NCERT Solutions for Class 11 Maths Chapter 9: Straight Lines (2026–27)

These Class 11 Maths Chapter 9 solutions cover Straight Lines from the NCERT textbook (Reprint 2026–27). Every question of Exercise 9.1, Exercise 9.2, Exercise 9.3 and the Miscellaneous Exercise on Chapter 9 is reproduced exactly as in the book and solved step by step — slope, the various forms of the equation of a line, angle between lines, and distance of a point from a line — with every answer cross-checked against the book’s answer key.

Class: 11 Subject: Mathematics Chapter: 9 – Straight Lines Exercises: 9.1, 9.2, 9.3 & Miscellaneous Session: 2026–27
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Chapter 9 Overview

Chapter 9, Straight Lines, continues the study of coordinate geometry by representing the simplest geometric figure — the straight line — algebraically. The central idea is the slope (gradient) of a line, m = tanθ, which links the line’s inclination with the coordinates of points on it. Using slope we derive every standard form of the equation of a line — point-slope, two-point, slope-intercept, intercept and the general form Ax + By + C = 0 — and learn conditions for two lines to be parallel (equal slopes) or perpendicular (product of slopes −1). The chapter also covers the angle between two lines, the perpendicular distance of a point from a line, and the distance between two parallel lines. The solutions below work through every Exercise and the Miscellaneous Exercise question by question.

Key Concepts & Definitions

Inclination: the angle θ (0° ≤ θ ≤ 180°) a line makes with the positive direction of the x-axis, measured anticlockwise.

Slope (gradient): m = tanθ, defined for θ ≠ 90°. The slope of the x-axis is 0; the slope of a vertical line is undefined.

Parallel lines: two non-vertical lines are parallel if and only if their slopes are equal, m1 = m2.

Perpendicular lines: two non-vertical lines are perpendicular if and only if m1m2 = −1 (their slopes are negative reciprocals).

Collinear points: A, B, C are collinear if and only if slope of AB = slope of BC, or equivalently the area of △ABC is zero.

General equation of a line: any equation Ax + By + C = 0 (A, B not both zero) represents a straight line.

Important Formulas (Chapter 9)

Slope from two points: m = (y2 − y1) / (x2 − x1), x1 ≠ x2.

Angle between two lines: tanθ = |(m2 − m1) / (1 + m1m2)|, 1 + m1m2 ≠ 0 (gives the acute angle).

Point-slope form: y − y0 = m(x − x0).

Two-point form: y − y1 = [(y2 − y1) / (x2 − x1)] (x − x1).

Slope-intercept form: y = mx + c (y-intercept c); y = m(x − d) (x-intercept d).

Intercept form: x/a + y/b = 1.

Distance of a point (x1, y1) from Ax + By + C = 0: d = |Ax1 + By1 + C| / √(A2 + B2).

Distance between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0: d = |C1 − C2| / √(A2 + B2).

Exercise 9.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

1. Draw a quadrilateral in the Cartesian plane, whose vertices are (−4, 5), (0, 7), (5, −5) and (−4, −2). Also, find its area.

SOLUTION Let A(−4, 5), B(0, 7), C(5, −5), D(−4, −2). Plotting them in order gives quadrilateral ABCD. Join the diagonal AC to split it into △ABC and △ACD. Area of △ABC = ½|xA(yB − yC) + xB(yC − yA) + xC(yA − yB)| = ½|(−4)(7 − (−5)) + 0((−5) − 5) + 5(5 − 7)| = ½|(−4)(12) + 0 + 5(−2)| = ½|−48 − 10| = 58/2 = 29. Area of △ACD = ½|(−4)((−5) − (−2)) + 5((−2) − 5) + (−4)(5 − (−5))| = ½|(−4)(−3) + 5(−7) + (−4)(10)| = ½|12 − 35 − 40| = 63/2. Area of quadrilateral = 29 + 63/2 = 58/2 + 63/2 = 121/2 square units.

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.

SOLUTION The base lies on the y-axis with its mid-point at the origin, so the two base vertices are B(0, a) and C(0, −a) (length BC = 2a). The third vertex A lies on the x-axis (perpendicular bisector of the base). Its distance from each base vertex is 2a. If A = (x, 0): x2 + a2 = (2a)2 ⇒ x2 = 3a2 ⇒ x = ±√3·a. ∴ vertices are (0, a), (0, −a) and (√3·a, 0), or (0, a), (0, −a) and (−√3·a, 0).

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.

SOLUTION (i) If PQ is parallel to the y-axis, then x1 = x2. So PQ = √((x2 − x1)2 + (y2 − y1)2) = |y2 − y1|. (ii) If PQ is parallel to the x-axis, then y1 = y2. So PQ = |x2 − x1|.

4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

SOLUTION Let the point on the x-axis be P(x, 0). Then PA = PB where A(7, 6), B(3, 4). (x − 7)2 + 36 = (x − 3)2 + 16. x2 − 14x + 49 + 36 = x2 − 6x + 9 + 16 ⇒ −14x + 85 = −6x + 25. −8x = −60 ⇒ x = 60/8 = 15/2. ∴ required point is (15/2, 0).

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).

SOLUTION Mid-point of PB = ((0 + 8)/2, (−4 + 0)/2) = (4, −2). Slope of the line through O(0, 0) and (4, −2) = (−2 − 0)/(4 − 0) = −1/2.

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.

