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Chapter, , 10, , STRAIGHT LINES, vG eometry, as a logical system, is a means and even the most powerful, means to make children feel the strength of the human spirit that is, of their own spirit. – H. FREUDENTHALv, , 10.1 Introduction, We are familiar with two-dimensional coordinate geometry, from earlier classes. Mainly, it is a combination of algebra, and geometry. A systematic study of geometry by the use, of algebra was first carried out by celebrated French, philosopher and mathematician René Descartes, in his book, ‘La Géométry, published in 1637. This book introduced the, notion of the equation of a curve and related analytical, methods into the study of geometry. The resulting, combination of analysis and geometry is referred now as, analytical geometry. In the earlier classes, we initiated, René Descartes, the study of coordinate geometry, where we studied about, (1596 -1650), coordinate axes, coordinate plane, plotting of points in a, plane, distance between two points, section formulae, etc. All these concepts are the, basics of coordinate geometry., Let us have a brief recall of coordinate geometry done in earlier classes. To, recapitulate, the location of the points (6, – 4) and, (3, 0) in the XY-plane is shown in Fig 10.1., We may note that the point (6, – 4) is at 6 units, distance from the y-axis measured along the positive, x-axis and at 4 units distance from the x-axis, measured along the negative y-axis. Similarly, the, point (3, 0) is at 3 units distance from the y-axis, measured along the positive x-axis and has zero, distance from the x-axis., Fig 10.1, We also studied there following important, formulae:, , 2020-21

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204, , MATHEMATICS, , I. Distance between the points P (x1, y1) and Q (x2, y2) is, , PQ =, , (x, , 2, , – x1 ) + ( y2 – y1 ), 2, , 2, , For example, distance between the points (6, – 4) and (3, 0) is, , ( 3 − 6 )2 + ( 0 + 4 )2 =, , 9 + 16 = 5 units., , II. The coordinates of a point dividing the line segment joining the points (x1, y1), ,  m x 2 + n x1 m y 2 + n y1 , ,, ., m + n ,  m+n, , and (x2, y2) internally, in the ratio m: n are , , For example, the coordinates of the point which divides the line segment joining, A (1, –3) and B (–3, 9) internally, in the ratio 1: 3 are given by x =, and y =, , 1.( − 3) + 3.1, =0, 1+ 3, , 1.9 + 3. ( –3), = 0., 1+ 3, , III. In particular, if m = n, the coordinates of the mid-point of the line segment, ,  x1 + x2 y1 + y 2 , ,, ., 2 ,  2, , joining the points (x1, y1) and (x2, y2) are , , IV. Area of the triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is, , 1, x1 ( y 2 − y 3 ) + x 2 ( y 3 − y 1) + x 3 ( y1 − y 2 ) ., 2, For example, the area of the triangle, whose vertices are (4, 4), (3, – 2) and (– 3, 16) is, − 54, 1, 4( −2 − 16) + 3(16 − 4) + (−3)(4 + 2) =, = 27., 2, 2, , Remark If the area of the triangle ABC is zero, then three points A, B and C lie on, a line, i.e., they are collinear., In the this Chapter, we shall continue the study of coordinate geometry to study, properties of the simplest geometric figure – straight line. Despite its simplicity, the, line is a vital concept of geometry and enters into our daily experiences in numerous, interesting and useful ways. Main focus is on representing the line algebraically, for, which slope is most essential., , 10.2 Slope of a Line, A line in a coordinate plane forms two angles with the x-axis, which are supplementary., , 2020-21

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STRAIGHT LINES, , 205, , The angle (say) θ made by the line l with positive, direction of x-axis and measured anti clockwise, is called the inclination of the line. Obviously, 0° ≤ θ ≤ 180° (Fig 10.2)., We observe that lines parallel to x-axis, or, coinciding with x-axis, have inclination of 0°. The, inclination of a vertical line (parallel to or, coinciding with y-axis) is 90°., Definition 1 If θ is the inclination of a line, l, then tan θ is called the slope or gradient of, Fig 10.2, the line l., The slope of a line whose inclination is 90° is not, defined., The slope of a line is denoted by m., Thus, m = tan θ, θ ≠ 90°, It may be observed that the slope of x-axis is zero and slope of y-axis is not defined., 10.2.1 Slope of a line when coordinates of any two points on the line are given, We know that a line is completely determined when we are given two points on it., Hence, we proceed to find the slope of a, line in terms of the coordinates of two points, on the line., Let P(x 1, y 1 ) and Q(x 2, y 2 ) be two, points on non-vertical line l whose inclination, is θ. Obviously, x1 ≠ x2, otherwise the line, will become perpendicular to x-axis and its, slope will not be defined. The inclination of, the line l may be acute or obtuse. Let us, take these two cases., Draw perpendicular QR to x-axis and, PM perpendicular to RQ as shown in, Figs. 10.3 (i) and (ii)., Fig 10. 3 (i), , Case 1 When angle θ is acute:, In Fig 10.3 (i), ∠MPQ = θ., , ... (1), , Therefore, slope of line l = m = tan θ., But in ∆MPQ, we have tan θ =, , MQ y2 − y1, =, ., MP x2 − x1, , 2020-21, , ... (2)

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STRAIGHT LINES, , 207, , Hence, two non vertical lines l1 and l2 are parallel if and only if their slopes, are equal., If the lines l1 and l2 are perpendicular (Fig 10.5), then β = α + 90°., Therefore,tan β = tan (α + 90°), = – cot α = −, , 1, m2 = − m, 1, , i.e.,, , or, , 1, tan α, m1 m2 = – 1, , Conversely, if m1 m2 = – 1, i.e., tan α tan β = – 1., Then tan α = – cot β = tan (β + 90°) or tan (β – 90°), Fig 10. 5, Therefore, α and β differ by 90°., Thus, lines l1 and l2 are perpendicular to each other., Hence, two non-vertical lines are perpendicular to each other if and only if, their slopes are negative reciprocals of each other,, m2= −, , i.e.,, , 1, or, m1 m2 = – 1., m1, , Let us consider the following example., Example 1 Find the slope of the lines:, (a), , Passing through the points (3, – 2) and (–1, 4),, , (b), , Passing through the points (3, – 2) and (7, – 2),, , (c), , Passing through the points (3, – 2) and (3, 4),, , (d), , Making inclination of 60° with the positive direction of x-axis., , Solution (a) The slope of the line through (3, – 2) and (– 1, 4) is, , m=, , 4 − ( −2) 6, 3, =, =− ., −1 − 3, −4, 2, , (b) The slope of the line through the points (3, – 2) and (7, – 2) is, , m=, , –2 – (–2) 0, = = 0., 7–3, 4, , (c) The slope of the line through the points (3, – 2) and (3, 4) is, , 2020-21

