, 4ax + 2ay + c = 0 and 5bx + 2by + d = 0, Since, these lines intersect in fourth quadrant and point, is equidistant from axes, then, Which lies on the line 2x + y = k

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8 . 6 2 | Straight Line, , The point can be of form (k, -k), k > 0, , On solving, we get, , ⇒ 4ak − 2ak + c =, 0 and 5bk − 2bk + d =, 0, , B ≡ (1, 2 ), , ⇒ 2ak + c =, 0 and 3bk + d =, 0, , We know that diagonals in rhombus bisect each other, so, D ≡ ( −3, − 6 ), , ⇒−, , c, d, =−, 2a, 3b, , Since BC || AD, , ⇒ 3bc − 2ad =, 0, Sol 18: (B) (2x - 3y + 4) + k (x – 2y + 3) = 0, Is a family of equation passes through T (1, 2), , ⇒ Equation of AD 7x − y + λ1 = 0 , Passes through, (-3, -6), ⇒ 7x − y + 15 =, 0, , From figure, , Similarly, AB || DC, , PT = P’T, ⇒ PT2 =, (P' T ), , 2, , ⇒ (h − 1 ) + (k − 2 ) =( 2 − 1 ) + ( 3 − 1 ), 2, , 2, , 2, , 2, , ⇒ (h − 1 ) + (k − 2 ) = 1 + 1 = 2, 2, , 2, , ⇒ ( x − 1) + ( y − 2) =, 2, 2, , 2, , Locus is a circle., Sol 19: (C) Equation of AB, x + y = 41, On line x = 1, there are 39 points inside ∆ OAB, Similarly, , ⇒ Equation of DC x – y + λ2 = 0, passes through, (-3, -6), ⇒ x − y −3 =, 0, 1,  7, 8, 4, ⇒ C  , −  and A  − , − , 3, 3, 3,  3, , JEE Advanced/Boards, Exercise 1, Sol 1: OA = 2AB = 4BC, A, , 1, , S, (7/2, 5/2), , G=, , 2, , G, , H, (2, 4), , 2+7 4 +5, ,, = (3, 3), 3, 3, , A = (3, 3) = (a, a), On line x = 2, there are 38 points inside ∆ DAB, Total points, = 39 + 38 + ….. + 2 + 1, =, , 39 ( 40 ), 2, , = 780, , Sol 20: (B) Let two sides AB and BC be x – y + = 0 and, 7x – y – 5 = 0 respectively., ,  3a, a, B =  ,a − , 2, 2, , , , , a a, a a, C =  a + + ,a − + , 2, 4, 2, 4, , , , a a a, a a a, ⇒  a + + + + ...,a − + − ....x , 2 4 8, 2 4 8, , 

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M a them a ti cs | 8.63, , A = (–2, 3), , , ,  a, , a , 2a , ,,  = 2a, , N= , 3, 1 − 1 1 + 1 , , , , , 2, 2, , α+β=, , … (i), , Eq. of AC is x + 2y = 4 , , … (ii), , 3 5, S=  , , 2 2, ⇒ Eq. of perpendicular bisector of AB and AC are, , 8a, =8, 3, , Sol 2:, , y = x + 1, , A(-3,1), , y = 2x –, , , , 1, , 2, , 2x+y=3, , Eq. (ii) and (iv) ⇒ F = (1, 3/2), , [Midpoint of A, C], , (2+3,2b-1), , ⇒ B = (2, –1) and C = (4, 0), , 7(2α + 3) – 4(2β – 1) = 1, 14α – 8β + 24 = 0 ⇒7α – 4β + 12 = 0, 2α + β – 3 = 0, α = 0, β = 3 ⇒ C = (3, 5), , 2, 3, , Eq. of median to AB is, , Sol 4:, , A(-8,5), , 7 2, 7, −, 4 = 4 3 ⇒ 4m − 7 = 13 = 1, 7m, 7 2, 26, 2, 7m + 4, 1+, 1+ ., 4, 4 3, ,  , m, , 8m – 14 = 7m + 4 ⇒ m = 18, , y −5, 2, =, Equation of AC =, x −3, 3, ⇒ 3y – 9 = 2x, , y −5, = 18, x −3, , ⇒ 18x – y = 49, 2x + y = 3;, x=, , y −0 1−0, =, ⇒ x + 4y = 4, x−4 0−4, , Eq. of median to AC is, 3, +1, y +1, = 2, ⇒ 5x + 2y = 8, 1−2, x−2, , m−, , Equation of BC =, , … (iv) respectively., [Midpoint of A, B], , 7x-4y=1, B m, , … (iii), , Eq. (i) and (iii) ⇒ E = (0, 1), , (,), , Slope of AC =, , ⇒ Eq. of AB is x + y = 1 , , 18x – y = 49, , 26, −11, ,y=, 10, 5, ,  26 −11 , B=  ,, ,  