How to Use This Page
This page provides complete, step-by-step NCERT solutions for every exercise in Chapter 9 — Straight Lines. Work through each problem yourself first, then check your method against the steps below. Pay close attention to which form of the line equation is used and why.
Key Concepts (Quick Recap)
- Slope of line through (x₁, y₁) and (x₂, y₂): m = (y₂ − y₁)/(x₂ − x₁)
- Slope-intercept form: y = mx + c
- Point-slope form: y − y₁ = m(x − x₁)
- Two-point form: (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁)
- Intercept form: x/a + y/b = 1
- General form: Ax + By + C = 0
- Angle between two lines with slopes m₁, m₂: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|
- Parallel lines: m₁ = m₂
- Perpendicular lines: m₁m₂ = −1
- Distance of point (x₁, y₁) from line 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
Q1. Draw the line passing through (2, 3) and (3, 4). Find its slope.
Slope m = (y₂ − y₁)/(x₂ − x₁) = (4 − 3)/(3 − 2) = 1/1 = 1
The line passes through the origin direction at 45°.
Q2. Find the slope of the line passing through the points (3, −2) and (7, 6).
m = (6 − (−2))/(7 − 3) = 8/4 = 2
Q3. Find the slope of the line passing through (4, −8) and (5, −2).
m = (−2 − (−8))/(5 − 4) = 6/1 = 6
Q4. Find the slope of the line which makes an angle of 30° with the positive direction of the x-axis.
m = tan 30° = 1/√3
Q5. Find the angle between the x-axis and the line joining the points (3, 1) and (4, 2).
m = (2 − 1)/(4 − 3) = 1
Angle θ = tan⁻¹(1) = 45°
Q6. Without using Pythagoras theorem, show that the points (4, 4), (3, 5) and (−1, −1) are the vertices of a right-angled triangle.
Find slopes of the sides:
- AB: m₁ = (5 − 4)/(3 − 4) = −1
- BC: m₂ = (−1 − 5)/(−1 − 3) = −6/(−4) = 3/2
- AC: m₃ = (−1 − 4)/(−1 − 4) = −5/(−5) = 1
Check: m₁ × m₃ = (−1)(1) = −1
Since m₁ × m₃ = −1, sides AB and AC are perpendicular. Hence the triangle is right-angled at A. Proved.
Q7. Find the slope of a line which passes through the origin and the midpoint of the line segment joining the points A(0, −4) and B(8, 0).
Midpoint of AB = ((0 + 8)/2, (−4 + 0)/2) = (4, −2).
Slope of line from origin (0, 0) to (4, −2):
m = (−2 − 0)/(4 − 0) = −2/4 = −1/2
Q8. Without using distance formula, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are the vertices of a parallelogram.
Find slopes of opposite sides:
- AB: m₁ = (0 − (−1))/(4 − (−2)) = 1/6
- CD: m₂ = (2 − 3)/(−3 − 3) = −1/(−6) = 1/6
- BC: m₃ = (3 − 0)/(3 − 4) = 3/(−1) = −3
- DA: m₄ = (−1 − 2)/(−2 − (−3)) = −3/1 = −3
Since m₁ = m₂ and m₃ = m₄, opposite sides are parallel. It is a parallelogram.
Exercise 9.2
Q1. Write the equation of the line passing through (2, 3) with slope 4.
Using point-slope form: y − y₁ = m(x − x₁)
y − 3 = 4(x − 2)
y − 3 = 4x − 8
4x − y − 5 = 0
Q2. Find the equation of the line passing through (1, 2) with slope m = −3.
y − 2 = −3(x − 1)
y − 2 = −3x + 3
3x + y − 5 = 0
Q3. Find the equation of the line passing through (4, −3) with slope −2.
y − (−3) = −2(x − 4)
y + 3 = −2x + 8
2x + y − 5 = 0
Q4. Find the equation of the line passing through (0, 0) with slope 5.
y − 0 = 5(x − 0)
y = 5x (or 5x − y = 0)
Q5. Find the equation of the line with slope 2 and y-intercept −5.
Using slope-intercept form y = mx + c:
y = 2x − 5
2x − y − 5 = 0
Q6. Find the equation of the line passing through (2, 2) and (4, 3).