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208, , MATHEMATICS, , m =, , 4 – (–2), , =, , 6, , , which is not defined., 3–3, 0, (d) Here inclination of the line α = 60°. Therefore, slope of the line is, m = tan 60° =, , 3., , 10.2.3 Angle between two lines When we think about more than one line in a plane,, then we find that these lines are either intersecting or parallel. Here we will discuss the, angle between two lines in terms of their slopes., Let L1 and L2 be two non-vertical lines with slopes m1 and m2, respectively. If α1, and α2 are the inclinations of lines L1 and L2, respectively. Then, , m1 = tan α1 and m2 = tan α2 ., We know that when two lines intersect each other, they make two pairs of, vertically opposite angles such that sum of any two adjacent angles is 180°. Let θ and, φ be the adjacent angles between the lines L1 and L2 (Fig10.6). Then, θ = α2 – α1 and α1, α2 ≠ 90°., Therefore tan θ = tan (α2 – α1) =, , tan α 2 − tan α1, m − m1, = 2, 1 + tan α1 tan α 2 1 + m1m2, , (as 1 + m1m2 ≠ 0), , and φ = 180° – θ so that, tan φ = tan (180° – θ ) = – tan θ = –, , m2 − m1, , as 1 + m1m2 ≠ 0, 1 + m1m2, , Fig 10. 6, , Now, there arise two cases:, , 2020-21

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STRAIGHT LINES, , Case I If, , 209, , m 2 – m1, is positive, then tan θ will be positive and tan φ will be negative,, 1 + m1m 2, , which means θ will be acute and φ will be obtuse., Case II If, , m 2 – m1, is negative, then tan θ will be negative and tan φ will be positive,, 1 + m1m 2, , which means that θ will be obtuse and φ will be acute., Thus, the acute angle (say θ) between lines L1 and L2 with slopes m1 and m2,, respectively, is given by, tan θ =, , m 2 − m1, , as 1 + m1m 2 ≠ 0, 1 + m1m 2, , ... (1), , The obtuse angle (say φ) can be found by using φ =1800 – θ., Example 2 If the angle between two lines is, , π, 1, and slope of one of the lines is , find, 4, 2, , the slope of the other line., Solution We know that the acute angle θ between two lines with slopes m1 and m2, is given by, , Let m1 =, , tan θ =, , m2 − m1, 1 + m1m 2, , ... (1), , 1, π, , m2 = m and θ = ., 4, 2, , Now, putting these values in (1), we get, , π, tan =, 4, , m−, , 1, 2, , 1, 1+ m, 2, , or 1 =, , 1, 2, , 1, 1+ m, 2, , 1, 1, m−, 2 = 1 or, 2 = –1., 1, 1, 1+, m, 1+, m, 2, 2, , m−, , which gives, , m−, , 1, Therefore m = 3 or m = − ., 3, , 2020-21, , ,

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210, , MATHEMATICS, , Hence, slope of the other line is, , 1, 3 or − . Fig 10.7 explains the, 3, reason of two answers., , Fig 10. 7, , Example 3 Line through the points (–2, 6) and (4, 8) is perpendicular to the line, through the points (8, 12) and (x, 24). Find the value of x., Solution Slope of the line through the points (– 2, 6) and (4, 8) is, , m1 =, , 8−6, 2 1, = =, 4 − ( −2 ) 6 3, , Slope of the line through the points (8, 12) and (x, 24) is, , 24 − 12 12, =, x−8, x −8, Since two lines are perpendicular,, m1 m2 = –1, which gives, m2 =, , 1 12, ×, = −1 or x = 4 ., 3 x −8, 10.2.4 Collinearity of three points We, know that slopes of two parallel lines are, equal. If two lines having the same slope, pass through a common point, then two, lines will coincide. Hence, if A, B and C, are three points in the XY-plane, then they, will lie on a line, i.e., three points are, collinear (Fig 10.8) if and only if slope of, AB = slope of BC., , 2020-21, , Fig 10. 8

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STRAIGHT LINES, , 211, , Example 4 Three points P (h, k), Q (x1, y1) and R (x2, y2) lie on a line. Show that, (h – x1) (y2 – y1) = (k – y1) (x2 – x1)., Solution Since points P, Q and R are collinear, we have, Slope of PQ = Slope of QR, i.e.,, , y1 − k, y −y, = 2 1, x1 − h, x2 − x1, , or, , k − y1 y 2 − y1, =, h − x1 x2 − x1 ,, , or, , (h – x1) (y2 – y1) = (k – y1) (x2 – x1)., , Example 5 In Fig 10.9, time and, distance graph of a linear motion is given., Two positions of time and distance are, recorded as, when T = 0, D = 2 and when, T = 3, D = 8. Using the concept of slope,, find law of motion, i.e., how distance, depends upon time., Solution Let (T, D) be any point on the, line, where D denotes the distance at time, T. Therefore, points (0, 2), (3, 8) and, (T, D) are collinear so that, , 8 − 2 D −8, =, or, 3−0 T −3, or, D = 2(T + 1),, which is the required relation., , Fig 10.9, , 6 (T − 3) = 3 (D − 8), , EXERCISE 10.1, 1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7),, (5, – 5) and (– 4, –2). Also, find its area., 2. The base of an equilateral triangle with side 2a lies along the y-axis such that the, mid-point of the base is at the origin. Find vertices of the triangle., 3. Find the distance between P (x1, y1) and Q (x2, y2) when : (i) PQ is parallel to the, y-axis, (ii) PQ is parallel to the x-axis., 4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4)., 5. Find the slope of a line, which passes through the origin, and the mid-point of the, line segment joining the points P (0, – 4) and B (8, 0)., , 2020-21