10 5 , , B, (-15, -19), , mAB =, , 24, 7, , mAC =, , 12, −4, =, −9, 3, , C, (1, -7), , 4, 24, m+, −m, 3, 7, =, 24m, 4m, 1+, 1−, 7, 3, , ⇒, , 3m + 4, 24 − 7m, =, 3 − 4m, 24m + 7, , −11, −1, y −1, −16, −4, Equation of AB =, = 5, =, =, 26, x+3, 28, 7, +3, 10, , ⇒ 72 – 117m + 28m2 = 72m2 + 117m + 28, , ⇒ 7y – 7 = –4x – 12 ⇒ 4x + 7y + 5 = 0, , ⇒ (11m – 2) (2m + 11) = 0, , Sol 3: x – y = 4 [Line perpendicular to AB], 2x – y = 5 [Line perpendicular to AC], , ⇒ 44m2 + 234m – 44 = 0, ⇒ 22m2 + 117m – 22 = 0, ⇒ m = 2/11, –11/2, y −5, 2, 11, =, or –, x+8, 11, 2

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M a them a ti cs | 8.67, , Slope of, , −2, x y, = –1, + =1⇒, 2, 2 2, , Range (–1, 1/5), (∴ Third line is go through (0, 0) and for triangle parallel, x y, to + = 1 meet at infinity), 2 2, , 1,  −1,  = (a, b), 5, , , 1 , ⇒  a + 2  = –1 + 52 = –1 + 15 = 24, b , , , Sol 14: 2x + 3xy – 2y – 10x + 15y – 28 = 0, 2, , 2, , y = 3x, , 1, 1, 4, Area =, 2, 1, −, 5, , , , = 1 1  12 − 18  − 3  4 + 1  + 1  18.4 + 1.12  , 2 , 5 , 5, 5, 5, , , , , =, , 1  42 3.21 72 + 12 , −, +, , , 2 5, 5, 5 , , =, , 63, 1  84 + 42 − 63 , ,  = 10, 2, 5, , , ⇒ a1 =, , 9 + 81 = 3 10, , ⇒ a2 =, , 36 9, =, +, 25 25, , ⇒ a3 =, ,  21  , 18 ,   +  12 −  =, 5 ,  5  , , (-1/5,18/5), C, A, , B, , (4,12), , (1,3), , 2x2+9x2–2(9x2)–10x+15(3x)–28 = 0, –7x2 + 35x – 28 = 0, x2 – 5x + 4 = 0 ⇒ x = 1, 4, ∴ y = 3, 12, , ∂p, = 4x + 3y – 10 = 0, dx, ∂p, = –4y + 3x + 15 = 0, dy, , 25y = 90 ⇒ y =, , x=, , 18, 5, , 54, 5 = −1, 5, 4, , 10 −, , 3, −1, ⇒ m1 = 3, m2 = 5 =, ,, −6, 2, 5, 18, 12 −, 42, 5, =, =2, ⇒ m3 =, 1, 21, 4+, 5, , 3−2, 1, =, 1+6, 7, 1, 3+, 2 = –7, ⇒ tanq2 =, 3, 1−, 2, 1, 50, +0=, ⇒ cotq1 + cotq2 + cotq3 = 17 +, 7, 7, ⇒ tanq1 =, , 3 1, 12 1, 18, 1, 5, , 45 =, , 2, , 3, 45, 5, , 2, , 21, 441 + (42)2, 5, =, 5, 25, , Incentre will be, 1 × 21 5 4 × 3 5 1, +, − × 3 10, 5, 5, 5, ,, 15 10 3 5 21 5, +, +, 5, 5, 5, 3 × 21 5 12 × 3 5 18, +, +, 10, 5, 5, 8, 24 5 + 15 10, 5,  11 5 − 10 33 5 + 18 10 , = , ,, ,  8 5 + 5 10 8 5 + 5 10 , , Radius = distance of incentre from any of the sides., 4 − 3x, 10, , =, , =, , 18 10 + 3 10, 10(8 5 + 5 10 ), , =, , 21, 8 5 + 5 10, , 3, 21(8 5 − 5 10 ), (8 5 − 5 10 ), =, 10, 70, , Sol 15: y = mx + c,  y − mx ,  y − mx , 3x2–y2– 2x ,  +4y ,  =0,  c ,  c , 3x2–y2 –, , 2xy 2mx2 4y 2 4mxy, +, =0, +, −, c, c, c, c

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M a them a ti cs | 8.71, , Sol 26:, , ⇒ 6m2 + 16m − 6 =, 0, ⇒ 3m2 + 8m − 3 =, 0, , P, , ⇒ 3m2 + 8m − 3 =, 0, ⇒ 3m2 + 9m − m − 3 =, 0, , Q, , ⇒ (3m− 1)(m+ 3) =, 0, ⇒m=, −3, 1/ 3, Equations, 3x + y=, + 7 0 or x − 3y −, =, 31 0, , P(3, 2), Q(3, –2), R(2, –3), S = (–2, 3), , 3 2, 3 −2, 1, 2 −3, Area of PQRS will be, A =, 2, −2 3, 3 2, 1, 1, = [–6–9+6–4–(6–4+6+9)] = [–13 – 17] = 15 units, 2, 2, , Sol 25:, A(1,-2), m=-1, , E, , D, , m=2, C, (,2-4), x-y+5=0, , (,-1-)B, x+2y=0, , (, , -10 5, ,, 3 3, , ), , Sol 27:, l3, , y+2, Eq. of AB =, = –1, x −1, , l4, , (0,1), A, , y+2=1–x, , (1,0), , (0,0), , (-1,0), , B, , C, , x+y+1=0, Eq. of AC =, , y=-x, , R, , D, l2, , y+2, =2, x −1, , y + 2 = 2x – 2 ⇒ y + 4 = 2x,  α + 1 −α − 3 , D→ , ,, ;, α ,  α, ,  β + 1 2β − 6 , E→ , ,, , 2 ,  2, , l1, , y – 1 = mx → 1, y = m(x + 1) →  2, y=, , −1x, → 3, m, , α +1  α +3, +,  +5 = 0, 2,  2 , , y=, , −1(x − 1), → 4, m, , α + 7 = 0 ⇒ α = –7, , 1 intersection  3 → mx + 1=, ,  2β − 6 , β +1, = −2 , , 2,  2 , , x=, , β + 1 = –4β + 12;, , β=, , 11, 5, , −m, 2, , 1+m, , ,y=, , 1, 1 + m2, , −(x − 1), 1 intersection  4 → mx +1 =, m, −x 1, +, m m, ,  11 2 , (–7, 6)  , ,  5 5, , mx + 1 =, , 28, −14, y −6, =, =, −46, 23, x+7, ,  m2 + 1 , 1−m, , x =, m,  m , , 23y + 14x = 138 – 98, 14x + 23y = 40, , −x, m, , x=, , 1−m, 1 + m2

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8 . 7 2 | Straight Line, , y=, , m − m2, 1 + m2, , +1 =, , m+1, , ⇒ x1 + x2 + x3 =, 5 , , … (iii), , m2 + 1, , ⇒ y1 + y 2 + y 3 =, 8 , , … (iv), ,  1−m m+1 , ,, B 2, ,  m + 1 m2 + 1 , , From (i), (ii), (iii) and (iv), we get, , 2 intersect 3 → m(x + 1) =, x, = –m, m, , mx +, x=, y=, , ⇒ 4 (5 − x1 ) − 3 ( 8 − y1 ) + 6 =, 0, , ⇒ 20 − 4x1 − 24 + 3y1 + 6 =, 0, , 2, , −m, , Now AH ⊥ BC, ,  −m2, m , ,, C , , 2, 2, 1 + m2,  1 + m 1 + m , m, , ⇒ mx +, , y1 − 2, x1 − 1, , −1, (x − 1), m, , x, 1, =, –m, m, m, , 4, =, −1, 3, , ⇒ 3x1 − 4y1 =, 11, , … (vi), , From (v) and (vi), we get, , 2, , 1+m, ,  41 38 , ,, ,  25 25 , , (x, ( x1,1, yy11)) = , ,  2 , y = m, 2 , 1 +m , , 2, , Radius = AO=, ,  (1 − m2 ) 2m , ,, ⇒ D, , 2, 2,  1 + m m + 1 ,  m−1, Area of square = ,  1 + m2, AB = BD, , 2, , =, , , (m − 1)2,  =, m2 + 1, , , 58, =, ( 25), , , 41  , 38 , 2 −,  + 3 −, , 25  , 25 , , , 2, , m, n, , =, ⇒ m 58, =, n 25 ⇒ m + n =, 83, , m2 + m4 = 2m(1 + m2), , Sol 29: (–6, 1) (6, 10), (9, 6), (–3, –3), , m2 = 2m, , Area =, , ×, , ⇒ 4y1 − 8y =, −3x1 + 3, , (1 − m2 ), , m=2, , … (v), , ⇒ −4x1 + 3y1 + 2 =0, , 1 + m2, , 2 intercept  4 ⇒ m(x + 1) =, , x=, , ⇒ 4 ( x 2 + x3 ) − 3 ( y 2 + y 3 ) + 6 =, 0, , −x, m, , (6, 10), , 1, p, =, 5, q, , A, (-6, 1), , p+q=6, , E, , B, C, (9, 6), , (1,0), , Sol 28:, , 2, H(1,2), , 1, , 5 3, Centroid (c) =  , , 3 3, Let coordinates of vertices of, ∆ABC be A (x, y), B(x2, y2), C (x3, y3), , 4x2 - 3y2 + 3=0 , , … (i), , 4x3 - 3y3 + 3=0 , , … (ii), , and, , x1 + x2 + x3, 3, , =, , D, (-3,-3), , P(2,3), , C, , 5, 3, , Area of rectangle – Area of ∆ =a / b, , A, , Area of rect =, B, , C, 4x-3y+3=0, , Eq. of CD, , 81 + 144, , y +3, 3, =, x+3, 4, , Point E = (1, 0), Eq. of AD =, −9, − 3 = x;, 4, , y +3, −4, =, x+3, 3, x=, , −21, 4, , 16 + 9 = 15. 5 = 75

