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Class 11 Maths NCERT Solutions

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Straight Lines Class 11 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 9, Straight Lines — 47 questions from Ex 9.1 to Ex 9.3, each worked through step by step in the CBSE marking pattern. Slope, angle of inclination, forms of the equation of a line, distance formulae, and the angle between two lines.

Class:11Subject:MathsChapter:9
4 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 11 Maths Chapter 9?

Chapter 9 carries 3 exercise questions, numbered Ex 9.1 to Ex 9.3. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

Coordinate geometry turns lines into equations. The slope m of a line, the angle between two lines via their slopes, and the relationship between perpendicular and parallel slopes unlock the whole chapter. From there, lines are written in point-slope, two-point, slope-intercept, and intercept forms, and closed with the perpendicular distance of a point from a line.

Board pattern

Two lines with slopes m₁ and m₂ are parallel when m₁ = m₂ and perpendicular when m₁m₂ = −1. Distance from (x₁, y₁) to Ax + By + C = 0 is |Ax₁ + By₁ + C|/√(A² + B²). For intercept sums/products, form the quadratic in a and b directly. In proof questions, substitute the point into the equation of the line rather than arguing verbally.
02

Exercise 9.1 — Slope of a Line

11Exercise questions

Step-by-step solution

  1. 1Label A(−4,5), B(0,7), C(5,−5), D(−4,−2).
  2. 2Shoelace sum₁ = xᵢyᵢ₊₁ = (−4)(7) + (0)(−5) + (5)(−2) + (−4)(5) = −58.
  3. 3sum₂ = yᵢxᵢ₊₁ = 5(0) + 7(5) + (−5)(−4) + (−2)(−4) = 63.
  4. 4Area = ½|sum₁ − sum₂| = ½|−58 − 63| = 121/2 = 60.5 sq units.

Final answer

Area = 121/2 = 60.5 sq units.

Step-by-step solution

  1. 1Base on y-axis with midpoint (0,0) → endpoints (0, a) and (0, −a).
  2. 2Third vertex on x-axis; height of the triangle = √(4a² − a²) = √3a.
  3. 3Vertices: (0, a), (0, −a), (√3a, 0) — also (−√3a, 0) on the other side.

Final answer

(0, a), (0, −a), (±√3a, 0).

Step-by-step solution

  1. 1(i) PQ ∥ y-axis ⇒ x₁ = x₂ → distance = |y₂ − y₁|.
  2. 2(ii) PQ ∥ x-axis ⇒ y₁ = y₂ → distance = |x₂ − x₁|.

Final answer

(i) |y₂ − y₁|, (ii) |x₂ − x₁|.

Step-by-step solution

  1. 1Let the point be (x, 0).
  2. 2(x − 7)² + 36 = (x − 3)² + 16 → −14x + 85 = −6x + 25.
  3. 38x = 60 → x = 15/2. Point: (15/2, 0).

Final answer

(15/2, 0).

Step-by-step solution

  1. 1Mid-point of PB = (4, −2).
  2. 2Slope through (0,0) and (4,−2) = (−2 − 0)/(4 − 0) = −1/2.

Final answer

−1/2.

Step-by-step solution

  1. 1Slope of (4,4)→(3,5) = (5−4)/(3−4) = −1.
  2. 2Slope of (4,4)→(−1,−1) = (−1−4)/(−1−4) = 1.
  3. 3Product = −1 → the two sides are perpendicular → right angled at (4, 4).

Final answer

Right angled at (4, 4) — shown.

Step-by-step solution

  1. 130° from the y-axis ⇒ 90° − 30° = 60° from the positive x-axis.
  2. 2Slope = tan 60° = √3.

Final answer

√3.

Step-by-step solution

  1. 1Slope of (−2,−1)→(4,0) = 1/6; slope of (3,3)→(−3,2) = (2−3)/(−3−3) = 1/6.
  2. 2Slope of (4,0)→(3,3) = −3; slope of (−2,−1)→(−3,2) = 3/(−1) = −3.
  3. 3Opposite sides are parallel → parallelogram.

Final answer

Opposite sides parallel — shown.

Step-by-step solution

  1. 1Slope = (−2 + 1)/(4 − 3) = −1.
  2. 2tan θ = −1 ⇒ θ = 135° (measured anticlockwise from the x-axis).

Final answer

135°.