Slope m = (3 − 2)/(4 − 2) = 1/2
Using point-slope: y − 2 = (1/2)(x − 2)
2y − 4 = x − 2
x − 2y + 2 = 0
Q7. Find the equation of the line passing through (1, −1) and (3, 5).
m = (5 − (−1))/(3 − 1) = 6/2 = 3
y − (−1) = 3(x − 1)
y + 1 = 3x − 3
3x − y − 4 = 0
Q8. Find the equation of the line with x-intercept 4 and y-intercept 3.
Using intercept form x/a + y/b = 1:
x/4 + y/3 = 1
3x + 4y = 12
3x + 4y − 12 = 0
Q9. Find the equation of the line making intercepts 2 and −3 on the x and y axes respectively.
x/2 + y/(−3) = 1
x/2 − y/3 = 1
3x − 2y = 6
3x − 2y − 6 = 0
Q10. Find the equation of the line passing through (3, 0) and (0, −2).
Using two-point form: m = (−2 − 0)/(0 − 3) = 2/3
y − 0 = (2/3)(x − 3)
3y = 2x − 6
2x − 3y − 6 = 0
Q11. Find the equation of the line perpendicular to x-axis and passing through (−4, 5).
A line perpendicular to x-axis is a vertical line: x = constant.
Since it passes through (−4, 5): x = −4 (or x + 4 = 0)
Q12. Find the equation of the line parallel to x-axis and passing through (3, −7).
A line parallel to x-axis is a horizontal line: y = constant.
Since it passes through (3, −7): y = −7 (or y + 7 = 0)
Exercise 9.3
Q1. Reduce the equation 3x + 4y − 12 = 0 to intercept form.
3x + 4y = 12
x/4 + y/3 = 1
x-intercept = 4, y-intercept = 3.
Q2. Reduce the equation 5x − 7y − 35 = 0 to intercept form and find the intercepts.
5x − 7y = 35
x/7 − y/5 = 1
x/7 + y/(−5) = 1
x-intercept = 7, y-intercept = −5.
Q3. Find the distance of the point (2, 3) from the line 3x + 4y + 7 = 0.
| d = | Ax₁ + By₁ + C | /√(A² + B²) |
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| d = | 3(2) + 4(3) + 7 | /√(9 + 16) = | 6 + 12 + 7 | /5 = 25/5 = 5 |
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Q4. Find the distance of the point (3, −5) from the line 4x − 3y − 20 = 0.
| d = | 4(3) − 3(−5) − 20 | /√(16 + 9) = | 12 + 15 − 20 | /5 = | 7 | /5 = 7/5 |
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Q5. Find the distance between the parallel lines 3x + 4y − 3 = 0 and 3x + 4y + 7 = 0.
| d = | C₁ − C₂ | /√(A² + B²) = | −3 − 7 | /√(9 + 16) = | −10 | /5 = 2 |
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Q6. Find the distance between the parallel lines 15x + 8y − 25 = 0 and 30x + 16y − 50 = 0.
First make the coefficients proportional. The second equation is 2 times the first:
30x + 16y − 50 = 0 → divide by 2: 15x + 8y − 25 = 0.
Both equations are the same line, so distance = 0.
Q7. Find the angle between the lines y = 2x + 3 and y = 3x + 1.
m₁ = 2, m₂ = 3.
| tan θ = | (m₂ − m₁)/(1 + m₁m₂) | = | (3 − 2)/(1 + 6) | = 1/7 |
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θ = tan⁻¹(1/7) ≈ 8.13°
Q8. Find the angle between the lines x − 2y + 3 = 0 and 2x + y − 5 = 0.
x − 2y + 3 = 0 → m₁ = 1/2
2x + y − 5 = 0 → m₂ = −2
| tan θ = | (−2 − 1/2)/(1 + (1/2)(−2)) | = | (−5/2)/(1 − 1) | = | −5/2 | /0 → undefined. |
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tan θ → ∞, so θ = 90°
The lines are perpendicular.
Q9. Find the equation of the line parallel to 3x − 4y + 7 = 0 and passing through (1, 2).
Parallel line has same slope: m = 3/4.
y − 2 = (3/4)(x − 1)
4y − 8 = 3x − 3
3x − 4y + 5 = 0
Q10. Find the equation of the line perpendicular to 2x + 5y − 3 = 0 and passing through (1, −1).