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M a them a ti cs | 8.75, , = (2pq – 1)(px + qy – 1), , ⇒ 4β – 8 = 3α –, , = (p2 + q2 – 1) (qx + py – 1), ,  27 36  25β − 36, 4, =, Mid point of AC=  ,  =, 25, 25, 3, ,  25α − 27, (Slope of ⊥ to AC), , Sol 4: (B) 3x + 4y = 9 × 4, ⇒ 12x + 16y = 36, 4x – 3y + 12 = 0 × 3, , ⇒ 100α – 108 = 75β – 108 ⇒β =, , ⇒12x – 9y = –36, , α=, , 25y = 72, y=, , 72, 25, , ⇒x =, , 4 × 72, 4 × 24, 21, 25, = 3−, = −, 25, 3, 25, , 4α, 3, , 3, β=2, 2, , Circumradius =, , 9−, , 9, 7, ⇒ 3α – 4β +, =0, 2, 2, , 9, 5, +4 =, 4, 2, , a 1 1, Sol 5: (D) 1 b 1 = 0, 1 1 c, , A(3,0), , a(bc – 1) – 1(c – 1) + 1(1 – b) = 0, (,), (0,4)B, , [, , 21 72, C - 25 , 25, , ], ,  21 72 , A(3, 0) B(0, 4]C  − 25 , 25 , , , , 1, DPAB =, 2, , 3, 0, 21, −, 25, , 0 1, 4 1, 72, 0, 25, , =, , , 1 , 72 , 84  , 3 4 −,  + 10 +, , 2 , 25 , 25  , , , =, , 84, 1  3 × 28 84 , +, ,  = 25, 2  25, 25 , , 72, −3, −72, 25, mAC =, =, =, 4, 21, 96, −, −3, 25, mBC =, , 72, 25 = 24 = 8, 21, 7, 21, 25, , 4−, , mAB = −, , 4, 3, , 3 , β−2, 3, Mid point of AB  ,2 ⇒, = (Slope of ⊥ AB), 3, 4, 2 , α−, 2, , abc – a – b – c + 2 = 0, a + b + c = abc + 2, , 1, 1, 1, +, +, 1−a 1−b 1−c, =, , 1 + bc − b − c + 1 + ab − a − b + 1 + ac − a − c, (1 − a)(1 − b)(1 − c), , =, , 3 + ab + bc + ca − 2a − 2b − 2c, 1 − a − b − c + ab + bc + ca − abc, , =, , 3 − 2(abc + 2) + ab + bc + ca, 1 − abc − (abc + 2) + ab + bc + ca, , =, , −1 − 2abc + ab + bc + ca, =1, −1 − 2abc + ab + bc + ca, , Sol 6: (D) A(a, 0) ; B(0, b) ; C (c, 0) ; D(0, d), , d, c, =, lines are not parallel i. e. not a trapezium, not, a, b, a parallelogram, OA OC = OB OD (For concyclic points), Assuming origin as centre, ac = bd i. e. ABCD are concyclic., Sol 7: (D) 7x2 + 8y2 – 4xy + 2x – 4y – 8 = 0, Homogenising,  3x − y , 7x2 + 8y2 – 4xy + 2x , ,  2 

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M a them a ti cs | 8.77, , ⇒ 2m +, , mAB. mBC = –1, , 2h 2h 2, 2a, m =, m, –, b, b, b, , mAB =, , 9 − 2x, x−4, , hm2 + m(a – b) – h = 0, , mBC =, , 13 − 2x, x−2, , Sol 13: (D), , (9 – 2x)(13 – x) = –1(x2 – 6x + 8), , ⇒ 2mb – 2am = –2h + 2hm, , 2, , 117 – 18x – 26x + 4x2 = –x2 + 6x – 8, , 3x-y=-3, , 5x2 – 50x + 125 = 0, , (0,3), , x2 – 10x + 25 = 0 ⇒ x = 5; y = 0, , 2x-3y = 6, , Area =, (-1,0), , 1, 1, × BC × AB =, 2, 2, , 9+9, , 1+1 = 3, , (3,0), (0,-2), , Multiple Correct Choice Type, , 3x+4y = 12, , Sol 16: (B, D) Let the co-ordinate of the vertex C is (h,, h – 2) then area of triangle will be, , From figure, α ∈ [–1, 3], , −5, 1, ⇒ 3, 2, h, , β ∈ [–2, 3], Sol 14: (C), , ⇒ (h – 2)8 = ± 40, , A, , ⇒ h – 2 = ±5, ⇒ h = 7 or –3, , x+2y=16, , 2x+3y=29, , 0, 1, 0, 1 =, 20, h− 2 1, , ∴ Co-ordinate are (7, 5) or (–3, –5), (,)B, , C(10-,12-), , (5, 6), , Sol 17: (B, C) y +, , 2α + 3β = 29, , y–, , 10 – α + 24 – 2β = 16, α + 2β = 18 , , …(i), , 2α + 4β = 36 , , …(ii), , 3x = 2, , x = 0, y = 2, -1/3, , On solving, we get, , m, , A, 4/3, , , 1/3, , 60, 2/3, A, , β = 7, α = 4, So we have, B(4, 7), y −7, Eq. of BC =, = –1 ⇒ x + y = 11, x−4, , 3, , A(4,1), , B, (x,10-2x), , 2/3, , 1, , Sol 15: (B), , 1+, C(2,-3), 2x+y=0, , 3x = 2, , ⇒, , −m, m, 3, , m+, =, , 1, , 1 − 3m, 3, ⇒, =, m, m+ 3, 1−, 3, , 3m + 1, 3 −m, , 3 – m –3m+m2 3 =m2/ 3 +m+3m+ 3