Step-by-step solution

  1. 1Let the slopes be m and 2m: tan θ = |(2m − m)/(1 + 2m²)| = |m/(1 + 2m²)| = 1/3.
  2. 2m/(1 + 2m²) = 1/3 → 2m² − 3m + 1 = 0 → (2m − 1)(m − 1) = 0.
  3. 3m = 1/2 or m = 1 → slopes are (1/2, 1) or (1, 2).

Final answer

(1/2, 1) or (1, 2).

Step-by-step solution

  1. 1By definition, m = (k − y₁)/(h − x₁).
  2. 2Cross-multiplying: k − y₁ = m(h − x₁).

Final answer

k − y₁ = m(h − x₁) — shown.

03

Exercise 9.2 — Various Forms of the Equation of a Line

19Exercise questions

Step-by-step solution

  1. 1x-axis: all points have y = 0 → y = 0.
  2. 2y-axis: all points have x = 0 → x = 0.

Final answer

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

Step-by-step solution

  1. 1Point-slope form: y − 3 = (1/2)(x + 4).
  2. 22y − 6 = x + 4 → x − 2y + 10 = 0.

Final answer

x − 2y + 10 = 0.

Step-by-step solution

  1. 1Point-slope form: y − 0 = m(x − 0).
  2. 2y = mx.

Final answer

y = mx.

Step-by-step solution

  1. 1m = tan 75° = 2 + √3.
  2. 2y − 2√3 = (2 + √3)(x − 2).
  3. 3y = (2 + √3)x − 4 → (2 + √3)x − y − 4 = 0.

Final answer

(2 + √3)x − y − 4 = 0.

Step-by-step solution

  1. 1Passes through (−3, 0) with slope −2.
  2. 2y − 0 = −2(x + 3) → y = −2x − 6 → 2x + y + 6 = 0.

Final answer

2x + y + 6 = 0.

Step-by-step solution

  1. 1Passes through (0, 2) with slope tan 30° = 1/√3.
  2. 2y = (1/√3)x + 2 → x − √3y + 2√3 = 0.

Final answer

x − √3 y + 2√3 = 0.

Step-by-step solution

  1. 1Slope = (−4 − 1)/(2 + 1) = −5/3.
  2. 2y − 1 = (−5/3)(x + 1) → 3y − 3 = −5x − 5.
  3. 35x + 3y + 2 = 0.

Final answer

5x + 3y + 2 = 0.

Step-by-step solution

  1. 1Mid-point of PQ = (0, 2).
  2. 2Median is the line through R(4, 5) and (0, 2): slope = 3/4.
  3. 3y − 2 = (3/4)(x − 0) → 3x − 4y + 8 = 0.

Final answer

3x − 4y + 8 = 0.

Step-by-step solution

  1. 1Slope of the given line = (6 − 5)/(−3 − 2) = −1/5.
  2. 2Perpendicular slope = 5.
  3. 3y − 5 = 5(x + 3) → 5x − y + 20 = 0.

Final answer

5x − y + 20 = 0.

Step-by-step solution

  1. 1Slope of the segment = 3 → the line has slope −1/3.
  2. 2Division point: ((n·1 + 1·2)/(n+1), (n·0 + 1·3)/(n+1)) = ((n+2)/(n+1), 3/(n+1)).
  3. 3y − 3/(n+1) = (−1/3)(x − (n+2)/(n+1)).
  4. 4(n + 1)(x + 3y) = n + 11.

Final answer

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

Step-by-step solution

  1. 1Equal intercepts: x/a + y/a = 1 → x + y = a.
  2. 2Through (2, 3): a = 5.
  3. 3x + y − 5 = 0.

Final answer

x + y − 5 = 0.

Step-by-step solution

  1. 1Let intercepts be a, b: a + b = 9 and 2/a + 2/b = 1.
  2. 22(a + b)/ab = 1 → ab = 18.
  3. 3a, b are roots of t² − 9t + 18 = 0 → a, b = 3 or 6.
  4. 4Lines: x/3 + y/6 = 1 or x/6 + y/3 = 1 → 2x + y = 6 or x + 2y = 6.

Final answer

2x + y − 6 = 0 or x + 2y − 6 = 0.

Step-by-step solution

  1. 1Slope = tan(2π/3) = −√3; through (0, 2): y = −√3x + 2.
  2. 2√3x + y − 2 = 0.
  3. 3Parallel line through (0, −2): √3x + y + 2 = 0.

Final answer

√3x + y − 2 = 0 and √3x + y + 2 = 0.

