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NCERT Solutions · Class 11 Maths · Chapter 9

Chapter 9: Straight Lines (Coordinate Geometry)

Q: The slope of one line is double that of another; the tangent of the angle between them is 1/3. Find the slopes.

(1, 2), (½, 1), (−1, −2), (−½, −1)

Step-by-step solutions to all exercises of Chapter 9, Straight Lines (NCERT Class 11 Maths, 2026-27): slope, angle between lines, collinearity, equations of a line in different forms, normal form, distance of a point from a line and the Miscellaneous Exercise. All 70 questions are answered, with the key answer highlighted.

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Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-11-maths/chapter-9-straight-lines

Slope $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = tan θ$ . Point-slope $y - y_{1} = m (x - x_{1})$ ; slope-intercept $y = m x + c$ ; intercept $\frac{x}{a} + \frac{y}{b} = 1$ ; normal $x cos ω + y sin ω = p$ . Distance from $(x_{1}, y_{1})$ to $A x + B y + C = 0$ : $\frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$ .

Exercise 9.1

Draw the quadrilateral with vertices (−4, 5), (0, 7), (5, −5), (−4, −2) and find its area.

Solution

Shoelace with A(−4, 5), B(0, 7), C(5, −5), D(−4, −2): Σ(x_(i)y_(i+1) − x_(i+1)y_(i)) = (−28 − 0) + (0 − 35) + (−10 − 20) + (−20 − 8) = −121. Area = 121/2 square units.

121/2 square units

The base of an equilateral triangle of side 2a lies on the y-axis with its midpoint at the origin. Find the vertices.

Solution

Base ends (0, a), (0, −a); the third vertex is on the x-axis at height √3a: (√3a, 0) or (−√3a, 0).

(0, a), (0, −a); (√3a, 0) or (−√3a, 0)

Find PQ when (i) PQ ∥ y-axis (ii) PQ ∥ x-axis.

Solution

(i) x₁ = x₂: PQ = |y₂ − y₁|. (ii) y₁ = y₂: |x₂ − x₁|.

|y₂ − y₁|; |x₂ − x₁|

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

Solution

(x − 7)² + 36 = (x − 3)² + 16 ⇒ −14x + 85 = −6x + 25 ⇒ x = 15/2: (15/2, 0).

(15/2, 0)

Find the slope of the line through the origin and the midpoint of P(0, −4) and B(8, 0).

Solution

Midpoint (4, −2): slope −2/4 = −1/2.

−1/2

Without Pythagoras, show (4, 4), (3, 5), (−1, −1) form a right triangle.

Solution

A(4, 4), B(3, 5), C(−1, −1): slope AB = 1/(−1) = −1, slope AC = (−5)/(−5) = 1. Product −1 ⇒ AB ⊥ AC: right angle at A.

right angle at A

Find the slope of the line making 30° with the positive y-axis measured anticlockwise.

Solution

Its inclination is 90° + 30° = 120°: slope tan 120° = −√3.

−√3

Without the distance formula, show (−2, −1), (4, 0), (3, 3), (−3, 2) form a parallelogram.

Solution

Slopes: AB = 1/6, CD = (2 − 3)/(−3 − 3) = 1/6 ⇒ AB ∥ CD; BC = 3/(−1) = −3, DA = (−1 − 2)/(−2 + 3) = −3 ⇒ BC ∥ DA. Both pairs of opposite sides parallel: a parallelogram.

a parallelogram

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

Solution

Slope (−2 + 1)/(4 − 3) = −1 = tan θ ⇒ θ = 135°.

θ = 135°

The slope of one line is double that of another; the tangent of the angle between them is 1/3. Find the slopes.

Solution

m and 2m: |(2m − m)/(1 + 2m²)| = 1/3 ⇒ 3|m| = 1 + 2m². m > 0: 2m² − 3m + 1 = 0 ⇒ m = 1 or ½; m < 0: 2m² + 3m + 1 = 0 ⇒ m = −1 or −½. Slopes (1, 2), (½, 1), (−1, −2), (−½, −1).

(1, 2), (½, 1), (−1, −2), (−½, −1)

A line passes through (x₁, y₁) and (h, k) and has slope m. Show k − y₁ = m(h − x₁).