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8 . 7 8 | Straight Line, , ⇒m=0, , B(0, b), , Bisector line is y = 2, ⇒ cos 60º =, , P, 4/ 3, , ;, , (, ), , 2, , P=, , A’(-a’, 0), , 3, ,  2 , ,2 , ⇒ Given point is ,  3 , , B’(0, b’), , Or the other possibility is m = ∞, ⇒ Foot of perpendicular = (0, 0), Sol 18: (A, B, D), , m(AB) =, , −b, a, , m(BA′) =, , b, a′, , Let orthocentre be at (α, β), , b−β, =∞⇒α=0, 0−α, , A(1,3), , P(x,y), (5,6)B, , C(-1,2), , y −6 3, AB =, x −5 4, 4y – 24 = 3x – 15 ⇒ 3x – 4y + 9 = 0, BC -, , β−0, a, =, α + a′, b, β=, , aa′, = b′, b, , [From (i)], , Sol 20: (A, B, C, D), C(0,a), , y −6, 2, =, x −5, 3, , B(x,x/3), 60, , 3y – 18 = 2x – 10 ⇒ 2x – 3y + 8 = 0, AC -, , (0,0), , y −3, 1, =, x −1, 2, , 2y – 6 = x – 1 ⇒ x – 2y + 5 = 0, C and P on same side, 3(–1) – 4(2) + 5 < 0 3x – 4y + 9 < 0, B and P on same side of AC, 5 – 2(6) + 5 < 0 x – 2y + 5 < 0, A and P on same side of BC, 2(1) – 3(3) + 8 > 0 2x – 3y + 8 > 0, So we can see [A,B,D] are correct., , x y, Sol 19: (B, C) + = 1A (a, 0) & B(0, b), a b, x y, = 1 C(–a′, 0) D(0, –b′), line +, a′ b′, AB A′B′ are concyclic, i. e. OAOA′ = OBOB′, aa′ = bb′, b′ =, , aa′, b, , A(0, a), , (0,-a), , 30, -30, x2 + x2/3 = a2, x=3a/2, , C either lies on y axis so C(0,a) or C(0,–a), or on y = −,  −1 , C  x,, , 3x , , ⇒x2 +, , 1, 3x, , x2, = a2, 3, , ±a 3, 2, a, y= , 2, x=, ,  3a −a , ⇒ (0, a), (0, –a), , , ,  2 2 , , , ,  − 3a a , , , ,  2 2, , 

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M a them a ti cs | 8.79, , Sol 21: (B, C) Vertices are (4, 3)(1, –1) & (7, k), , Sol 23: (B, C) (x + 2) (y + 2) = 0; x + y + 2 = 0, Circumcentre lies on mid-point of hypotenuse i. e. AB, , A(4,3), , ( −2,0) + (0, −2), = (–1, –1), 2, x=-2, , y, , C(7,k), , (1,-1)B, , A, ,  2+k , C =  4,, , 3 , , , x, , That occurs only in isosceles ∆, , B, , AC = AB, , y=-2, , C, , (1 – 4)2 + (3 + 1)2 = (k – 3)2 + 9, , y=-x-2, , Sol 24: (B, C) x + y = 1; x = 7y; x = 3y, , 16 = (k – 3)2, , Centroid and In centre always Lie inside of the triangle., , ⇒ k – 3 = ±4, ⇒ k = 7, –1 or, , (7/8, 1/8), , 25 = (k + 1) + 36, 2, , ≠ not possible (AB = BC), , (3/4, 1/4), , (k – 3) + 9 = (k + 1) + 36, , or, , 2, , 2, , (BC = AC), , (0,0), , Sol 25: (B, C), , 18 – 6k = 37 + 2k, , y+2x=5, , −19, ⇒ 8k = –19 ⇒ k =, 8, , m, , For k = 7AB = 5, AC = 5, BC = 10, [∆ is not possible], , tan θ =, , −19, 8, , So k = –1 or, , m+2, 1, =±, 1 − 2m, 2, , Sol 22: (A, C) (7, 4), , 2m + 4 = ±(1 – 2m), , y – 4 = m(x – 7), , ⇒ 2m + 4 = 2m – 1 ⇒(m = ∞), , Centre ≡ (3, –2), , ⇒, , (3 − 7)m + 4 + 2, 1 + m2, , m − ( −2), 1 + m( −2), , ⇒ x = 2 , , 2m + 4 = 1 – 2m, , =4, , ⇒ (–4m + 6) = 16(1 + m ), 2, , (B), , 2, , ⇒ 9 + 4m2 – 12m = 4 + 4m2, 5, or m = ∞, 12, 5, ⇒ y–4=, (x – 7), 12, , ⇒ m=, , ⇒5x – 12y + 13 = 0 and x = 7, , 4m = –3 ⇒ m = −, y −3, −3, =, x−2, 4, , 3x + 4y = 18, , (C), , 3, 4