Step-by-step solution

  1. 1Slope of the joining radius = 9/(−2) = −9/2.
  2. 2Line is perpendicular → slope = 2/9 (and passes through (−2, 9)).
  3. 3y − 9 = (2/9)(x + 2) → 2x − 9y + 85 = 0.

Final answer

2x − 9y + 85 = 0.

Step-by-step solution

  1. 1Slope = (125.134 − 124.942)/(110 − 20) = 0.192/90 = 16/7500.
  2. 2L − 124.942 = (16/7500)(C − 20).
  3. 3L = (16/7500)C + 124.8993 (approx).

Final answer

L = (16/7500)C + 124.8993.

Step-by-step solution

  1. 1Points (14, 980) and (16, 1220): slope = 240/2 = 120.
  2. 2q = 120(p − 14) + 980 = 120p − 700.
  3. 3At p = 17: q = 2040 − 700 = 1340 litres.

Final answer

1340 litres.

Step-by-step solution

  1. 1Let the segment cut the axes at (α, 0) and (0, β).
  2. 2Mid-point: (α/2, β/2) = (a, b) → α = 2a, β = 2b.
  3. 3Line: x/α + y/β = 1 → x/(2a) + y/(2b) = 1 → x/a + y/b = 2.

Final answer

x/a + y/b = 2 — shown.

Step-by-step solution

  1. 1Let the segment have intercepts (X, 0) and (0, Y).
  2. 2R = (1·0 + 2·X, 1·Y + 2·0)/3 = (2X/3, Y/3) = (h, k) → X = 3h/2, Y = 3k.
  3. 3Line: x/(3h/2) + y/(3k) = 1 → 2x/(3h) + y/(3k) = 1.
  4. 42kx + hy = 3hk.

Final answer

2kx + hy = 3hk (i.e. 2x/(3h) + y/(3k) = 1).

Step-by-step solution

  1. 1Line through (3, 0) and (−2, −2): slope = 2/5 → y = (2/5)(x − 3).
  2. 22x − 5y − 6 = 0.
  3. 3Put (8, 2): 16 − 10 − 6 = 0 → the point lies on the line → collinear.

Final answer

Collinear — proved.

04

Exercise 9.3 — Distance of a Point from a Line

17Exercise questions

Step-by-step solution

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

Final answer

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

Step-by-step solution

  1. 1(i) 3x + 2y = 12 → x/4 + y/6 = 1: intercepts 4 and 6.
  2. 2(ii) 4x − 3y = 6 → x/(3/2) + y/(−2) = 1: intercepts 3/2 and −2.
  3. 3(iii) y = −2/3: parallel to x-axis, no x-intercept; y-intercept −2/3.

Final answer

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

Step-by-step solution

  1. 1Line: 12x − 5y + 82 = 0.
  2. 2d = |12(−1) − 5(1) + 82|/√(144 + 25) = |65|/13 = 5.

Final answer

5 units.

Step-by-step solution

  1. 1Line: 4x + 3y − 12 = 0; a point (x, 0): d = |4x − 12|/5 = 4.
  2. 2|4x − 12| = 20 → 4x = 32 or 4x = −8.
  3. 3Points: (8, 0) and (−2, 0).

Final answer

(8, 0) and (−2, 0).

Step-by-step solution

  1. 1(i) d = |−34 − 31|/√(225 + 64) = 65/17.
  2. 2(ii) Lines: lx + ly + p = 0 and lx + ly − r = 0 → d = |p + r|/(l√2).

Final answer

(i) 65/17; (ii) |p + r|/(l√2).

Step-by-step solution

  1. 1Parallel lines share the constant-structure: 3x − 4y + k = 0.
  2. 2At (−2, 3): −6 − 12 + k = 0 → k = 18.
  3. 33x − 4y + 18 = 0.

Final answer

3x − 4y + 18 = 0.

Step-by-step solution

  1. 1Given line slope = 1/7 → perpendicular slope = −7.
  2. 2Passes through (3, 0): y = −7(x − 3).
  3. 37x + y − 21 = 0.

Final answer

7x + y − 21 = 0.

Step-by-step solution

  1. 1Slopes: m₁ = −√3, m₂ = −1/√3.
  2. 2tan θ = |(m₁ − m₂)/(1 + m₁m₂)| = |(−√3 + 1/√3)/2| = (2/√3)/2 = 1/√3.
  3. 3θ = 30°.

Final answer

30°.