Solution

m = (k − y₁)/(h − x₁) ⇒ k − y₁ = m(h − x₁).

k − y₁ = m(h − x₁)

Exercise 9.2

Find the equation of the line: the axes

Solution

x-axis y = 0; y-axis x = 0.

y = 0; x = 0

Find the equation of the line: through (−4, 3), slope ½

Solution

y − 3 = ½(x + 4) ⇒ x − 2y + 10 = 0.

x − 2y + 10 = 0

Find the equation of the line: through (0, 0), slope m

Solution

y = mx.

y = mx

Find the equation of the line: through (2, 2√3) at 75° to the x-axis

Solution

slope tan 75° = 2 + √3: y − 2√3 = (2 + √3)(x − 2) ⇒ (2 + √3)x − y − 4 = 0 (= (√3 + 1)x − (√3 − 1)y = 4(√3 − 1)).

(2 + √3)x − y − 4 = 0

Find the equation of the line: meeting the x-axis 3 units left of O, slope −2

Solution

y = −2(x + 3) ⇒ 2x + y + 6 = 0.

2x + y + 6 = 0

Find the equation of the line: meeting the y-axis 2 units above O at 30° to the x-axis

Solution

y = x/√3 + 2 ⇒ x − √3y + 2√3 = 0.

x − √3y + 2√3 = 0

Find the equation of the line: through (−1, 1), (2, −4)

Solution

slope −5/3: 5x + 3y + 2 = 0.

5x + 3y + 2 = 0

Find the equation of the line: the median through R of ΔPQR with P(2, 1), Q(−2, 3), R(4, 5).

Solution

midpoint of PQ (0, 2); slope (5 − 2)/4 = 3/4: y − 5 = ¾(x − 4) ⇒ 3x − 4y + 8 = 0.

3x − 4y + 8 = 0

Line through (−3, 5) perpendicular to the line through (2, 5) and (−3, 6).

Solution

Slope of the given line −1/5 ⇒ required slope 5: y − 5 = 5(x + 3) ⇒ 5x − y + 20 = 0.

5x − y + 20 = 0

A line perpendicular to the segment joining (1, 0), (2, 3) divides it in 1 : n. Find it.

Solution

Point ((n + 2)/(n + 1), 3/(n + 1)); segment slope 3, so the line's slope −1/3: 3(n + 1)y − 9 = −(n + 1)x + (n + 2) ⇒ (1 + n)x + 3(1 + n)y = n + 11.

(1 + n)x + 3(1 + n)y = n + 11

Line with equal intercepts through (2, 3).

Solution

x/a + y/a = 1 ⇒ 2 + 3 = a ⇒ x + y = 5.

x + y = 5

Line through (2, 2) whose intercepts add to 9.

Solution

x/a + y/(9 − a) = 1 with (2, 2): 2(9 − a) + 2a = a(9 − a) ⇒ a² − 9a + 18 = 0 ⇒ a = 3 or 6: 2x + y = 6 or x + 2y = 6.

2x + y = 6; x + 2y = 6

Line through (0, 2) making 2π/3 with the x-axis; and the parallel line crossing the y-axis 2 units below O.

Solution

Slope −√3: √3x + y − 2 = 0; parallel through (0, −2): √3x + y + 2 = 0.

√3x + y − 2 = 0; √3x + y + 2 = 0

The perpendicular from the origin meets a line at (−2, 9). Find the line.

Solution

Slope of OP = −9/2 ⇒ line slope 2/9: y − 9 = (2/9)(x + 2) ⇒ 2x − 9y + 85 = 0.

2x − 9y + 85 = 0

Rod length L = 124.942 cm at C = 20 and 125.134 at C = 110 (linear). Express L in terms of C.

Solution

Slope (125.134 − 124.942)/90 = 0.192/90: L = (0.192/90)(C − 20) + 124.942.

L = (0.192/90)(C − 20) + 124.942

980 L/week sell at ₹14 and 1220 at ₹16 (linear). How many at ₹17?

Solution

Slope 240/2 = 120 L per rupee: 980 + 120 × 3 = 1340 litres.

1340 litres

P(a, b) is the midpoint of the segment between the axes. Show the line is x/a + y/b = 2.

Solution

Intercepts 2a and 2b: x/(2a) + y/(2b) = 1 ⇒ x/a + y/b = 2.

x/a + y/b = 2

R(h, k) divides the segment between the axes in 1 : 2. Find the line.

Solution

A(a, 0), B(0, b): R = (2a/3, b/3) ⇒ a = 3h/2, b = 3k: x/(3h/2) + y/(3k) = 1 ⇒ 2kx + hy = 3hk.