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8 . 8 0 | Straight Line, , Sol 26: (A, C),   , , 5, y, y=-2x, , (1,0), /4, /6, , x’, L1, , y – a = m(x – 1), , L3, , P, , − sin75º, m = – tan 75º =, cos75º, =, , ( 3 − 1), , = – (2 + 3), , x, , O, , (-2,-2), , y = m(x – 1), , −( 3 + 1), , y=x, , L2, , /2, , /2, , 45, , 2 2, , =, , Q, , R, , y=-2, , (1,-2), , y’, , Sol 2: (A) Solving equations L1 and L2., , y, x, 1, ⇒ = =, –36 + 10 –25 + 12 2 – 15, , y = – (2 + 3) (x – 1), , ∴, , (2 + 3) x + y = (2 + 3) (C), , L1, L2, L3 are concurrent if point (2, 1) lies on L2, , x + (2 − 3) y = 1, , ∴, , Sol 27: (B, C) y – y1 = m(x – x1), , (B) Either L1 is parallel to L2 or L3 is parallel to L2, then, , y = y1 + m(x – x1), (B) Set of parallel lines, (C) All these lines pass through x = x1, , x = 2, y = 1, 6 – k – 1= 0 ⇒ k = 5, (A) → (S), , , ⇒, , 1, 3, 3 –k, =, or =, 3 –k, 5, 2, , k = – 9 , , or k =, , –6, 5, , (B) → (p, q), (C) L1, L2, L3 form a triangle, if they are not concurrent,, or not parallel., ∴, , k ≠ 5, – 9, –, , 6, 5, ⇒k=, 5, 6, , (c) → (r), (D) L1, L2, L3 do not form a triangle. If, , Previous Years’ Questions, Sol 1: (C) It is not necessary that the bisector of an, angle will divide the triangle into two similar triangle,, therefore, statement-II is false., Now we verify statement-I, ∆ OPQ, OR is the internal bisector of ∠POQ, ∴, , PR OP, =, RQ OQ, , ⇒, , PR, 22 + 22, =, RQ, 12 + 22, , k = 5, – 9, –, , 6, 5, , (D) → (p, q, s), Sol 3: (A, C) Given lines px + qy + r = 0, qx + ry + p = 0, and rx + py + q = 0 are concurrent., ∴ , , p q r, q r p =0, r p q, , Applying R1 → R1 + R2 + R3 and taking common from R1, ⇒, , 1 1 1, (p + q + r) q r p, r p q

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M a them a ti cs | 8.81, , ⇒, , (p + q + r) (p2 + q2 + r2 – pq – qr – pr) = 0, , ⇒, , p3 + q3 + r3 – 3pqr = 0., , Given plane is, x – 2y +z = d , , Sol 4: (A, C) Since, 3x + 2y ≥ 0, , ….(i), , Where (1, 3) (5, 0) and (– 1, 2) satisfy Equation (i), ∴, , …….(ii), , Equations (i) and (ii) are parallel., Clearly, A = 1, Now, distance between plane, , Option (a) is true., , , =, , Again 2x + y – 13 ≥ 0, is not satisfied by (1, 3),, , ⇒ , , ∴ Option (b) is false., 2x – 3y – 12 ≤ 0,, , d, =, 1+ 4 +1, , 6, , |d| = 6, , Sol 7: x2 + y2 ≤ 6 and 2x – 3y = 1 is shown as, , is satisfied for all points,, ∴ Option (c) is true., , L, , And – 2x + y ≥ 0,, is not satisfied by (5, 0),, ∴, , 1/2, , Option (d) is false,, , 1/3, , Sol 5: (B, C) Let equation of line L1 be y =mx. Intercepts, made by L1 and L2 on the circle will be equal ie, L1 and L2, are at the same distance from the centre of the circle., Centre of the given circle is (1/2, – 3/2). Therefore,, m 3, +, |1 / 2–3/ 2–1|, = 2 2, 1+1, m2 + 1, ⇒, , 2, 2, , =, , |m+3|, 2 m +1, , 8m + 8 = m2 + 6m + 9, , ⇒, , 7m2 – 6m – 1 = 0, , ⇒, , ∴ For any point in shaded part L > 0 and for any point, inside the circle S < 0.,  3,  2,  , L : 2x – 3y – 1,  4, 9, 3, L : 4 – – 1=, >0, 4, 4, , Now, for, , and S : x2 + y 2 – 6, S : 4 +, , 2, , ⇒, , For the point to lie in the shaded part, origin and the, point lie on opposite side of straight line L., , 9, –6<0, 