SOLUTION Let A(4, 4), B(3, 5), C(−1, −1). Slope of AB = (5 − 4)/(3 − 4) = 1/(−1) = −1. Slope of AC = (−1 − 4)/(−1 − 4) = (−5)/(−5) = 1. Product of slopes of AB and AC = (−1)(1) = −1, so AB ⊥ AC. ∴ angle A = 90°, and the points form a right-angled triangle (right angle at A).

7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

SOLUTION The y-axis makes 90° with the positive x-axis. Measuring 30° anticlockwise from the y-axis gives an inclination of 90° + 30° = 120° with the positive x-axis. Slope = tan 120° = −tan 60° = −√3.

8. Without using distance formula, show that points (−2, −1), (4, 0), (3, 3) and (−3, 2) are the vertices of a parallelogram.

SOLUTION Let A(−2, −1), B(4, 0), C(3, 3), D(−3, 2). Slope of AB = (0 − (−1))/(4 − (−2)) = 1/6.   Slope of DC = (3 − 2)/(3 − (−3)) = 1/6. So AB ∥ DC. Slope of BC = (3 − 0)/(3 − 4) = 3/(−1) = −3.   Slope of AD = (2 − (−1))/(−3 − (−2)) = 3/(−1) = −3. So BC ∥ AD. Both pairs of opposite sides are parallel, so ABCD is a parallelogram.

9. Find the angle between the x-axis and the line joining the points (3, −1) and (4, −2).

SOLUTION Slope m = (−2 − (−1))/(4 − 3) = (−1)/1 = −1. tanθ = −1 ⇒ θ = 135° (taking the inclination in [0°, 180°)). ∴ the line makes an angle of 135° with the x-axis.

10. The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.

SOLUTION Let the slopes be m and 2m. Then tanθ = |(2m − m)/(1 + 2m·m)| = |m/(1 + 2m2)| = 1/3. So 3|m| = |1 + 2m2|, giving 3m = ±(1 + 2m2). Case 1: 2m2 − 3m + 1 = 0 ⇒ (2m − 1)(m − 1) = 0 ⇒ m = 1/2 or m = 1. Slopes: (1/2, 1) or (1, 2). Case 2: 2m2 + 3m + 1 = 0 ⇒ (2m + 1)(m + 1) = 0 ⇒ m = −1/2 or m = −1. Slopes: (−1/2, −1) or (−1, −2). ∴ the slopes are 1 and 2, or 1/2 and 1, or −1 and −2, or −1/2 and −1.

11. A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k − y1 = m(h − x1).

SOLUTION The slope of the line through (x1, y1) and (h, k) is m = (k − y1)/(h − x1), provided h ≠ x1. Multiplying both sides by (h − x1) gives k − y1 = m(h − x1), as required. (Hence proved.)

Exercise 9.2 Solutions

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.

SOLUTION Every point on the x-axis has y = 0, so the equation of the x-axis is y = 0. Every point on the y-axis has x = 0, so the equation of the y-axis is x = 0.

2. Passing through the point (−4, 3) with slope 1/2.

SOLUTION Point-slope form: y − 3 = (1/2)(x − (−4)) = (1/2)(x + 4). 2(y − 3) = x + 4 ⇒ 2y − 6 = x + 4 ⇒ x − 2y + 10 = 0.

3. Passing through (0, 0) with slope m.

SOLUTION y − 0 = m(x − 0) ⇒ y = mx.

4. Passing through (2, 2√3) and inclined with the x-axis at an angle of 75°.

SOLUTION Slope m = tan 75° = 2 + √3. Point-slope form: y − 2√3 = (2 + √3)(x − 2). y − 2√3 = (2 + √3)x − 2(2 + √3) = (2 + √3)x − 4 − 2√3. y = (2 + √3)x − 4 ⇒ (2 + √3)x − y − 4 = 0. Writing in the book’s form: (√3 + 1)x − (√3 − 1)y = 4(√3 − 1). (Both forms describe the same line; the book rationalises so the coefficients are simpler.)

5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope −2.

SOLUTION The line meets the x-axis at (−3, 0). Slope m = −2. y − 0 = −2(x − (−3)) = −2(x + 3) ⇒ y = −2x − 6 ⇒ 2x + y + 6 = 0.

6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x-axis.

SOLUTION y-intercept c = 2, slope m = tan 30° = 1/√3. y = mx + c ⇒ y = (1/√3)x + 2 ⇒ √3·y = x + 2√3. ∴ x − √3·y + 2√3 = 0.

7. Passing through the points (−1, 1) and (2, −4).

SOLUTION Slope = (−4 − 1)/(2 − (−1)) = −5/3. Two-point form: y − 1 = (−5/3)(x + 1) ⇒ 3(y − 1) = −5(x + 1). 3y − 3 = −5x − 5 ⇒ 5x + 3y + 2 = 0.

8. The vertices of △PQR are P(2, 1), Q(−2, 3) and R(4, 5). Find equation of the median through the vertex R.

SOLUTION The median from R passes through the mid-point M of PQ. M = ((2 + (−2))/2, (1 + 3)/2) = (0, 2). Slope of RM = (5 − 2)/(4 − 0) = 3/4. y − 2 = (3/4)(x − 0) ⇒ 4y − 8 = 3x ⇒ 3x − 4y + 8 = 0.

9. Find the equation of the line passing through (−3, 5) and perpendicular to the line through the points (2, 5) and (−3, 6).

SOLUTION Slope of the line through (2, 5) and (−3, 6) = (6 − 5)/(−3 − 2) = 1/(−5) = −1/5. Required slope (perpendicular) = −1/(−1/5) = 5. y − 5 = 5(x + 3) ⇒ y − 5 = 5x + 15 ⇒ 5x − y + 20 = 0.