2kx + hy = 3hk

Prove (3, 0), (−2, −2), (8, 2) are collinear using equations of lines.

Solution

Line through (3, 0), (−2, −2): slope 2/5: 2x − 5y − 6 = 0. At (8, 2): 16 − 10 − 6 = 0: it lies on it: collinear.

collinear

Exercise 9.3

Slope–intercept form, slopes and y-intercepts: (i) x + 7y = 0 (ii) 6x + 3y − 5 = 0 (iii) y = 0.

Solution

(i) y = −x/7: m = −1/7, c = 0. (ii) y = −2x + 5/3: m = −2, c = 5/3. (iii) y = 0x + 0: m = 0, c = 0.

(i) m = −1/7, c = 0; (ii) m = −2, c = 5/3; (iii) m = 0, c = 0

Intercept form and intercepts: (i) 3x + 2y − 12 = 0 (ii) 4x − 3y = 6 (iii) 3y + 2 = 0.

Solution

(i) x/4 + y/6 = 1: 4, 6. (ii) x/(3/2) + y/(−2) = 1: 3/2, −2. (iii) y = −2/3: y-intercept −2/3, no x-intercept (parallel to the x-axis).

(i) 4, 6; (ii) 3/2, −2; (iii) y-intercept −2/3, no x-intercept

Distance of (−1, 1) from 12(x + 6) = 5(y − 2).

Solution

12x − 5y + 82 = 0: |−12 − 5 + 82|/13 = 5 units.

5 units

Points on the x-axis at distance 4 from x/3 + y/4 = 1.

Solution

4x + 3y − 12 = 0: |4x − 12|/5 = 4 ⇒ x = 8 or −2: (8, 0), (−2, 0).

(8, 0), (−2, 0)

Distance between (i) 15x + 8y − 34 = 0, 15x + 8y + 31 = 0 (ii) l(x + y) + p = 0, l(x + y) − r = 0.

Solution

(i) |−34 − 31|/17 = 65/17. (ii) |p + r|/√(l² + l²) = |p + r|/(l√2).

65/17; |p + r|/(l√2)

Line ∥ 3x − 4y + 2 = 0 through (−2, 3).

Solution

3x − 4y + k = 0: −6 − 12 + k = 0 ⇒ 3x − 4y + 18 = 0.

3x − 4y + 18 = 0

Line ⊥ x − 7y + 5 = 0 with x-intercept 3.

Solution

7x + y + k = 0 through (3, 0): k = −21: 7x + y = 21.

7x + y = 21

Angles between √3x + y = 1 and x + √3y = 1.

Solution

m₁ = −√3, m₂ = −1/√3: tan θ = |(−1/√3 + √3)/(1 + 1)| = 1/√3 ⇒ 30° and 150°.

30° and 150°

The line through (h, 3), (4, 1) meets 7x − 9y − 19 = 0 at right angles. Find h.

Solution

Slopes −2/(4 − h) and 7/9: product −1 ⇒ 14 = 9(4 − h) ⇒ h = 22/9.

h = 22/9

Prove the line through (x₁, y₁) parallel to Ax + By + C = 0 is A(x − x₁) + B(y − y₁) = 0.

Solution

Slope −A/B: y − y₁ = (−A/B)(x − x₁) ⇒ A(x − x₁) + B(y − y₁) = 0 (also true for B = 0: x = x₁).

A(x − x₁) + B(y − y₁) = 0

Two lines through (2, 3) meet at 60°; one has slope 2. Find the other.

Solution

√3 = |(m − 2)/(1 + 2m)| ⇒ m = (2 + √3)/(1 − 2√3) or (2 − √3)/(1 + 2√3). Lines: (2 + √3)x + (2√3 − 1)y = 8√3 + 1 or (√3 − 2)x + (1 + 2√3)y = 8√3 − 1.

(2 + √3)x + (2√3 − 1)y = 8√3 + 1; (√3 − 2)x + (1 + 2√3)y = 8√3 − 1

Right bisector of the segment joining (3, 4) and (−1, 2).

Solution

Midpoint (1, 3); segment slope ½ ⇒ bisector slope −2: 2x + y = 5.

2x + y = 5

Foot of the perpendicular from (−1, 3) to 3x − 4y − 16 = 0.