16, , ⇒, ,  3,  2,  lies in shaded part.,  4, , (7m + 1) (m – 1) = 0, , For, , –1, =, ,m 1, 7, , 5 3,  ,  L : 5 – 9 – 1 < 0 , 2 4, , For, , 1 1, 1 3,  ,–  L : + – 1 > 0, 2 4, 4 4, , ∴, , 1 1,  –  lies in the shaded part., 4 4, , For, , 1 1 , 1 3,  ,  L : – – 1 < 0 (neglect), 4 4, 8 4, , ⇒, , Only 2 points lie in the shaded part., , 2, , ⇒=, m, , Thus, two chords are x + 7y = 0 and y – x = 0 Therefore,, (b) and (c) are correct answers., Sol 6: Equation of plane containing the given lines is, x –1 y –2 z –3, 2, 3, 4 =0, 3, 4, 5, , ⇒, , (x – 1)(– 1) – (y – 2)(– 2) + (z – 3)(– 1) = 0, , ⇒, , – x +1 + 2y – 4 – z + 3 = 0, , ⇒, , x + 2y – z = 0 , , ……(i), , (neglect), , Sol 8: Note: d : (P, Q) = | x1 – x2 | + | y1 – y2 |. It is new, method representing distance between two points, P and Q and in future very important in coordinate, geometry.

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8 . 8 2 | Straight Line, , Now, let P(x, y) be any point in the first quadrant. We, have, + | y | = x + y ( x,, , d(P, 0) = | x – 0 | + | y – 0 | = | x |, y > given), d (P, A) = |X – 3| + |Y – 2| (given), ,  1, m , P≡, ,, , m+1 m+1,  3, 3m , Q≡, ,, , m, +, 1, m, +1, , y, , d(P, 0) = D(P, A), ⇒, , x + y = |x – 3| + |y – 2|, , L1, , Case I : When 0 < x < 3, 0 < y < 2, , R, L2, , In this case I Eq.(i) becomes, x + y= 3 – x + 2 – y, , x + y = 3 – x + y – 2 , 1, 2, , (1/2,2), , 1, , m–2, , m+1, , …. (i), , And equation of L2 : y + 3x =, , 3m + 9, , m+1, , ….(ii), , Sol 10: Let the square S is to be bounded by the lines x, 1, 1, and y = ±, =±, 2, 2, 2, 2, , , 1 1, =, a2  x1 –  +  – y1 , We have,, 2 2, , , , Infinite segment, x=1/2, y=2, , x+y=5/2,, Finite segment, , 2, , Now, equation of L1 : y – 2x =, , By eliminating ‘m’ from Equ. (i) and (ii), we get locus of, R as x – 3y + 5 = 0, which represents a straight line., , Case III : When x ≥ 3, 0 < y < 2, y, , y, A(x1,1/2) 1/2, , x, O, , 1/2, , (5/2, 0), , 3, D(-1/2,y2), , Now, Eq.(i) becomes x + y = x – 3 + 2 – y, ⇒ 2y = – 1, , or, , y= –, , 1, 2, , Case IV : When x ≥ 3, y ≥ 2, In this case (i) changes to x + y = x – 3 + y – 2, , c, , = x12 – y12 – x1 – y1 +, , Hence, this solution set is {(x, y)} | x = 12, y ≥ 2} ∪ {(x, y)} |, , Sol 9: Let the equation of straight line L be, , 1/2, , x, B(1/2, y1), , O, , y’, , Which is not possible., , The graph is given in adjoining figure., , a, , b, C(x2-1/2) -1/2, , ⇒0=–5, , x + y = 5/2, 0 < x < 3, 0 < y > 2 }, , d, , x’, -1/2, , Hence, no solution., , y = mx, , x+y=3, , x+y=1, , Now, Eq. (i) becomes, , ⇒x=, , x, , O, , Case II : When 0 < x < 3, y ≥ 2, , ⇒ 2x = 1 , , L, , P, , 5, ⇒x+y=, 2, , ⇒ 2x + 2y = 5 , , Q, , 1, 2, , Similarly,, =, b2 x22 – y12 – x2 + y1 +, c2 = x22 – y 22 + x2 + y 2 +, , 1, 2, , d2 = x12 – y 22 + x1 + y 2 +, , 1, 2, , ∴, , 1, 2, , a2 + b2 + c2 + d2= 2(x12 + y12 + x22 + y 22 ) + 2, , Therefore 0 ≤ x12 , x22 , y12 , y 22 ≤, , 1, 4