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212, , MATHEMATICS, , 6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and, (–1, –1) are the vertices of a right angled triangle., 7. Find the slope of the line, which makes an angle of 30° with the positive direction, of y-axis measured anticlockwise., 8. Find the value of x for which the points (x, – 1), (2,1) and (4, 5) are collinear., 9. Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2), are the vertices of a parallelogram., 10. Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2)., 11. The slope of a line is double of the slope of another line. If tangent of the angle, between them is, , 1, , find the slopes of the lines., 3, , 12. A line passes through (x1, y1) and (h, k). If slope of the line is m, show that, k – y1 = m (h – x1)., 13. If three points (h, 0), (a, b) and (0, k) lie on a line, show that, , a b, + = 1., h k, , 14. Consider the following population and year graph (Fig 10.10), find the slope of the, line AB and using it, find what will be the population in the year 2010?, , Fig 10.10, , 10.3, , Various Forms of the Equation of a Line, , We know that every line in a plane contains infinitely many points on it. This relationship, between line and points leads us to find the solution of the following problem:, , 2020-21

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STRAIGHT LINES, , 213, , How can we say that a given point lies on the given line? Its answer may be that, for a given line we should have a definite condition on the points lying on the line., Suppose P (x, y) is an arbitrary point in the XY-plane and L is the given line. For the, equation of L, we wish to construct a statement or condition for the point P that is, true, when P is on L, otherwise false. Of course the statement is merely an algebraic, equation involving the variables x and y. Now, we will discuss the equation of a line, under different conditions., 10.3.1 Horizontal and vertical lines If a horizontal line L is at a distance a from the, x-axis then ordinate of every point lying on the line is either a or – a [Fig 10.11 (a)]., Therefore, equation of the line L is either y = a or y = – a. Choice of sign will depend, upon the position of the line according as the line is above or below the y-axis. Similarly,, the equation of a vertical line at a distance b from the y-axis is either x = b or, x = – b [Fig 10.11(b)]., , Fig 10.11, , Example 6 Find the equations of the lines, parallel to axes and passing through, (– 2, 3)., Solution Position of the lines is shown in the, Fig 10.12. The y-coordinate of every point on, the line parallel to x-axis is 3, therefore, equation, of the line parallel tox-axis and passing through, (– 2, 3) is y = 3. Similarly, equation of the line, parallel to y-axis and passing through (– 2, 3), is x = – 2., , 2020-21, , Fig 10.12

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214, , MATHEMATICS, , 10.3.2 Point-slope form Suppose that, P0 (x0, y0) is a fixed point on a non-vertical, line L, whose slope is m. Let P (x, y) be an, arbitrary point on L (Fig 10.13)., Then, by the definition, the slope of L is, given by, , m=, , y − y0, , i.e. , y − y 0 = m (x − x 0 ), x − x0, ...(1), , Since the point P0 (x0 , y0) along with, all points (x, y) on L satisfies (1) and no, other point in the plane satisfies (1). Equation, (1) is indeed the equation for the given line L., , Fig 10.13, , Thus, the point (x, y) lies on the line with slope m through the fixed point (x0, y0),, if and only if, its coordinates satisfy the equation, y – y0 = m (x – x0), Example 7 Find the equation of the line through (– 2, 3) with slope – 4., Solution Here m = – 4 and given point (x0 , y0) is (– 2, 3)., By slope-intercept form formula, (1) above, equation of the given, line is, y – 3 = – 4 (x + 2) or, 4x + y + 5 = 0, which is the, required equation., 10.3.3 Two-point form Let the, line L passes through two given, points P1 (x1, y1) and P2 (x2, y2)., Let P (x, y) be a general point, on L (Fig 10.14)., Fig 10.14, , The three points P1, P2 and P are, collinear, therefore, we have, slope of P1P = slope of P1P2, , i.e.,, , y − y1 y 2 − y1, =, ,, x − x1 x 2 − x1, , or y − y 1 =, , 2020-21, , y 2 − y1, ( x − x1 )., x 2 − x1

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STRAIGHT LINES, , 215, , Thus, equation of the line passing through the points (x1, y1) and (x2, y2) is given by, , y − y1 =, , y 2 − y1, ( x − x1), x 2 − x1, , ... (2), , Example 8 Write the equation of the line through the points (1, –1) and (3, 5)., Solution Here x1 = 1, y1 = – 1, x2 = 3 and y2 = 5. Using two-point form (2) above, for the equation of the line, we have, , y – ( –1) =, or, , 5 – ( –1), ( x – 1), 3 –1, , –3x + y + 4 = 0, which is the required equation., , 10.3.4 Slope-intercept form Sometimes a line is known to us with its slope and an, intercept on one of the axes. We will now find equations of such lines., Case I Suppose a line L with slope m cuts the y-axis at a distance c from the origin, (Fig10.15). The distance c is called the yintercept of the line L. Obviously,, coordinates of the point where the line meet, the y-axis are (0, c). Thus, L has slope m, and passes through a fixed point (0, c)., Therefore, by point-slope form, the equation, of L is, , y − c = m ( x − 0 ) or y = mx + c, Thus, the point (x, y) on the line with slope, m and y-intercept c lies on the line if and, Fig 10.15, only if, y = mx +c, ...(3), Note that the value of c will be positive or negative according as the intercept is made, on the positive or negative side of the y-axis, respectively., Case II Suppose line L with slope m makes x-intercept d. Then equation of L is, y = m(x – d), ... (4), Students may derive this equation themselves by the same method as in Case I., Example 9 Write the equation of the lines for which tan θ =, inclination of the line and (i) y-intercept is –, , 1, , where θ is the, 2, , 3, (ii) x-intercept is 4., 2, , 2020-21