Solution

t = −(−3 − 12 − 16)/25 = 31/25: (−1 + 93/25, 3 − 124/25) = (68/25, −49/25).

(68/25, −49/25)

The perpendicular from O to y = mx + c meets it at (−1, 2). Find m and c.

Solution

OP slope −2 ⇒ m = ½; 2 = −½ + c ⇒ m = ½, c = 5/2.

m = ½, c = 5/2

p, q are the perpendiculars from O to x cos θ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k. Prove p² + 4q² = k².

Solution

k²

ΔABC with A(2, 3), B(4, −1), C(1, 2): find the altitude from A (equation and length).

Solution

BC: slope −1, x + y − 3 = 0. Altitude slope 1 through A: x − y + 1 = 0; length |2 + 3 − 3|/√2 = √2.

x − y + 1 = 0; √2

If p is the perpendicular from O to the line with intercepts a and b, show 1/p² = 1/a² + 1/b².

Solution

x/a + y/b − 1 = 0: p = 1/√(1/a² + 1/b²) ⇒ 1/p² = 1/a² + 1/b².

1/p² = 1/a² + 1/b²

Miscellaneous Exercise on Chapter 9

For (k − 3)x − (4 − k²)y + k² − 7k + 6 = 0, find k if it is (a) ∥ x-axis (b) ∥ y-axis (c) through the origin.

Solution

(a) x-coefficient 0: k = 3. (b) y-coefficient 0: 4 − k² = 0 ⇒ k = ±2. (c) constant 0: k² − 7k + 6 = 0 ⇒ k = 1 or 6.

(a) k = 3; (b) k = ±2; (c) k = 1 or 6

Lines whose intercepts have sum 1 and product −6.

Solution

a + b = 1, ab = −6 ⇒ (a, b) = (3, −2) or (−2, 3): 2x − 3y = 6 and −3x + 2y = 6.

2x − 3y = 6; −3x + 2y = 6

Points on the y-axis at distance 4 from x/3 + y/4 = 1.

Solution

|3y − 12|/5 = 4 ⇒ y = 32/3 or −8/3: (0, 32/3), (0, −8/3).

(0, 32/3), (0, −8/3)

Distance of O from the line through (cos θ, sin θ) and (cos φ, sin φ).

Solution

Both points are on the unit circle: the line is a chord subtending (φ − θ) at O; half the chord = |sin((φ − θ)/2)|, so the distance = √(1 − sin²((φ − θ)/2)) = |cos((φ − θ)/2)|.

|cos((φ − θ)/2)|

Line ∥ y-axis through the intersection of x − 7y + 5 = 0 and 3x + y = 0.

Solution

y = −3x ⇒ x + 21x + 5 = 0 ⇒ x = −5/22: x = −5/22.

x = −5/22

Line ⊥ x/4 + y/6 = 1 through the point where it meets the y-axis.

Solution

Meets the y-axis at (0, 6); slope of the given line −3/2 ⇒ 2/3: y − 6 = (2/3)x ⇒ 2x − 3y + 18 = 0.

2x − 3y + 18 = 0

Area of the triangle formed by y − x = 0, x + y = 0, x − k = 0.

Solution

Vertices (0, 0), (k, k), (k, −k): area = ½ × 2|k| × |k| = k² square units.

k² square units

Find p so that 3x + y − 2 = 0, px + 2y − 3 = 0, 2x − y − 3 = 0 are concurrent.

Solution

The first and third meet at (1, −1) (adding: 5x = 5); then p − 2 − 3 = 0 ⇒ p = 5.

p = 5

If y = m₁x + c₁, y = m₂x + c₂, y = m₃x + c₃ are concurrent, show m₁(c₂ − c₃) + m₂(c₃ − c₁) + m₃(c₁ − c₂) = 0.

Solution

The first two meet at x = (c₂ − c₁)/(m₁ − m₂), y = (m₁c₂ − m₂c₁)/(m₁ − m₂). Putting this in the third: m₁c₂ − m₂c₁ = m₃(c₂ − c₁) + c₃(m₁ − m₂), which rearranges to the given condition.

See the working above.

Lines through (3, 2) making 45° with x − 2y = 3.

Solution

|(m − ½)/(1 + m/2)| = 1 ⇒ m = 3 or −1/3: 3x − y = 7 and x + 3y = 9.

3x − y = 7; x + 3y = 9

Line through the intersection of 4x + 7y − 3 = 0 and 2x − 3y + 1 = 0 with equal intercepts.