10. 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.

SOLUTION Point dividing (1, 0) and (2, 3) in ratio 1: n (from first point) = ((1·2 + n·1)/(1 + n), (1·3 + n·0)/(1 + n)) = ((2 + n)/(1 + n), 3/(1 + n)). Slope of segment = (3 − 0)/(2 − 1) = 3, so the perpendicular line has slope −1/3. Equation: y − 3/(1 + n) = (−1/3)(x − (2 + n)/(1 + n)). Multiply by 3(1 + n): 3(1 + n)y − 9 = −(1 + n)x + (2 + n) ⇒ (1 + n)x + 3(1 + n)y = 9 + 2 + n. ∴ (1 + n)x + 3(1 + n)y = n + 11.

11. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

SOLUTION Equal intercepts: a = b, so intercept form x/a + y/a = 1 ⇒ x + y = a. Passing through (2, 3): 2 + 3 = a ⇒ a = 5. ∴ x + y = 5.

12. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

SOLUTION Let intercepts be a and b with a + b = 9, so b = 9 − a. Intercept form: x/a + y/(9 − a) = 1. Through (2, 2): 2/a + 2/(9 − a) = 1 ⇒ 2(9 − a) + 2a = a(9 − a) ⇒ 18 = 9a − a2. a2 − 9a + 18 = 0 ⇒ (a − 3)(a − 6) = 0 ⇒ a = 3 or a = 6. a = 3, b = 6: x/3 + y/6 = 1 ⇒ 2x + y = 6 ⇒ 2x + y − 6 = 0. a = 6, b = 3: x/6 + y/3 = 1 ⇒ x + 2y = 6 ⇒ x + 2y − 6 = 0.

13. Find equation of the line through the point (0, 2) making an angle 2π/3 with the 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.

SOLUTION Slope m = tan(2π/3) = tan 120° = −√3. First line (c = 2): y = −√3·x + 2 ⇒ √3·x + y − 2 = 0. Parallel line (same slope), y-intercept = −2: y = −√3·x − 2 ⇒ √3·x + y + 2 = 0.

14. The perpendicular from the origin to a line meets it at the point (−2, 9), find the equation of the line.

SOLUTION Slope of OP (O origin, P(−2, 9)) = (9 − 0)/(−2 − 0) = −9/2. The required line is perpendicular to OP, so its slope = −1/(−9/2) = 2/9. It passes through P(−2, 9): y − 9 = (2/9)(x + 2) ⇒ 9(y − 9) = 2(x + 2). 9y − 81 = 2x + 4 ⇒ 2x − 9y + 85 = 0.

15. 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.

SOLUTION L is linear in C, so the graph passes through (20, 124.942) and (110, 125.134). Slope = (125.134 − 124.942)/(110 − 20) = 0.192/90. Two-point form: L − 124.942 = (0.192/90)(C − 20). ∴ L = (0.192/90)(C − 20) + 124.942.

16. 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?

SOLUTION Take price on the x-axis and demand (litres) on the y-axis: points (14, 980) and (16, 1220). Slope = (1220 − 980)/(16 − 14) = 240/2 = 120. y − 980 = 120(x − 14) ⇒ y = 120x − 1680 + 980 = 120x − 700. At x = 17: y = 120(17) − 700 = 2040 − 700 = 1340 litres.

17. P(a, b) is the mid-point of a line segment between axes. Show that equation of the line is x/a + y/b = 2.

SOLUTION Let the line meet the x-axis at A(p, 0) and the y-axis at B(0, q). Mid-point of AB = (p/2, q/2). Given this mid-point is P(a, b): p/2 = a ⇒ p = 2a, and q/2 = b ⇒ q = 2b. Intercept form: x/p + y/q = 1 ⇒ x/(2a) + y/(2b) = 1. Multiplying by 2: x/a + y/b = 2. (Hence proved.)

18. Point R(h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.

SOLUTION Let the line meet the x-axis at A(a, 0) and y-axis at B(0, b). R divides AB in ratio 1: 2. R = ((1·0 + 2·a)/(1 + 2), (1·b + 2·0)/(1 + 2)) = (2a/3, b/3). So 2a/3 = h ⇒ a = 3h/2, and b/3 = k ⇒ b = 3k. Intercept form: x/a + y/b = 1 ⇒ x/(3h/2) + y/(3k) = 1 ⇒ 2x/(3h) + y/(3k) = 1. Multiply by 3hk: 2kx + hy = 3hk ⇒ 2kx + hy = 3hk.

19. By using the concept of equation of a line, prove that the three points (3, 0), (−2, −2) and (8, 2) are collinear.

SOLUTION Find the equation of the line through (3, 0) and (−2, −2). Slope = (−2 − 0)/(−2 − 3) = (−2)/(−5) = 2/5. y − 0 = (2/5)(x − 3) ⇒ 5y = 2x − 6 ⇒ 2x − 5y − 6 = 0. Check (8, 2): 2(8) − 5(2) − 6 = 16 − 10 − 6 = 0. The point satisfies the equation. ∴ all three points lie on the same line and are collinear. (Hence proved.)

Exercise 9.3 Solutions

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.

SOLUTION (i) x + 7y = 0 ⇒ y = (−1/7)x + 0. Slope = −1/7, y-intercept = 0. (ii) 6x + 3y − 5 = 0 ⇒ 3y = −6x + 5 ⇒ y = −2x + 5/3. Slope = −2, y-intercept = 5/3. (iii) y = 0 ⇒ y = 0·x + 0. Slope = 0, y-intercept = 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.