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M a them a ti cs | 8.83, , Sol 11: Let the vertices of a triangle be, O(0, 0) A(a, 0), and B(b, c) equation of altitude BD is x =b., , 0 ≤ x12 + x22 + y12 + y 22 ≤ 1, , ⇒, , 0 ≤ 2(x12 + x22 + y12 + y 22 ) ≤ 2, , But 2 ≤, , 2(x12, , +, , x22, , +, , y12, , +, , y 22 ) + 2, , Slope of OB is, , ≤4, , c, ., b, , Slope of AF is –, , Alternate Solution, 2, c=, x22 + y 22 , , ….(i), , b, ., c, , Now, the equation of altitude AF is, y, , y, , B(b,c), , A(x1,1), a, , d, , F, , B(1,y1), , P, , (0,y2)D, b, , c, x’, , 90, , O, , x’, , O, , x, , C(x2,0), , 90 E, , D(b,0) A(a,0), y’, , b, y – 0 = – (x – a), c, , y’, , =, b2 (1 – x2 )2 + y12 , , …. (ii), , Suppose, BD and OE intersect at P., , =, a2 (1 – y1 )2 + (1 – x1 )2 , , …..(iii), , ,  a – b , Coordinates of P are b,b , ,  c , , , 2, , d=, , x12, , 2, , + (1 – y 2 ) , , .....(iv), , On adding Eqs. (i), (ii), (iii) and (iv), we get, 2, , 2, , 2, , 2, , a +b +c +d =, , {x12, , 2, , + (1 – x1 ) }, , + {y12, , 2, , + (1 – y1 ) }, , + {x22 + (1 – x2 )2 } + {y 22 + (1 – y 2 )2 }, Where x1, y1, x2, y2 all vary in the interval [0, 1]., Now, consider the function y = x2 + (1 – x)2, 0 ≤ x ≤ 1, , dy, =, 2x – 2(1 – x) . For maximum, Differentiating ⇒ ⇒, dx, dy, =0., or minimum, dx, ⇒, , 2x – 2(1 – x ) = 0 ⇒ 2x – 2 + 2x = 0, , ⇒, , 4x = 2, , Again,, ⇒, , d2 y, dx2, , ⇒x=, , x, , Let m1 be the slope of OP =, , a–b, c, , and m2 be the slope of AB =, , c, b–a, ,  a – b  c , Now, m1m2 =, =, ,  –1,  c  b – a , , We get that the line through O and P is perpendicular, to AB., 1, then, Sol 12: Since, line L make 60° with line 3 x + y =, , 1, 2, , =2+2 = 4, , Which is positive., , 1, and its minimum value, Hence, y is minimum at x =, 2, 1, − 3 + tan 60°, is ., m1 = 0, =, 4, 1 − − 3 ( tan 60° ), Clearly, value is maximum when x = 1., , (, , ∴ Minimum value of a2 + b2 + c2 + d2 =, , 1 1 1 1, + + + =2, 2 2 2 2, , And maximum value is 1 + 1 + 1 + 1 = 4, , m2 =, , ), , − 3 − tan 60°, −2 3 −2 3, = = =, 1−3, −2, 1 + − 3 tan 60°, , (, , ), , Equation of line having slope, (3, -2), , 3, , 3 and passes through

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8 . 8 4 | Straight Line, , y +=, 2, , ⇒ y + 2=, , 3 ( x − 3), , α −β, , ⇒2≤, , 3x − 3 3, , 2, , +, , α+β, 2, , ≤4, , ⇒ y − 3x + 2 + 3 3 =, 0, Sol 13: (A) The point of intersection of lines ax + by + c, = 0 and bx + ay + c = 0 is, ,  −c, −c , P, ,, , a+b a+b, Given that distance of point P from (1, 1) is less than, 2 2, 2, , 2, , , c  , c , 1 +,  + 1 +,  <2 2, a+b , a+b, , , a+b+c, ⇒ 2, <2 2, a+b, , Since, Point P lies in the first quadrant α ; β > 0, Case 1:, , a+b+c, a+b+c, <2, ⇒, < 2 ⇒ −2 <, a+b, a+b, , 2≤, , ⇒ a + b + c < 2a + 2b ⇒ a + b − c > 0, , ⇒, , Sol 14: Let P be ( α , β ) , then, d1 (P ) =, d2 (P ) =, , α −β, 2, α+β, 2, , ⇒ 2 ≤ d1 (P ) + d2 (P ) ≤ 4, , α≥β, , α −β, 2, , 2, , ≤4, , 2≤α≤2 2, , Case 2:, , 2≤, , α+β, , +, , α<β, , −α + β, 2, , +, , α+β, 2, , ≤4⇒, , 2 ≤β≤2 2, , ( ) − ( 2) = 8 – 2, , Area = 2 2, = 6 sq. unit, , 2, , 2