Solution

Intersection (1/13, 5/13); x + y = a with a = 6/13: 13x + 13y = 6.

13x + 13y = 6

Show that the line through the origin making θ with y = mx + c is y/x = (m ± tan θ)/(1 ∓ m tan θ).

Solution

If its slope is m′ = y/x: tan θ = |(m′ − m)/(1 + mm′)| ⇒ m′ − m = ±tan θ(1 + mm′) ⇒ m′ = (m ± tan θ)/(1 ∓ m tan θ).

m′ = (m ± tan θ)/(1 ∓ m tan θ)

In what ratio does x + y = 4 divide the line joining (−1, 1) and (5, 7)?

Solution

Values of x + y − 4: at (−1, 1): −4; at (5, 7): 8. Ratio = 4 : 8 = 1 : 2 (internally).

1 : 2

Distance of 4x + 7y + 5 = 0 from (1, 2) along 2x − y = 0.

Solution

The line 2x − y = 0 passes through (1, 2); it meets 4x + 7y + 5 = 0 where 4x + 14x + 5 = 0 ⇒ (−5/18, −5/9). Distance √((23/18)² + (23/9)²) = 23√5/18.

23√5/18

In which direction must a line be drawn through (−1, 2) so that it meets x + y = 4 at distance 3?

Solution

Point (−1 + 3 cos θ, 2 + 3 sin θ) on x + y = 4 ⇒ cos θ + sin θ = 1 ⇒ θ = 0 or 90°: the line is parallel to the x-axis or to the y-axis.

parallel to the x-axis or to the y-axis

The hypotenuse of a right triangle has ends (1, 3), (−4, 1). Find the legs parallel to the axes.

Solution

The right-angle vertex is (1, 1) or (−4, 3): legs x = 1, y = 1 or x = −4, y = 3.

x = 1, y = 1; x = −4, y = 3

Image of (3, 8) in x + 3y = 7.

Solution

t = −2(3 + 24 − 7)/10 = −4: (3 − 4, 8 − 12) = (−1, −4).

(−1, −4)

y = 3x + 1 and 2y = x + 3 are equally inclined to y = mx + 4. Find m.

Solution

(3 − m)/(1 + 3m) = ±(m − ½)/(1 + m/2). The + case: 7m² − 2m − 7 = 0 ⇒ m = (1 ± 5√2)/7; the − case has no real root.

m = (1 ± 5√2)/7

The sum of the distances of P(x, y) from x + y − 5 = 0 and 3x − 2y + 7 = 0 is always 10. Show P moves on a line.

Solution

|x + y − 5|/√2 + |3x − 2y + 7|/√13 = 10. In any region where the expressions keep their signs, removing the moduli gives (x + y − 5)/√2 ± (3x − 2y + 7)/√13 = 10, a linear equation in x and y: a straight line.

a straight line

Line equidistant from 9x + 6y − 7 = 0 and 3x + 2y + 6 = 0.

Solution

Write the second as 9x + 6y + 18 = 0: the midway line 9x + 6y + (−7 + 18)/2 = 0 ⇒ 18x + 12y + 11 = 0.

18x + 12y + 11 = 0

A ray through (1, 2) reflects at A on the x-axis and passes through (5, 3). Find A.

Solution

Join (1, 2) to the image (5, −3) of (5, 3) in the x-axis; the reflection point A lies on this join: y − 2 = −(5/4)(x − 1); y = 0 ⇒ x = 13/5: A(13/5, 0).

A(13/5, 0)

Prove the product of the perpendiculars from (±√(a² − b²), 0) to (x/a) cos θ + (y/b) sin θ = 1 is b².

Solution

With c = √(a² − b²): p₁p₂ = |(c cos θ/a − 1)(−c cos θ/a − 1)|/(cos²θ/a² + sin²θ/b²) = [(a² − c²cos²θ)/a²]/[(b²cos²θ + a²sin²θ)/(a²b²)]. Since a² − c²cos²θ = a²sin²θ + b²cos²θ, this is b².

b²

A person at the crossing of 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 wants to reach 6x − 7y + 8 = 0 in the least time. Find the path.

Solution

The crossing is (−1/17, 22/17). Shortest path = perpendicular: slope −7/6: 7x + 6y = 7(−1/17) + 6(22/17) = 125/17 ⇒ 119x + 102y = 125.

119x + 102y = 125

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