Slope of given line: m₁ = −2/5.
Perpendicular slope: m₂ = 5/2.
y − (−1) = (5/2)(x − 1)
2y + 2 = 5x − 5
5x − 2y − 7 = 0
Q11. Find the equation of the line passing through the intersection of 2x + 3y − 4 = 0 and 3x − 4y + 5 = 0 and the point (1, 2).
Find intersection of the two lines:
2x + 3y = 4 …(1) 3x − 4y = −5 …(2)
Multiply (1) by 4 and (2) by 3: 8x + 12y = 16 and 9x − 12y = −15.
Adding: 17x = 1 → x = 1/17.
From (1): 2/17 + 3y = 4 → 3y = 66/17 → y = 22/17.
Intersection point: (1/17, 22/17).
Slope from (1/17, 22/17) to (1, 2):
m = (2 − 22/17)/(1 − 1/17) = (12/17)/(16/17) = 12/16 = 3/4
Equation: y − 2 = (3/4)(x − 1) → 4y − 8 = 3x − 3 → 3x − 4y + 5 = 0
Q12. Find the distance of the line 4x + 3y − 12 = 0 from the origin.
| d = | 4(0) + 3(0) − 12 | /√(16 + 9) = | −12 | /5 = 12/5 |
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Q13. Find the distance of the line 6x − 8y + 15 = 0 from the point (2, 3).
| d = | 6(2) − 8(3) + 15 | /√(36 + 64) = | 12 − 24 + 15 | /10 = | 3 | /10 = 3/10 |
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Q14. Find the equation of the line passing through (1, 2) and perpendicular to x + 2y + 3 = 0.
Slope of given line: m₁ = −1/2.
Perpendicular slope: m₂ = 2.
y − 2 = 2(x − 1)
y − 2 = 2x − 2
2x − y = 0 (or y = 2x)
Miscellaneous Exercise
Q1. Find the value of p so that the lines 3x + py − 7 = 0 and 2x − 5y + 3 = 0 are parallel.
For parallel lines, slopes are equal.
m₁ = −3/p, m₂ = 2/5.
−3/p = 2/5 → −15 = 2p → p = −15/2
Q2. Find the value of q so that the lines 2x + 3y − 4 = 0 and qx − 6y + 7 = 0 are perpendicular.
For perpendicular lines: m₁ × m₂ = −1.
m₁ = −2/3, m₂ = q/6.
(−2/3)(q/6) = −1 → −2q/18 = −1 → q/9 = 1 → q = 9
Q3. Find the equation of the line passing through (3, 4) and making an angle of 45° with the line 3x − y + 5 = 0.
Slope of given line: m₁ = 3.
| tan 45° = | (m − 3)/(1 + 3m) | = 1 |
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Case 1: (m − 3)/(1 + 3m) = 1 → m − 3 = 1 + 3m → −2m = 4 → m = −2.
Case 2: (m − 3)/(1 + 3m) = −1 → m − 3 = −1 − 3m → 4m = 2 → m = 1/2.
For m = −2: y − 4 = −2(x − 3) → 2x + y − 10 = 0
For m = 1/2: y − 4 = (1/2)(x − 3) → x − 2y + 5 = 0
The equations are 2x + y − 10 = 0 and x − 2y + 5 = 0.
Q4. Find the equation of the line passing through the point (2, 3) and at a distance of 2 units from the origin.
Let the line be y − 3 = m(x − 2), i.e., mx − y + (3 − 2m) = 0.
| Distance from origin: | 3 − 2m | /√(m² + 1) = 2 |
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(3 − 2m)² = 4(m² + 1)
9 − 12m + 4m² = 4m² + 4
−12m = −5 → m = 5/12
Equation: (5/12)x − y + 3 − 5/6 = 0 → 5x − 12y + 36 − 10 = 0 → 5x − 12y + 26 = 0
| Also check the case where distance formula gives same | value | : |
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The two lines are 5x − 12y + 26 = 0 and 5x − 12y + 2 = 0.
Q5. The line 2x + 3y = 12 meets the x-axis at A and the y-axis at B. The line through (4, 4) parallel to AB meets the x-axis at C and the y-axis at D. Find the area of the parallelogram ABCD.
A = (6, 0), B = (0, 4). Slope = −2/3.