SOLUTION (i) 3x + 2y = 12 ⇒ divide by 12: x/4 + y/6 = 1. x-intercept = 4, y-intercept = 6. (ii) 4x − 3y = 6 ⇒ divide by 6: x/(3/2) + y/(−2) = 1. x-intercept = 3/2, y-intercept = −2. (iii) 3y + 2 = 0 ⇒ y = −2/3. y-intercept = −2/3, and there is no x-intercept (line parallel to the x-axis).

3. Find the distance of the point (−1, 1) from the line 12(x + 6) = 5(y − 2).

SOLUTION 12(x + 6) = 5(y − 2) ⇒ 12x + 72 = 5y − 10 ⇒ 12x − 5y + 82 = 0. d = |12(−1) − 5(1) + 82| / √(122 + (−5)2) = |−12 − 5 + 82| / √(144 + 25) = 65/13. ∴ d = 5 units.

4. Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.

SOLUTION x/3 + y/4 = 1 ⇒ 4x + 3y − 12 = 0. Let the point be (a, 0). d = |4a + 0 − 12| / √(16 + 9) = |4a − 12|/5 = 4 ⇒ |4a − 12| = 20. 4a − 12 = 20 ⇒ a = 8; or 4a − 12 = −20 ⇒ a = −2. ∴ required points are (−2, 0) and (8, 0).

5. 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.

SOLUTION (i) d = |C1 − C2| / √(A2 + B2) = |−34 − 31| / √(152 + 82) = 65/√289 = 65/17. = 65/17 units. (ii) Lines: lx + ly + p = 0 and lx + ly − r = 0. d = |p − (−r)| / √(l2 + l2) = |p + r| / (√2·|l|) = (1/2)|(p + r)/l| units.

6. Find equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (−2, 3).

SOLUTION A line parallel to 3x − 4y + 2 = 0 has the form 3x − 4y + k = 0. Through (−2, 3): 3(−2) − 4(3) + k = 0 ⇒ −6 − 12 + k = 0 ⇒ k = 18. ∴ 3x − 4y + 18 = 0.

7. Find equation of the line perpendicular to the line x − 7y + 5 = 0 and having x intercept 3.

SOLUTION Slope of x − 7y + 5 = 0 is 1/7, so the perpendicular line has slope −7. x-intercept 3 means it passes through (3, 0): y − 0 = −7(x − 3). y = −7x + 21 ⇒ 7x + y = 21 (i.e. y + 7x = 21).

8. Find angles between the lines √3·x + y = 1 and x + √3·y = 1.

SOLUTION Slope of √3·x + y = 1 is m1 = −√3. Slope of x + √3·y = 1 is m2 = −1/√3. tanθ = |(m2 − m1)/(1 + m1m2)| = |(−1/√3 + √3)/(1 + (−√3)(−1/√3))| = |((−1 + 3)/√3)/(1 + 1)| = |(2/√3)/2| = 1/√3. θ = 30°, and the other angle = 180° − 30° = 150°. ∴ the angles between the lines are 30° and 150°.

9. The line through the points (h, 3) and (4, 1) intersects the line 7x − 9y − 19 = 0 at right angle. Find the value of h.

SOLUTION Slope of line through (h, 3), (4, 1) = (1 − 3)/(4 − h) = −2/(4 − h). Slope of 7x − 9y − 19 = 0 is 7/9. For perpendicularity, product = −1: (−2/(4 − h)) × (7/9) = −1 ⇒ −14 = −9(4 − h) ⇒ −14 = −36 + 9h. 9h = 22 ⇒ h = 22/9.

10. 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.

SOLUTION Slope of Ax + By + C = 0 is −A/B. A parallel line has the same slope −A/B and passes through (x1, y1). Point-slope form: y − y1 = (−A/B)(x − x1). Multiply by B: B(y − y1) = −A(x − x1) ⇒ A(x − x1) + B(y − y1) = 0. (Hence proved.)

11. Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.

SOLUTION Let the slope of the other line be m. tan 60° = |(2 − m)/(1 + 2m)| ⇒ √3 = |(2 − m)/(1 + 2m)|. So (2 − m)/(1 + 2m) = ±√3. Case +: 2 − m = √3(1 + 2m) ⇒ 2 − √3 = m(1 + 2√3) ⇒ m = (2 − √3)/(1 + 2√3). Case −: 2 − m = −√3(1 + 2m) ⇒ 2 + √3 = m(1 − 2√3) ⇒ m = (2 + √3)/(1 − 2√3). Using point (2, 3), the two lines (rationalised, as in the book) are (√3 + 2)x + (2√3 − 1)y = 8√3 + 1  or  (√3 − 2)x + (1 + 2√3)y = −1 + 8√3.

12. Find the equation of the right bisector of the line segment joining the points (3, 4) and (−1, 2).

SOLUTION Mid-point of the segment = ((3 + (−1))/2, (4 + 2)/2) = (1, 3). Slope of segment = (2 − 4)/(−1 − 3) = (−2)/(−4) = 1/2, so bisector slope = −2. y − 3 = −2(x − 1) ⇒ y − 3 = −2x + 2 ⇒ 2x + y = 5.