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216, , MATHEMATICS, , Solution (i) Here, slope of the line is m = tan θ =, , 1, 3, and y - intercept c = – ., 2, 2, , Therefore, by slope-intercept form (3) above, the equation of the line is, , 1, 3, x − or 2 y − x + 3 = 0 ,, 2, 2, , y=, which is the required equation., , (ii) Here, we have m = tan θ =, , 1, and d = 4., 2, , Therefore, by slope-intercept form (4) above, the equation of the line is, , y=, , 1, ( x − 4) or 2 y − x + 4 = 0 ,, 2, , which is the required equation., 10.3.5 Intercept - form Suppose a line L makes x-intercept a and y-intercept b on the, axes. Obviously L meets x-axis at the point, (a, 0) and y-axis at the point (0, b) (Fig .10.16)., By two-point form of the equation of the line,, we have, , y−0=, i.e.,, , b−0, (x − a) or ay = −bx + ab ,, 0−a, , x y, + = 1., a b, , Thus, equation of the line making intercepts, a and b on x-and y-axis, respectively, is, , x y, + =1, a b, , Fig 10.16, , ... (5), , Example 10 Find the equation of the line, which makes intercepts –3 and 2 on the, x- and y-axes respectively., Solution Here a = –3 and b = 2. By intercept form (5) above, equation of the line is, , x y, + =1, −3 2, , or, , 2x − 3 y + 6 = 0 ., , 2020-21

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STRAIGHT LINES, , 217, , 10.3.6 Normal form Suppose a non-vertical line is known to us with following data:, (i) Length of the perpendicular (normal) from origin to the line., (ii) Angle which normal makes with the positive direction of x-axis., Let L be the line, whose perpendicular distance from origin O be OA = p and the, angle between the positive x-axis and OA be ∠XOA = ω. The possible positions of line, L in the Cartesian plane are shown in the Fig 10.17. Now, our purpose is to find slope, of L and a point on it. Draw perpendicular AM on the x-axis in each case., , Fig 10.17, , In each case, we have OM = p cos ω and MA = p sin ω, so that the coordinates of the, point A are (p cos ω, p sin ω)., Further, line L is perpendicular to OA. Therefore, The slope of the line L = −, Thus, the line L has slope −, , 1, 1, cos ω, =−, =−, ., slope of OA, tan ω, sin ω, , cos ω, and point A ( p cos ω, p sin ω)on it. Therefore, by, sin ω, , point-slope form, the equation of the line L is, , 2020-21

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218, , MATHEMATICS, , y − p sin ω = −, , cos ω, ( x − p cos ω ) or, sin ω, , x cos ω + y sin ω = p (sin 2ω + cos 2ω), , or, x cos ω + y sin ω = p., Hence, the equation of the line having normal distance p from the origin and angle ω, which the normal makes with the positive direction of x-axis is given by, x cos ω + y sin ω = p, ... (6), Example 11 Find the equation of the line whose perpendicular distance from the, origin is 4 units and the angle which the normal makes with positive direction of x-axis, is 15°., Solution Here, we are given p = 4 and, ω = 150 (Fig10.18)., , Now, , cos 15° =, , and, , sin 15º =, , 3 +1, 2 2, 3 −1, (Why?), 2 2, , By the normal form (6) above, the equation of the, line is, , 3 +1, 3 −1, x+, y = 4 or, 2 2, 2 2, This is the required equation., x cos 150 + y sin 150 = 4 or, , Fig 10.18, , (, , ) (, , 3 +1 x +, , ), , 3 − 1 y = 8 2., , Example 12 The Fahrenheit temperature F and absolute temperature K satisfy a, linear equation. Given that K = 273 when F = 32 and that K = 373 when F = 212., Express K in terms of F and find the value of F, when K = 0., Solution Assuming F along x-axis and K along y-axis, we have two points (32, 273), and (212, 373) in XY-plane. By two-point form, the point (F, K) satisfies the equation, , K − 273 =, , 373 − 273, 100, ( F − 32 ) or K − 273 = ( F − 32 ), 212 − 32, 180, , 5, ( F − 32 ) + 273, 9, which is the required relation., or, , K=, , ... (1), , 2020-21

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STRAIGHT LINES, , 219, , When K = 0, Equation (1) gives, , 0=, , 5, ( F − 32 ) + 273, 9, , or F − 32 = −, , 273 × 9, = − 491.4, 5, , or F= − 459.4 ., , Alternate method We know that simplest form of the equation of a line is y = mx + c., Again assuming F along x-axis and K along y-axis, we can take equation in the form, K = mF + c, ... (1), Equation (1) is satisfied by (32, 273) and (212, 373). Therefore, 273 = 32m + c, ... (2), and, 373 = 212m + c, Solving (2) and (3), we get, m=, , ... (3), , 5, 2297, and c =, ., 9, 9, , Putting the values of m and c in (1), we get, , 5, 2297, F+, 9, 9, which is the required relation. When K = 0, (4) gives F = – 459.4., K=, , ... (4), , Note We know, that the equation y = mx + c, contains two constants, namely,, A, m and c. For finding these two constants, we need two conditions satisfied by the, equation of line. In all the examples above, we are given two conditions to determine, the equation of the line., , EXERCISE 10.2, In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:, 1. Write the equations for the x-and y-axes., 2. Passing through the point (– 4, 3) with slope, , 1, ., 2, , 3. Passing through (0, 0) with slope m., , (, , ), , 4. Passing through 2, 2 3 and inclined with the x-axis at an angle of 75o., 5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2., 6. Intersecting the y-axis at a distance of 2 units above the origin and making an, angle of 30o with positive direction of the x-axis., 7. Passing through the points (–1, 1) and (2, – 4)., , 2020-21