Line through (4, 4) with slope −2/3: y − 4 = (−2/3)(x − 4) → 2x + 3y = 20.
C = (10, 0), D = (0, 20/3).
AB = √(36 + 16) = √52. Height from C to line AB:
| d = | 2(10) + 3(0) − 12 | /√13 = 8/√13. |
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Area = base × height = √52 × 8/√13 = 2√13 × 8/√13 = 16 sq units
Q6. Find the image of the point (3, 8) with respect to the line x + 3y = 7.
Let the image be (h, k). The midpoint ((3+h)/2, (8+k)/2) lies on x + 3y = 7:
(3 + h)/2 + 3(8 + k)/2 = 7 → 3 + h + 24 + 3k = 14 → h + 3k = −13 …(1)
The line joining (3, 8) and (h, k) is perpendicular to x + 3y = 7 (slope = −1/3).
So slope of joining line = 3: (k − 8)/(h − 3) = 3 → k − 8 = 3h − 9 → 3h − k = 1 …(2)
From (1): h = −13 − 3k. Substituting in (2): 3(−13 − 3k) − k = 1 → −39 − 9k − k = 1 → −10k = 40 → k = −4.
h = −13 − 3(−4) = −13 + 12 = −1.
Image is (−1, −4).
Q7. If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find m.
m₁ = 3, m₂ = 1/2.
| tan θ₁ | = | tan θ₂ |
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| (m − 3)/(1 + 3m) | = | (m − 1/2)/(1 + m/2) |
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Case 1: (m − 3)/(1 + 3m) = (m − 1/2)/(1 + m/2)
Cross multiply: (m − 3)(1 + m/2) = (m − 1/2)(1 + 3m)
Expanding and simplifying yields m = ±1.
m = 1 or m = −1
Q8. Find the equation of the line through the intersection of 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 and parallel to 6x − 7y + 8 = 0.
Line through intersection: (2x − 3y + 4) + λ(3x + 4y − 5) = 0
(2 + 3λ)x + (−3 + 4λ)y + (4 − 5λ) = 0
Parallel to 6x − 7y + 8 = 0, so slope must be 6/7.
(−3 + 4λ)/(2 + 3λ) = −7/6
Wait — parallel means same ratio of coefficients: (2 + 3λ)/6 = (−3 + 4λ)/(−7)
Cross multiply: −7(2 + 3λ) = 6(−3 + 4λ) → −14 − 21λ = −18 + 24λ → 4 = 45λ → λ = 4/45.
Substituting: (2 + 4/15)x + (−3 + 16/45)y + (4 − 4/9) = 0
= (34/15)x + (−119/45)y + (32/9) = 0
Multiplying by 45: 102x − 119y + 160 = 0
Q9–Q16 follow the same method — using the distance formula, angle conditions, or intersection methods as demonstrated above. Key results:
| Q9. Distance of (1, 1) from 8x + 15y + 25 = 0: ** | 8 + 15 + 25 | /17 = 48/17** |
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Q10. Line through (2, −3) perpendicular to 2x − 4y + 7 = 0: slope = −2, equation: 2x + y − 1 = 0
Q11. Foot of perpendicular from (1, 2) on 2x − 3y + 4 = 0: (−4/13, 18/13)
| Q12. Distance between 4x + 3y − 6 = 0 and 8x + 6y + 7 = 0 (rewrite second as 4x + 3y + 7/2): d = | −6 − 7/2 | /5 = 19/10 |
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| Q13. Angle between x + y + 1 = 0 and x − y + 1 = 0: m₁ = −1, m₂ = 1, tan θ = | 2/0 | → 90° |
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Q14. Line through (0, 0) parallel to 3x + 4y − 7 = 0: 3x + 4y = 0
Q15. Line through (1, 1) perpendicular to 5x − 12y + 7 = 0: slope = −12/5, equation: 12x + 5y − 17 = 0
Q16. Equation of the line through (2, 3) making equal intercepts: a = a, so x/a + y/a = 1 → x + y = a → a = 5. x + y = 5
Internal Links
- [Chapter 9 Guide — Straight Lines](/school/cbse/class/11/maths/cbse-straight-lines)
- [All Class 11 Maths Solutions](/school/cbse/class/11/maths/solutions)
- [Class 11 Maths Subject Guide](/school/cbse/class/11/maths)