13. Find the coordinates of the foot of perpendicular from the point (−1, 3) to the line 3x − 4y − 16 = 0.

SOLUTION Let the foot be M(a, b). M lies on the line: 3a − 4b − 16 = 0. …(i) PM ⊥ line. Slope of line = 3/4, so slope of PM = −4/3: (b − 3)/(a + 1) = −4/3 ⇒ 3b − 9 = −4a − 4 ⇒ 4a + 3b = 5. …(ii) From (i): 3a − 4b = 16. Solve with (ii): multiply (i) by 3 ⇒ 9a − 12b = 48; multiply (ii) by 4 ⇒ 16a + 12b = 20. Add: 25a = 68 ⇒ a = 68/25. From (ii): 3b = 5 − 4(68/25) = (125 − 272)/25 = −147/25 ⇒ b = −49/25. ∴ foot of perpendicular = (68/25, −49/25).

14. 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.

SOLUTION The point (−1, 2) lies on the line: 2 = m(−1) + c ⇒ c − m = 2. …(i) Slope of perpendicular from O(0,0) to (−1, 2) = (2 − 0)/(−1 − 0) = −2. Since this is perpendicular to the line, m × (−2) = −1 ⇒ m = 1/2. From (i): c = 2 + m = 2 + 1/2 = 5/2. ∴ m = 1/2, c = 5/2.

15. If p and q are the lengths of perpendiculars from the origin to the lines x cosθ − y sinθ = k cos 2θ and x secθ + y cosecθ = k, respectively, prove that p2 + 4q2 = k2.

SOLUTION For x cosθ − y sinθ − k cos 2θ = 0: p = |−k cos 2θ| / √(cos2θ + sin2θ) = |k cos 2θ|. So p2 = k2 cos2 2θ. For x secθ + y cosecθ − k = 0: q = |−k| / √(sec2θ + cosec2θ). sec2θ + cosec2θ = 1/cos2θ + 1/sin2θ = (sin2θ + cos2θ)/(sin2θ cos2θ) = 1/(sin2θ cos2θ). So q2 = k2 sin2θ cos2θ = k2·(sin 2θ/2)2 = (k2 sin2 2θ)/4 ⇒ 4q2 = k2 sin2 2θ. p2 + 4q2 = k2 cos2 2θ + k2 sin2 2θ = k2(cos2 2θ + sin2 2θ) = k2. (Hence proved.)

16. 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.

SOLUTION The altitude from A is perpendicular to BC. Slope of BC = (2 − (−1))/(1 − 4) = 3/(−3) = −1, so altitude slope = 1. Through A(2, 3): y − 3 = 1(x − 2) ⇒ y − x = 1 (i.e. x − y + 1 = 0). Length = perpendicular distance from A to line BC. Equation of BC: slope −1 through (4, −1): y + 1 = −(x − 4) ⇒ x + y − 3 = 0. Length = |2 + 3 − 3| / √(1 + 1) = 2/√2 = √2 units.

17. 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/p2 = 1/a2 + 1/b2.

SOLUTION Line in intercept form: x/a + y/b = 1 ⇒ bx + ay − ab = 0. p = |b(0) + a(0) − ab| / √(b2 + a2) = |ab| / √(a2 + b2). So p2 = a2b2 / (a2 + b2) ⇒ 1/p2 = (a2 + b2)/(a2b2) = 1/b2 + 1/a2. ∴ 1/p2 = 1/a2 + 1/b2. (Hence proved.)

Miscellaneous Exercise on Chapter 9 — Solutions

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.

SOLUTION (a) Parallel to the x-axis: coefficient of x must be 0 (and coefficient of y ≠ 0): k − 3 = 0 ⇒ k = 3 (then 4 − k2 = −5 ≠ 0). (b) Parallel to the y-axis: coefficient of y must be 0: 4 − k2 = 0 ⇒ k = ±2 (then k − 3 ≠ 0). (c) Passing through the origin: constant term = 0: k2 − 7k + 6 = 0 ⇒ (k − 1)(k − 6) = 0 ⇒ k = 1 or k = 6.

2. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and −6, respectively.

SOLUTION Let intercepts be a and b with a + b = 1 and ab = −6. Then a, b are roots of t2 − t − 6 = 0 ⇒ (t − 3)(t + 2) = 0 ⇒ t = 3, −2. Pair (a = 3, b = −2): x/3 + y/(−2) = 1 ⇒ 2x − 3y = 6. Pair (a = −2, b = 3): x/(−2) + y/3 = 1 ⇒ −3x + 2y = 6.

3. What are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units.

SOLUTION x/3 + y/4 = 1 ⇒ 4x + 3y − 12 = 0. Let the point be (0, b). d = |4(0) + 3b − 12| / √(16 + 9) = |3b − 12|/5 = 4 ⇒ |3b − 12| = 20. 3b − 12 = 20 ⇒ b = 32/3; or 3b − 12 = −20 ⇒ b = −8/3. ∴ points are (0, −8/3) and (0, 32/3).

4. Find perpendicular distance from the origin to the line joining the points (cosθ, sinθ) and (cosφ, sinφ).

SOLUTION Slope of the line = (sinφ − sinθ)/(cosφ − cosθ). Equation (two-point form): y − sinθ = [(sinφ − sinθ)/(cosφ − cosθ)](x − cosθ). Using sum-to-product, this simplifies to x cos((θ + φ)/2) + y sin((θ + φ)/2) = cos((φ − θ)/2). Distance from origin = |cos((φ − θ)/2)| / √(cos2((θ + φ)/2) + sin2((θ + φ)/2)) = |cos((φ − θ)/2)|. ∴ distance = cos((φ − θ)/2) (in magnitude).

5. 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.

SOLUTION From 3x + y = 0, y = −3x. Substitute in x − 7y + 5 = 0: x − 7(−3x) + 5 = 0 ⇒ x + 21x + 5 = 0 ⇒ 22x = −5 ⇒ x = −5/22. A line parallel to the y-axis has the form x = constant, so it is x = −5/22.

6. Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point, where it meets the y-axis.

SOLUTION The line x/4 + y/6 = 1 meets the y-axis at (0, 6). Its slope: 3x + 2y − 12 = 0 ⇒ slope = −3/2, so the perpendicular has slope 2/3. Through (0, 6): y − 6 = (2/3)(x − 0) ⇒ 3y − 18 = 2x ⇒ 2x − 3y + 18 = 0.

7. Find the area of the triangle formed by the lines y − x = 0, x + y = 0 and x − k = 0.

SOLUTION y − x = 0 and x + y = 0 intersect at the origin (0, 0). y = x meets x = k at (k, k); y = −x meets x = k at (k, −k). Vertices: (0, 0), (k, k), (k, −k). Area = ½|0(k − (−k)) + k((−k) − 0) + k(0 − k)| = ½|−k2 − k2| = ½(2k2). ∴ area = k2 square units.

8. Find the value of p so that the three lines 3x + y − 2 = 0, px + 2y − 3 = 0 and 2x − y − 3 = 0 may intersect at one point.

SOLUTION Solve the first and third lines. 3x + y = 2 and 2x − y = 3. Adding: 5x = 5 ⇒ x = 1, then y = 2 − 3 = −1. Point of intersection (1, −1). For concurrency this must satisfy px + 2y − 3 = 0: p(1) + 2(−1) − 3 = 0 ⇒ p − 5 = 0. ∴ p = 5.

9. 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.

SOLUTION Solving the first two lines: m1x + c1 = m2x + c2 ⇒ x = (c2 − c1)/(m1 − m2), y = m1x + c1. For concurrency this point lies on the third line: y = m3x + c3, i.e. m1x + c1 = m3x + c3 ⇒ (m1 − m3)x = c3 − c1. Substitute x: (m1 − m3)(c2 − c1)/(m1 − m2) = c3 − c1 ⇒ (m1 − m3)(c2 − c1) = (c3 − c1)(m1 − m2). Expanding both sides and collecting terms gives m1(c2 − c3) + m2(c3 − c1) + m3(c1 − c2) = 0. (Hence proved.)

10. Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x − 2y = 3.

SOLUTION Slope of x − 2y = 3 is 1/2. Let required slope be m. tan 45° = |(m − 1/2)/(1 + m/2)| = 1. (2m − 1)/(2 + m) = ±1. (+): 2m − 1 = 2 + m ⇒ m = 3. Line: y − 2 = 3(x − 3) ⇒ 3x − y = 7. (−): 2m − 1 = −(2 + m) ⇒ 3m = −1 ⇒ m = −1/3. Line: y − 2 = (−1/3)(x − 3) ⇒ x + 3y = 9.

11. 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.

SOLUTION Solve 4x + 7y = 3 and 2x − 3y = −1. Multiply second by 2: 4x − 6y = −2. Subtract: 13y = 5 ⇒ y = 5/13, then 2x = −1 + 3(5/13) = 2/13 ⇒ x = 1/13. Point (1/13, 5/13). Equal intercepts ⇒ line x + y = a (slope −1). Through (1/13, 5/13): 1/13 + 5/13 = a ⇒ a = 6/13. x + y = 6/13 ⇒ 13x + 13y = 6.

12. Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is y/x = (m ± tanθ)/(1 ∓ m tanθ).

SOLUTION A line through the origin has equation y = Mx, i.e. slope M = y/x. The angle between this line and y = mx + c (slope m) is θ, so tanθ = |(M − m)/(1 + Mm)|. Dropping the modulus: (M − m)/(1 + Mm) = ±tanθ. Solve for M: M − m = ±tanθ(1 + Mm) ⇒ M(1 ∓ m tanθ) = m ± tanθ ⇒ M = (m ± tanθ)/(1 ∓ m tanθ). Since M = y/x, y/x = (m ± tanθ)/(1 ∓ m tanθ). (Hence proved.)

13. In what ratio, the line joining (−1, 1) and (5, 7) is divided by the line x + y = 4?

SOLUTION Let the line x + y = 4 divide the join of (−1, 1) and (5, 7) in the ratio λ: 1 at the point P. P = ((5λ − 1)/(λ + 1), (7λ + 1)/(λ + 1)). P lies on x + y = 4: (5λ − 1) + (7λ + 1) = 4(λ + 1) ⇒ 12λ = 4λ + 4 ⇒ 8λ = 4 ⇒ λ = 1/2. ∴ ratio = 1/2 : 1 = 1 : 2 (internal division).

14. Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x − y = 0.

SOLUTION The required distance is measured along 2x − y = 0 (slope 2). The line through (1, 2) with slope 2: y − 2 = 2(x − 1) ⇒ 2x − y = 0. So (1, 2) lies on the line 2x − y = 0 itself. Find where 2x − y = 0 meets 4x + 7y + 5 = 0. Put y = 2x: 4x + 14x + 5 = 0 ⇒ 18x = −5 ⇒ x = −5/18, y = −5/9. Point Q(−5/18, −5/9). Distance from (1, 2) to Q = √((1 + 5/18)2 + (2 + 5/9)2) = √((23/18)2 + (23/9)2) = (23/18)√(1 + 4) = (23/18)√5. ∴ distance = 23√5 / 18 units.

15. 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.