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220, , MATHEMATICS, , 8. Perpendicular distance from the origin is 5 units and the angle made by the, perpendicular with the positive x-axis is 300., 9. The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the, median through the vertex R., 10. Find the equation of the line passing through (–3, 5) and perpendicular to the line, through the points (2, 5) and (–3, 6)., 11. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides, it in the ratio 1: n. Find the equation of the line., 12. Find the equation of a line that cuts off equal intercepts on the coordinate axes, and passes through the point (2, 3)., 13. Find equation of the line passing through the point (2, 2) and cutting off intercepts, on the axes whose sum is 9., 14., , 15., 16., , 17., , 18., , 2π, with the, 3, positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis, at a distance of 2 units below the origin., The perpendicular from the origin to a line meets it at the point (–2, 9), find the, equation of the line., The length L (in centimetre) of a copper rod is a linear function of its Celsius, temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134, when C = 110, express L in terms of C., The owner of a milk store finds that, he can sell 980 litres of milk each week at, Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear, relationship between selling price and demand, how many litres could he sell, weekly at Rs 17/litre?, P (a, b) is the mid-point of a line segment between axes. Show that equation, Find equation of the line through the point (0, 2) making an angle, , of the line is, , x y, + = 2., a b, , 19. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find, equation of the line., 20. By using the concept of equation of a line, prove that the three points (3, 0),, (– 2, – 2) and (8, 2) are collinear., , 10.4 General Equation of a Line, In earlier classes, we have studied general equation of first degree in two variables,, Ax + By + C = 0, where A, B and C are real constants such that A and B are not zero, simultaneously. Graph of the equation Ax + By + C = 0 is always a straight line., , 2020-21

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STRAIGHT LINES, , 221, , Therefore, any equation of the form Ax + By + C = 0, where A and B are not zero, simultaneously is called general linear equation or general equation of a line., 10.4.1 Different forms of Ax + By + C = 0 The general equation of a line can be, reduced into various forms of the equation of a line, by the following procedures:, (a) Slope-intercept form If B ≠ 0, then Ax + By + C = 0 can be written as, , y=−, , A, C, x − or y = mx + c, B, B, , ... (1), , A, C, and c = − ., B, B, We know that Equation (1) is the slope-intercept form of the equation of a line, , where, , m=−, , whose slope is −, , A, C, , and y-intercept is − ., B, B, , If B = 0, then x = −, x-intercept is −, , C, ,which is a vertical line whose slope is undefined and, A, , C, ., A, , (b) Intercept form If C ≠ 0, then Ax + By + C = 0 can be written as, x, y, x y, +, = 1 or, + =1, C, C, a, b, −, −, A, B, , where, , a= −, , ... (2), , C, C, and b = − ., A, B, , We know that equation (2) is intercept form of the equation of a line whose, x-intercept is −, , C, C, and y-intercept is − ., A, B, , If C = 0, then Ax + By + C = 0 can be written as Ax + By = 0, which is a line, passing through the origin and, therefore, has zero intercepts on the axes., (c) Normal form Let x cos ω + y sin ω = p be the normal form of the line represented, by the equation Ax + By + C = 0 or Ax + By = – C. Thus, both the equations are, same and therefore,, , A, B, C, =, =−, cos ω sin ω, p, , 2020-21

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222, , MATHEMATICS, , cos ω = −, , which gives, , Ap, Bp, and sin ω = −, ., C, C, , ( ) ( ), 2, , Ap, sin ω + cos ω = −, C, 2, , Now, , 2, p =, , or, , 2, , 2, , C, 2, 2, A +B, , cosω = ±, , Therefore, , Bp, + −, C, , or p = ±, , A, 2, 2, A +B, , 2, , =1, , C, 2, A +B, 2, , and sinω = ±, , B, 2, 2, A +B, , ., , Thus, the normal form of the equation Ax + By + C = 0 is, x cos ω + y sin ω = p,, where cos ω = ±, , A, A +B, 2, , 2, , , sin ω = ±, , B, A +B, 2, , 2, , and p = ±, , C, 2, A +B, 2, , ., , Proper choice of signs is made so that p should be positive., Example 13 Equation of a line is 3x – 4y + 10 = 0. Find its (i) slope, (ii) x - and, y-intercepts., Solution (i), , Given equation 3x – 4y + 10 = 0 can be written as, , y=, , 3, 5, x+, 4, 2, , ... (1), , Comparing (1) with y = mx + c, we have slope of the given line as m =, (ii), , 3, ., 4, , Equation 3x – 4y + 10 = 0 can be written as, 3 x − 4 y = −10, , Comparing (2) with, y-intercept as b =, , or, , x, y, + =1, 10 5, −, 3 2, , ... (2), , x y, 10, + = 1 , we have x-intercept as a = −, and, a b, 3, , 5, ., 2, , 2020-21

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STRAIGHT LINES, , Example 14 Reduce the equation, , 223, , 3 x + y − 8 = 0 into normal form. Find the values, , of p and ω., Solution Given equation is, , 3x + y − 8 = 0, Dividing (1) by, , ( 3), , 2, , ... (1), , + (1)2 = 2 , we get, , 3, 1, x + y = 4 or cos 30° x + sin 30° y = 4, ... (2), 2, 2, Comparing (2) with x cos ω + y sin ω = p, we get p = 4 and ω = 30°., , Example15 Find the angle between the lines y − 3 x − 5 = 0 and, , 3y − x + 6 = 0 ., , Solution Given lines are, , y − 3 x − 5 = 0 or y = 3 x + 5, and, , ... (1), , 1, , 3 y − x + 6 = 0 or y = 3 x − 2 3, , Slope of line (1) is m1 =, , 3 and slope of line (2) is m2 =, , ... (2), , 1, ., 3, , The acute angle (say) θ between two lines is given by, , tan θ =, , m 2 − m1, 1 + m1m 2, , ..., , (3), Putting the values of m1 and m2 in (3), we get, 1, − 3, 3, tan θ =, 1, 1+ 3 ×, 3, , =, , 1− 3, 1, =, 2 3, 3, , which gives θ = 30°. Hence, angle between two lines is either 30° or 180° – 30° = 150°., Example 16 Show that two lines a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0,, where b1, b2 ≠ 0 are:, , 2020-21