Step-by-step solution

  1. 1Slope of the joining line = (1 − 3)/(4 − h) = −2/(4 − h).
  2. 2Slope of the given line = 7/9; perpendicular ⇒ product = −1.
  3. 3(−2/(4 − h))(7/9) = −1 → 14 = 9(4 − h) → 9h = 22 → h = 22/9.

Final answer

h = 22/9.

Step-by-step solution

  1. 1A line parallel to Ax + By + C = 0 is Ax + By + K = 0.
  2. 2Through (x₁, y₁): Ax₁ + By₁ + K = 0 → K = −(Ax₁ + By₁).
  3. 3Ax + By − (Ax₁ + By₁) = 0 → A(x − x₁) + B(y − y₁) = 0.

Final answer

A(x − x₁) + B(y − y₁) = 0 — proved.

Step-by-step solution

  1. 1tan 60° = |(m − 2)/(1 + 2m)| → (m − 2)/(1 + 2m) = ±√3.
  2. 2Case +: m − 2 = √3 + 2√3m → m = (2 − √3)/(2√3 + 1) = (5√3 − 8)/11.
  3. 3Case −: m − 2 = −√3 − 2√3m → m = −(8 + 5√3)/11.
  4. 4Lines: (8 − 5√3)x + 11y = 49 − 10√3 and (8 + 5√3)x + 11y = 49 + 10√3.

Final answer

(8 ± 5√3)x + 11y = 49 ± 10√3.

Step-by-step solution

  1. 1Mid-point = (1, 3); slope of segment = (2 − 4)/(−1 − 3) = 1/2.
  2. 2Perpendicular slope = −2.
  3. 3y − 3 = −2(x − 1) → 2x + y − 5 = 0.

Final answer

2x + y − 5 = 0.

Step-by-step solution

  1. 1Perpendicular slope = −4/3; line through (−1, 3): 3y − 9 = −4x − 4 → 4x + 3y − 5 = 0.
  2. 2Solve 4x + 3y = 5 with 3x − 4y = 16.
  3. 3x = 68/25, y = −49/25.

Final answer

(68/25, −49/25).

Step-by-step solution

  1. 1Radius to (−1, 2) has slope −2; the line is perpendicular → m·(−2) = −1 → m = 1/2.
  2. 2(−1, 2) lies on the line: 2 = (1/2)(−1) + c → c = 5/2.

Final answer

m = 1/2, c = 5/2.

Step-by-step solution

  1. 1p = |k cos 2θ|/√(cos²θ + sin²θ) = k|cos 2θ|.
  2. 2q = |k|/√(sec²θ + cosec²θ) = k sin θ cos θ (taking positive lengths).
  3. 3p² + 4q² = k²cos²2θ + 4k²sin²θcos²θ = k²(cos²2θ + sin²2θ) = k².

Final answer

p² + 4q² = k² — proved.

Step-by-step solution

  1. 1Slope of BC = (2 + 1)/(1 − 4) = −1 → altitude slope = 1.
  2. 2Altitude through (2, 3): y − 3 = x − 2 → x − y + 1 = 0.
  3. 3Line BC: x + y − 3 = 0; length = |2 + 3 − 3|/√2 = √2.

Final answer

x − y + 1 = 0, length √2.

Step-by-step solution

  1. 1Line: x/a + y/b = 1 → bx + ay − ab = 0.
  2. 2p = |−ab|/√(a² + b²).
  3. 31/p² = (a² + b²)/(a²b²) = 1/a² + 1/b².

Final answer

1/p² = 1/a² + 1/b² — proved.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Slope-intercept form

Point-slope form

Intercept form

Angle between two lines

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • Slope is finite unless the line is vertical, and a vertical line has no slope-intercept form — check the coefficient of y before dividing.
  • For a line through two points use the determinant form; it never needs you to compute a slope, so it is safer for vertical pairs.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Maths Chapter 9 (Straight Lines)?

There are 3 exercise questions in this chapter, numbered Ex 9.1 to Ex 9.3. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Straight Lines Class 11 Maths?

The formulas this chapter's questions actually turn on are: Slope-intercept form, Point-slope form, Intercept form, Angle between two lines. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Straight Lines important for JEE Main?

Important — the line forms and the distance and angle formulae are directly examined in boards and in JEE Main, and they are prerequisites for conic sections.

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Previous: Sequences and SeriesNext: Conic SectionsStraight Lines revision notesAll Class 11 Maths NCERT Solutions

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