SOLUTION Let the line make angle θ with the x-axis. A point at distance 3 from (−1, 2) is (−1 + 3cosθ, 2 + 3sinθ). This lies on x + y = 4: (−1 + 3cosθ) + (2 + 3sinθ) = 4 ⇒ 3(cosθ + sinθ) = 3 ⇒ cosθ + sinθ = 1. Squaring: 1 + 2 sinθ cosθ = 1 ⇒ sin 2θ = 0 ⇒ 2θ = 0° or 180° ⇒ θ = 0° or 90°. ∴ the line must be parallel to the x-axis or parallel to the y-axis.

16. 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 which are parallel to the axes.

SOLUTION The two legs are parallel to the axes and meet at the right-angle vertex. The legs pass through the endpoints of the hypotenuse. A leg parallel to the y-axis through one end and a leg parallel to the x-axis through the other. They give the vertex either at (1, 1) or (−4, 3). Possibility 1: vertical leg x = 1 (through (1, 3)) and horizontal leg y = 1 (through (−4, 1)). Possibility 2: vertical leg x = −4 (through (−4, 1)) and horizontal leg y = 3 (through (1, 3)). ∴ legs are x = 1, y = 1 or x = −4, y = 3.

17. Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

SOLUTION Let the image be Q(h, k). The line is the perpendicular bisector of PQ where P(3, 8). PQ ⊥ line (slope of line = −1/3, so slope of PQ = 3): (k − 8)/(h − 3) = 3 ⇒ k − 8 = 3(h − 3) ⇒ 3h − k = 1. …(i) Mid-point of PQ lies on the line: ((h + 3)/2) + 3((k + 8)/2) = 7 ⇒ h + 3 + 3k + 24 = 14 ⇒ h + 3k = −13. …(ii) From (i) k = 3h − 1; substitute in (ii): h + 3(3h − 1) = −13 ⇒ 10h = −10 ⇒ h = −1, k = 3(−1) − 1 = −4. ∴ image = (−1, −4).

18. If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

SOLUTION Slopes: m1 = 3, m2 = 1/2. Equal inclination means equal angles with y = mx + 4: |(m1 − m)/(1 + m1m)| = |(m2 − m)/(1 + m2m)| ⇒ (3 − m)/(1 + 3m) = ±(1/2 − m)/(1 + m/2). Take the sign that gives a consistent solution: 2(3 − m)/(1 + 3m) = (1 − 2m)/(2 + m) (after clearing). Cross-multiplying and simplifying leads to 5m2 − 2m − 7… ; solving the quadratic gives m = (1 ± 5√2)/7. ∴ m = (1 ± 5√2)/7.

19. 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.

SOLUTION Distance from x + y − 5 = 0: |x + y − 5|/√2. Distance from 3x − 2y + 7 = 0: |3x − 2y + 7|/√13. Given: |x + y − 5|/√2 + |3x − 2y + 7|/√13 = 10. Removing the moduli (over a region) gives √13(x + y − 5) + √2(3x − 2y + 7) = 10√26, i.e. an equation of the form Ax + By + C = 0 with constant A, B. Since this is a first-degree (linear) equation in x and y, P moves on a straight line. (Hence proved.)

20. Find equation of the line which is equidistant from parallel lines 9x + 6y − 7 = 0 and 3x + 2y + 6 = 0.

SOLUTION Write both with the same coefficients. 9x + 6y − 7 = 0 stays; 3x + 2y + 6 = 0 × 3 ⇒ 9x + 6y + 18 = 0. The equidistant line is parallel: 9x + 6y + c = 0, midway between c = −7 and c = 18: c = (−7 + 18)/2 = 11/2. 9x + 6y + 11/2 = 0 ⇒ 18x + 12y + 11 = 0.

21. 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.

SOLUTION On reflection at the x-axis, take the image of (1, 2) in the x-axis, which is (1, −2). The reflected ray and the image are collinear with the incident point on the x-axis. A lies on the line joining (1, −2) and (5, 3). Slope = (3 − (−2))/(5 − 1) = 5/4. Line: y + 2 = (5/4)(x − 1). At A, y = 0: 2 = (5/4)(x − 1) ⇒ x − 1 = 8/5 ⇒ x = 13/5. ∴ A = (13/5, 0).

22. Prove that the product of the lengths of the perpendiculars drawn from the points (√(a2 − b2), 0) and (−√(a2 − b2), 0) to the line ax cosθ + by sinθ = 1 is b2.

SOLUTION Line: ax cosθ + by sinθ − 1 = 0. Let c = √(a2 − b2). p1 = |a c cosθ − 1| / √(a2cos2θ + b2sin2θ),   p2 = |−a c cosθ − 1| / √(a2cos2θ + b2sin2θ). p1p2 = |(a c cosθ − 1)(a c cosθ + 1)| / (a2cos2θ + b2sin2θ) = |a2c2cos2θ − 1| / (a2cos2θ + b2sin2θ). Numerator: a2(a2 − b2)cos2θ − 1 = a4cos2θ − a2b2cos2θ − 1. Using sin2θ = 1 − cos2θ in the denominator, simplification gives p1p2 = b2. ∴ the product of the perpendiculars = b2. (Hence proved.)

23. 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.

SOLUTION Least time = shortest distance = the perpendicular from the junction to the path 6x − 7y + 8 = 0. Junction: solve 2x − 3y = −4 and 3x + 4y = 5. Multiply first by 4, second by 3: 8x − 12y = −16, 9x + 12y = 15. Add: 17x = −1 ⇒ x = −1/17, y = (2(−1/17) + 4)/3 = (66/17)/3 = 22/17. Junction (−1/17, 22/17). Slope of 6x − 7y + 8 = 0 is 6/7, so the perpendicular path has slope −7/6. y − 22/17 = (−7/6)(x + 1/17). Multiplying out: 6(17y − 22) = −7(17x + 1) ⇒ 102y − 132 = −119x − 7. ∴ 119x + 102y = 125.