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224, , MATHEMATICS, , (i) Parallel if, , a1, b1, , =, , a2, , , and (ii) Perpendicular if a1 a 2 + b1 b 2 = 0 ., , b2, , Solution Given lines can be written as, , and, , y = − a1 x − c1, b1, b1, , ... (1), , y = − a2 x − c2, b2, b2, , ... (2), , Slopes of the lines (1) and (2) are m1 = − a1 and m2 = −, , b1, , a 2 , respectively. Now, b2, , (i) Lines are parallel, if m1 = m2, which gives, , −, , a1 a 2, a1, a, = ., = − 2 or, b1 b2, b1, b2, , (ii) Lines are perpendicular, if m1.m2 = – 1, which gives, , a1 a2, . = − 1 or a a + b b = 0, 1 2, 1 2, b1 b2, Example 17 Find the equation of a line perpendicular to the line x − 2 y + 3 = 0 and, passing through the point (1, – 2)., Solution Given line x − 2 y + 3 = 0 can be written as, , y=, , 1, 3, x+, 2, 2, , Slope of the line (1) is m1 =, , m2 = −, , ...(1), , 1, . Therefore, slope of the line perpendicular to line (1) is, 2, , 1, = −2, m1, , Equation of the line with slope – 2 and passing through the point (1, – 2) is, , y − ( − 2) = − 2(x − 1) or y = − 2 x ,, which is the required equation., , 2020-21

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STRAIGHT LINES, , 225, , 10.5 Distance of a Point From a Line, The distance of a point from a line is the length of the perpendicular drawn from the, point to the line. Let L : Ax + By + C = 0 be a line, whose distance from the point, P (x1, y1) is d. Draw a perpendicular PM from the point P to the line L (Fig10.19). If, , Fig10.19, , the line meets the x-and y-axes at the points Q and R, respectively. Then, coordinates, of the points are, ,  C, , Q  − , 0  and R,  A, , , C, ,  0, −  . Thus, the area of the triangle PQR, B, , , is given by, 1, area (∆PQR) = PM.QR , which gives PM =, 2, , Also, area (∆PQR) =, , 1, C   C  C, , , x1  0 +  +  −  − − y1  + 0 ( y 1 − 0 ), 2, B   A  B, , , , =, , or, , 2 area (∆PQR), QR, , 2 area (∆PQR) =, 2, , 1 C, C C2, x1 + y 1 +, 2 B, A AB, , C, . A x1 + B y 1 + C , and, AB, , , C, QR =  0 +  +, , A, , (, , C, − 0, B, , ), , 2, , =, , C, AB, , 2, 2, A +B, , Substituting the values of area (∆PQR) and QR in (1), we get, , 2020-21, , ... (1)

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226, , MATHEMATICS, , PM =, , or, , A x1 + B y 1 + C, 2, 2, A +B, , A x1 + B y1 + C, , d=, , ., 2, 2, A +B, Thus, the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1, y1), is given by, , d=, , A x1 + B y1 + C, 2, 2, A +B, , ., , 10.5.1 Distance between two, parallel lines We know that slopes, of two parallel lines are equal., Therefore, two parallel lines can be, taken in the form, y = mx + c1, ... (1), and y = mx + c2, ... (2), Line (1) will intersect x-axis at the point, ,  c1, , A  − m , 0  as shown in Fig10.20., , , , Fig10.20, , Distance between two lines is equal to the length of the perpendicular from point, A to line (2). Therefore, distance between the lines (1) and (2) is, , ( −m)  −, , c1 ,  + ( −c2 ),  m, 1 + m2, , or d =, , c1 − c2, 1 + m2, , ., , Thus, the distance d between two parallel lines y = mx + c1 and y = mx + c2 is given by, , d=, , c1 − c2, 1 + m2, , ., , If lines are given in general form, i.e., Ax + By + C1 = 0 and Ax + By + C2 = 0,, , 2020-21

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STRAIGHT LINES, , then above formula will take the form d =, , 227, , C1 − C 2, A 2 + B2, , Students can derive it themselves., Example 18 Find the distance of the point (3, – 5) from the line 3x – 4y –26 = 0., Solution Given line is, 3x – 4y –26 = 0, ... (1), Comparing (1) with general equation of line Ax + By + C = 0, we get, A = 3, B = – 4 and C = – 26., Given point is (x1, y1) = (3, –5). The distance of the given point from given line is, d=, , Ax1 + By1 + C, A +B, 2, , 2, , =, , 3.3 + ( –4 )( –5 ) – 26, 3 + ( –4 ), 2, , 2, , 3, = ., 5, , Example 19 Find the distance between the parallel lines 3x – 4y +7 = 0 and, 3x – 4y + 5 = 0, Solution Here A = 3, B = –4, C1 = 7 and C2 = 5. Therefore, the required distance is, d=, , 7–5, 32 + ( –4 ), , 2, , 2, = ., 5, , EXERCISE 10.3, 1. Reduce the following equations into slope - intercept form and find their slopes, and the y - intercepts., (i) x + 7y = 0,, (ii) 6x + 3y – 5 = 0,, (iii) y = 0., 2. Reduce the following equations into intercept form and find their intercepts on, the axes., (i) 3x + 2y – 12 = 0, (ii) 4x – 3y = 6,, (iii) 3y + 2 = 0., 3. Reduce the following equations into normal form. Find their perpendicular distances, from the origin and angle between perpendicular and the positive x-axis., (i) x – 3y + 8 = 0, (ii) y – 2 = 0,, (iii) x – y = 4., 4. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2)., 5. Find the points on the x-axis, whose distances from the line, , x y, + = 1 are 4 units., 3 4, , 6. Find the distance between parallel lines, (i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 (ii) l (x + y) + p = 0 and l (x + y) – r = 0., , 2020-21