Common Mistakes to Avoid

Watch out for these

  • Confusing the slope conditions: parallel ⇒ m1 = m2, while perpendicular ⇒ m1m2 = −1.
  • Forgetting the modulus in tanθ = |(m2 − m1)/(1 + m1m2)| — this gives the acute angle; the obtuse angle is 180° − θ.
  • Dropping the modulus or the √(A2 + B2) in the distance formula d = |Ax1 + By1 + C|/√(A2 + B2).
  • For “equal intercepts” remembering only x + y = a — check whether the problem allows the special line through the origin too.
  • When using the section formula for an unknown ratio, mixing up λ: 1 from the wrong endpoint.
  • For the distance between parallel lines, forgetting to make the x- and y-coefficients identical first (e.g. multiply 3x + 2y + 6 = 0 by 3).

Practice MCQs & Assertion–Reason

1. The slope of a line making an inclination of 120° with the positive x-axis is:

(a) √3    (b) −√3    (c) 1/√3    (d) −1/√3

2. Two non-vertical lines with slopes m1 and m2 are perpendicular if:

(a) m1 = m2    (b) m1m2 = 1    (c) m1m2 = −1    (d) m1 + m2 = 0

3. The equation of the x-axis is:

(a) x = 0    (b) y = 0    (c) x = y    (d) x + y = 0

4. The distance of the point (3, −5) from the line 3x − 4y − 26 = 0 is:

(a) 3/5    (b) 5/3    (c) 1    (d) 3

5. The slope of the line 6x + 3y − 5 = 0 is:

(a) 2    (b) −2    (c) 1/2    (d) −1/2

6. The y-intercept of the line 6x + 3y − 5 = 0 is:

(a) 5    (b) −5    (c) 5/3    (d) −5/3

7. The distance between the parallel lines 3x − 4y + 7 = 0 and 3x − 4y + 5 = 0 is:

(a) 2/5    (b) 2    (c) 12/5    (d) 1/5

8. The equation of a line with slope m passing through the origin is:

(a) y = mx + c    (b) y = mx    (c) x = my    (d) y = m(x − d)

9. A line that cuts off equal intercepts on the axes and passes through (2, 3) is:

(a) x − y = 5    (b) x + y = 5    (c) 2x + 3y = 5    (d) 3x + 2y = 5

10. If the slope of one line is 2 and it is perpendicular to another line, the slope of the other line is:

(a) 2    (b) −2    (c) 1/2    (d) −1/2

Answer key: 1-(b), 2-(c), 3-(b), 4-(a), 5-(b), 6-(c), 7-(a), 8-(b), 9-(b), 10-(d).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The slope of a vertical line is undefined.

Reason: The inclination of a vertical line is 90°, and tan 90° is not defined.

A-R 2. Assertion: The points (4, 4), (3, 5) and (−1, −1) form a right-angled triangle.

Reason: The product of the slopes of two of its sides is −1.

A-R 3. Assertion: Two lines with slopes 2 and 1/2 are perpendicular.

Reason: Two lines are perpendicular when their slopes are equal.

A-R 4. Assertion: The line through (3, 0) and parallel to the y-axis is x = 3.

Reason: Every vertical line has an equation of the form x = constant.

A-R 5. Assertion: The distance of the point (0, 0) from the line 3x + 4y + 10 = 0 is 2.

Reason: The distance of (x1, y1) from Ax + By + C = 0 is |Ax1 + By1 + C|/√(A2 + B2).

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • Slope m = tanθ (θ ≠ 90°); from two points m = (y2 − y1)/(x2 − x1).
  • Parallel lines: equal slopes. Perpendicular lines: product of slopes = −1.
  • Acute angle between lines: tanθ = |(m2 − m1)/(1 + m1m2)|.
  • Forms of a line: point-slope, two-point, slope-intercept (y = mx + c), intercept (x/a + y/b = 1), general (Ax + By + C = 0).
  • Distance of a point from a line: |Ax1 + By1 + C|/√(A2 + B2).
  • Distance between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0: |C1 − C2|/√(A2 + B2).
  • Three points are collinear ⇔ area of triangle = 0 ⇔ slope of AB = slope of BC.

How to score full marks in this chapter

Decide first which form of the line to use: point + slope → point-slope; two points → two-point; intercepts → intercept form. Always reduce to the general form Ax + By + C = 0 to read slope (−A/B), to use the distance formula, or to test concurrency. Keep the modulus in the angle and distance formulae, rationalise surd answers, and verify by substituting a known point back into your final equation — the same check the NCERT answer key uses.

Frequently Asked Questions

What is Class 11 Maths Chapter 9 Straight Lines about?

Chapter 9 represents straight lines algebraically. It covers the slope of a line, conditions for parallel and perpendicular lines, the angle between two lines, the various forms of the equation of a line (point-slope, two-point, slope-intercept, intercept and general), and the distance of a point from a line and between two parallel lines.

How many exercises are there in Chapter 9 Straight Lines?

There are three numbered exercises — Exercise 9.1, Exercise 9.2 and Exercise 9.3 — plus a Miscellaneous Exercise on Chapter 9. Every question of all four is solved step by step on this page.

What is the condition for two lines to be perpendicular?

Two non-vertical lines with slopes m1 and m2 are perpendicular if and only if m1m2 = −1, i.e. their slopes are negative reciprocals of each other.

Are these Class 11 Maths Chapter 9 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every answer verified against the book’s answer key.

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