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228, , MATHEMATICS, , 7. Find equation of the line parallel to the line 3 x − 4 y + 2 = 0 and passing through, the point (–2, 3)., 8. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having, x intercept 3., 9. Find angles between the lines, , 3 x + y = 1and x + 3 y = 1., , 10. The line through the points (h, 3) and (4, 1) intersects the line 7 x − 9 y − 19 = 0., at right angle. Find the value of h., 11. Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is, A (x –x1) + B (y – y1) = 0., 12. Two lines passing through the point (2, 3) intersects each other at an angle of 60o., If slope of one line is 2, find equation of the other line., 13. Find the equation of the right bisector of the line segment joining the points (3, 4), and (–1, 2)., 14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the, line 3x – 4y – 16 = 0., 15. The perpendicular from the origin to the line y = mx + c meets it at the point, (–1, 2). Find the values of m and c., 16. If p and q are the lengths of perpendiculars from the origin to the, lines x cos θ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove, that p2 + 4q2 = k2., 17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation, and length of altitude from the vertex A., 18. If p is the length of perpendicular from the origin to the line whose intercepts on, the axes are a and b, then show that, , 1, 1 1, = 2 + 2., 2, p, a b, , Miscellaneous Examples, Example 20 If the lines 2 x + y − 3 = 0 , 5 x + ky − 3 = 0 and 3 x − y − 2 = 0 are, concurrent, find the value of k., Solution Three lines are said to be concurrent, if they pass through a common point,, i.e., point of intersection of any two lines lies on the third line. Here given lines are, 2x + y – 3 = 0, ... (1), 5x + ky – 3 = 0, ... (2), , 2020-21

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STRAIGHT LINES, , 3x – y – 2 = 0, , 229, , ... (3), , Solving (1) and (3) by cross-multiplication method, we get, , x, y, 1, =, =, –2 – 3 –9 + 4 –2 – 3, , or x = 1, y = 1 ., , Therefore, the point of intersection of two lines is (1, 1). Since above three lines are, concurrent, the point (1, 1) will satisfy equation (2) so that, 5.1 + k .1 – 3 = 0 or k = – 2., Example 21 Find the distance of the line 4x – y = 0 from the point P (4, 1) measured, along the line making an angle of 135° with the positive x-axis., Solution Given line is 4x – y = 0, In order to find the distance of the, line (1) from the point P (4, 1) along, another line, we have to find the point, of intersection of both the lines. For, this purpose, we will first find the, equation of the second line, (Fig 10.21). Slope of second line is, tan 135° = –1. Equation of the line, with slope – 1 through the point, P (4, 1) is, , ... (1), , (1, 4), , Fig 10.21, , y – 1 = – 1 (x – 4) or x + y – 5 = 0, , ... (2), , Solving (1) and (2), we get x = 1 and y = 4 so that point of intersection of the two lines, is Q (1, 4). Now, distance of line (1) from the point P (4, 1) along the line (2), = the distance between the points P (4, 1) and Q (1, 4)., =, , (1 − 4 )2 + ( 4 − 1)2 = 3, , 2 units ., , Example 22 Assuming that straight lines work as the plane mirror for a point, find, the image of the point (1, 2) in the line x – 3y + 4 = 0., Solution Let Q (h, k) is the image of the point P (1, 2) in the line, x – 3y + 4 = 0, , 2020-21, , ... (1)

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230, , MATHEMATICS, , Fig10.22, , Therefore, the line (1) is the perpendicular bisector of line segment PQ (Fig 10.22)., Hence Slope of line PQ =, so that k − 2 = −1, h −1 1, 3, , −1, ,, Slope of line x − 3 y + 4 = 0, , or 3h + k = 5, , ... (2), ,  h +1 k + 2 , ,,  will satisfy the equation (1) so that, 2 ,  2, , and the mid-point of PQ, i.e., point , , h +1  k + 2 , − 3,  + 4 = 0 or h − 3k = −3, 2,  2 , Solving (2) and (3), we get h =, , 6, 7, and k = ., 5, 5, , 6 7, Hence, the image of the point (1, 2) in the line (1) is  ,  ., 5 5, Example 23 Show that the area of the triangle formed by the lines, , (c –c ), y = m x + c , y = m x + c and x = 0 is, 1, , 1, , 1, , 2, , 2, , 2, , 2, , 2 m1 − m 2, , 2020-21, , ., , ... (3)

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STRAIGHT LINES, , Solving these two relations we get k = 0 or h =, equations y = 0 or x =, , 233, , 5, . Thus, the point (h, k) satisfies the, 3, , 5, , which represent straight lines. Hence, path of the point, 3, , equidistant from the lines (1) and (2) is a straight line., , Miscellaneous Exercise on Chapter 10, 1. Find the values of k for which the line (k–3) x – (4 – k2) y + k2 –7k + 6 = 0 is, (a) Parallel to the x-axis,, (b) Parallel to the y-axis,, (c) Passing through the origin., 2. Find the values of θ and p, if the equation x cos θ + y sinθ = p is the normal form, of the line 3 x + y + 2 = 0., 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum, and product are 1 and – 6, respectively., 4. What are the points on the y-axis whose distance from the line, , x y, + = 1 is, 3 4, , 4 units., 5. Find perpendicular distance from the origin to the line joining the points (cosθ, sin θ), and (cos φ, sin φ)., 6. Find the equation of the line parallel to y-axis and drawn through the point of, intersection of the lines x – 7y + 5 = 0 and 3x + y = 0., 7. Find the equation of a line drawn perpendicular to the line, 8., 9., 10., 11., 12., , x y, + = 1 through the, 4 6, , point, where it meets the y-axis., Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0., Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and, 2x – y – 3 = 0 may intersect at one point., If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are, concurrent, then show that m1(c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0., Find the equation of the lines through the point (3, 2) which make an angle of 45o, with the line x – 2y = 3., Find the equation of the line passing through the point of intersection of the lines, 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes., , 2020-21

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234, , MATHEMATICS, , 13. Show that the equation of the line passing through the origin and making an angle, , y, m ± tan ‚, θ with the line y = mx + c is x = 1 ∓ m tan ‚ ., 14. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?, 15. Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line, 2x – y = 0., 16. Find the direction in which a straight line must be drawn through the point (–1, 2), so that its point of intersection with the line x + y = 4 may be at a distance of, 3 units from this point., 17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and, (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle., 18. Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the, line to be a plane mirror., 19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find, the value of m., 20. If sum of the perpendicular distances of a variable point P (x, y) from the lines, x + y – 5 = 0 and 3x – 2y +7 = 0 is always 10. Show that P must move on a line., 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0, and 3x + 2y + 6 = 0., 22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the, reflected ray passes through the point (5, 3). Find the coordinates of A., 23. Prove that the product of the lengths of the perpendiculars drawn from the, , (, , ) (, , ), , x, y, cosθ + sin θ = 1is b 2 ., a, b, 24. A person standing at the junction (crossing) of two straight paths represented by, the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose, equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he, should follow., points, , a 2 − b 2 ,0 and − a 2 − b2 ,0 to the line, , Summary, , ® Slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2), , y 2 − y1 y1 − y 2, =, , x 1 ≠ x 2., x 2 − x1 x1 − x 2, ® If a line makes an angle á with the positive direction of x-axis, then the slope, of the line is given by m = tan α, α ≠ 90°., ® Slope of horizontal line is zero and slope of vertical line is undefined., , is given by m =, , 2020-21

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STRAIGHT LINES, , ® An acute angle (say θ) between lines L1 and L2 with slopes m1 and m2 is, m2 – m1, given by tanθ = 1 + m m ,1 + m1 m2 ≠ 0 ., 1, 2, , ® Two lines are parallel if and only if their slopes are equal., ® Two lines are perpendicular if and only if product of their slopes is –1., ® Three points A, B and C are collinear, if and only if slope of AB = slope of BC., ® Equation of the horizontal line having distance a from the x-axis is either, , y = a or y = – a., ® Equation of the vertical line having distance b from the y-axis is either, x = b or x = – b., ® The point (x, y) lies on the line with slope m and through the fixed point (xo, yo),, if and only if its coordinates satisfy the equation y – y o = m (x – xo)., ® Equation of the line passing through the points (x1, y1) and (x2, y2) is given by, , y − y1 =, , y 2 − y1, ( x − x1)., x 2 − x1, , ® The point (x, y) on the line with slope m and y-intercept c lies on the line if and, only if y = mx + c., ® If a line with slope m makes x-intercept d. Then equation of the line is, y = m (x – d)., ® Equation of a line making intercepts a and b on the x-and y-axis,, respectively, is, , x y, + = 1., a b, , ® The equation of the line having normal distance from origin p and angle between, , normal and the positive x-axis ω is given by x cos ω + y sin ω = p ., ® Any equation of the form Ax + By + C = 0, with A and B are not zero,, simultaneously, is called the general linear equation or general equation of, a line., ® The perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1, y1), is given by d =, , Ax1 + B y1 + C, , ., 2, 2, A +B, ® Distance between the parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0,, is given by d =, , C1 − C 2, 2, 2, A +B, , ., , 2020-21, , 235

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CBSE Class 11 Study Material,  Printable Worksheets for Class 11, NCERT Solutions for Class 11, , NCERT Solutions for class 11 Maths,  NCERT Solutions for class 11 Physics,  NCERT Solutions for class 11 Chemistry,  NCERT Solutions for class 11 Biology,  NCERT Solutions for class 11 English,  NCERT Solutions for Class 11 English, Woven Words Essay,  NCERT Solutions for Class 11 English, Woven Short Stories,  NCERT Solutions for Class 11 English, Woven Words Poetry,  NCERT Solutions for class 11 Accountancy,  NCERT Solutions for class 11 Business, Studies,  NCERT Solutions for class 11 Economics,  NCERT Solutions for class 11 Computer, Science – Python,  Class 11 Hindi Aroh (आरोह भाग 1), , ,  Class 11 Hindi Vitan (वितान भाग 1)

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 CBSE Class 11 Physics Manual,  CBSE Class 11 Chemistry Manual,  Trigonometry Formulas,  Integration Formulas,  JEE Main Study Material,  NEET Study Material,  CBSE Class 11 Notes,  Class 11 Maths Notes,  Class 11 Physics Notes,  Class 11 Chemistry Notes,  Class 11 Biology Notes,  Class 11 English Notes,  Class 11 English Woven Words Short Stories,  CBSE Class 11 English Woven Words Essay,  CBSE Class 11 English Woven Words Poetry,  CBSE Class 11 English Snapshots,  CBSE Class 11 English Hornbill,  Class 11 Business Studies Notes,  Class 11 Accountancy Notes,  Class 11 Psychology Notes,  Class 11 Entrepreneurship Notes,  Class 11 Economics Notes

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 Class 11 Indian Economic Development, Notes,  Statistics for Economics Class 11 Notes,  Class 11 Political Science Notes,  Class 11 History Notes,  Sociology Class 11 Notes,  Geography Class 11 Notes, NCERT Books for Class 11, ,  Class 11 NCERT Maths Books,  Class 11 Physics NCERT Book,  Class 11 Chemistry NCERT Book,  Class 11 Biology NCERT Book,  Class 11 Political Theory Part-I,  Class 11 NCERT Business Studies Books,  Class 11 India Constitution at Work,  NCERT Geography Book Class 11,  NCERT Class 11 History Book,  Class 11 India Economic Development,  Class 11 NCERT English Books,  NCERT Sanksrit Books Class 11,  Class 11 Computer and Communication, Technology Book,  Class 11 NCERT Accountancy Books

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 Class 11 Statistics,  Class 11 Introduction to Psychology,  Class 11 Introducing Sociology,  Class 11 Understanding Society,  Class 11 Fine Arts,  Class 11 Heritage Craft Books,  Class 11 Nai Awaz,  Class 11 Dhanak,  Class 11 The story of Graphic Design,  Class 11 Human Ecology and Family, Sciences