NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 – Straight Lines (2025-26)

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 covers advanced topics related to straight lines including slope-intercept form, intercept form, distance of a point from a line, distance between parallel lines, and equations of lines under various geometric conditions such as parallel, perpendicular, and right bisector problems. The exercise also includes important proof-based questions that test students' conceptual understanding.

Exercise 10.3 is an important part of the CBSE Class 11 Maths syllabus 2025–26 and carries significant weightage in board examinations. Thorough practice of these 17 questions helps students develop problem-solving skills required for board exams and builds a strong analytical base for competitive exams like JEE Main and JEE Advanced.

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 – Straight Lines (2025-26)

Straight Lines
Medium
Q.
Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.
Straight Lines
Medium
Q.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).
Straight Lines
Difficult
Q.
Find the equation of the line through the intersection of lines 3x + 4y = 7 and x – y + 2 = 0 and whose slope is 5.
Straight Lines
Medium
Q.
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  1p2=1a2+1b2.
Straight Lines
Medium
Q.
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.
Straight Lines
Difficult
Q.
If p and q are the lengths of perpendiculars from the origin to the lines x cosθ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p2 + 4q2 = k2.
Straight Lines
Medium
Q.
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.
Straight Lines
Medium
Q.
Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.
Straight Lines
Difficult
Q.
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.
Straight Lines
Medium
Q.
By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear.
Straight Lines
Medium
Q.
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.
Straight Lines
Medium
Q.
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.
Straight Lines
Difficult
Q.
Find angles between the lines 3 x + y = 1and x + 3y = 1.
Straight Lines
Medium
Q.
Find equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (–2, 3).
Straight Lines
Difficult
Q.
Find the points on the x-axis, whose distances from the line  x3+y4=1  are 4 units.
Straight Lines
Medium
Q.
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.
Straight Lines
Medium
Q.
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.
Straight Lines
Medium
Q.
Find the equation of the line through the intersection of the lines 2x + 3y – 4 = 0 and x – 5y = 7 that has its x-intercept equal to – 4.

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 – Straight Lines (2025-26)

Question 1. Reduce the following equations into slope-intercept form and find their slopes and y-intercepts.

i. x + 7y = 0

The equation is x + 7y = 0. We can write it as y = −(1/7)x + 0 …(1) This is of the form y = mx + c, where m = −1/7 and c = 0.

Therefore, the slope is −1/7 and the y-intercept is 0.

ii. 6x + 3y − 5 = 0

The given equation is 6x + 3y − 5 = 0. y = (1/3)(−6x + 5) ⟹ y = −2x + 5/3 …(1) This is of the form y = mx + c, where m = −2 and c = 5/3.

Therefore, the slope is −2 and the y-intercept is 5/3.

iii. y = 0

The given equation is y = 0. We can write it as y = 0·x + 0 …(1) This is of the form y = mx + c, where m = 0 and c = 0.

Therefore, the slope is 0 and the y-intercept is 0.

Question 2. Reduce the following equations into intercept form and find their intercepts on the axes.

i. 3x + 2y − 12 = 0

The given equation is 3x + 2y − 12 = 0. 3x + 2y = 12 ⟹ (3x/12) + (2y/12) = 1 ⟹ x/4 + y/6 = 1 …(1) This is of the form x/a + y/b = 1, where a = 4 and b = 6.

Therefore, the x-intercept is 4 and the y-intercept is 6.

ii. 4x − 3y = 6

The given equation is 4x − 3y = 6. (4x/6) − (3y/6) = 1 ⟹ x/(3/2) + y/(−2) = 1 …(2)

Therefore, the x-intercept is 3/2 and the y-intercept is −2.

iii. 3y + 2 = 0

The given equation is 3y + 2 = 0. 3y = −2 ⟹ y/(−2/3) = 1 …(3) This is of the form x/a + y/b = 1, where a = 0 and b = −2/3.

Therefore, the y-intercept is −2/3. It has no x-intercept.

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

The equation of the line is 12(x + 6) = 5(y − 2). ⟹ 12x + 72 = 5y − 10 ⟹ 12x − 5y + 82 = 0

Comparing with Ax + By + C = 0, we get A = 12, B = −5, C = 82.

d = |Ax₁ + By₁ + C| / √(A² + B²) = |12(−1) + (−5)(1) + 82| / √(144 + 25) = |−12 − 5 + 82| / √169 = |65| / 13

Therefore, the distance = 5 units.

Question 4. Find the points on the x-axis at distance 4 units from the line x/3 + y/4 = 1.

The given line is x/3 + y/4 = 1, or 4x + 3y − 12 = 0. A = 4, B = 3, C = −12. Let (a, 0) be the point on the x-axis.

4 = |4a + 3(0) − 12| / √(16 + 9) ⟹ 4 = |4a − 12| / 5 ⟹ |4a − 12| = 20 ⟹ ±(4a − 12) = 20

Case 1: 4a − 12 = 20 ⟹ 4a = 32 ⟹ a = 8 Case 2: −(4a − 12) = 20 ⟹ 4a = −8 ⟹ a = −2

Therefore, the required points are (−2, 0) and (8, 0).

Question 5. Find the distance between the following parallel lines.

i. 15x + 8y − 34 = 0 and 15x + 8y + 31 = 0

A = 15, B = 8, C₁ = −34, C₂ = 31.

d = |C₁ − C₂| / √(A² + B²) = |−34 − 31| / √(225 + 64) = |−65| / √289 = 65/17

Therefore, the distance = 65/17 units.

ii. l(x + y) + p = 0 and l(x + y) − r = 0

The lines are lx + ly + p = 0 and lx + ly − r = 0. A = l, B = l, C₁ = p, C₂ = −r.

d = |C₁ − C₂| / √(A² + B²) = |p + r| / √(l² + l²) = |p + r| / (l√2)

Therefore, d = (1/√2) · |p + r| / l units.

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

Slope of 3x − 4y + 2 = 0 is m = 3/4. Parallel line has same slope.

Equation through (−2, 3): y − 3 = (3/4)(x + 2) 4y − 12 = 3x + 6

Therefore, the equation of the required line is 3x − 4y + 18 = 0.

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

Slope of x − 7y + 5 = 0 is 1/7. Slope of perpendicular line = −1/(1/7) = −7.

Equation with slope −7 passing through (3, 0): y = −7(x − 3) ⟹ y = −7x + 21

Therefore, the equation of the required line is 7x + y = 21.

Question 8. Find the angle between the lines √3x + y = 1 and x + √3y = 1.

y = −√3x + 1 ⟹ m₁ = −√3 y = −(1/√3)x + 1/√3 ⟹ m₂ = −1/√3

tan θ = |(m₁ − m₂)/(1 + m₁m₂)| = |(−√3 + 1/√3)/(1 + (−√3)(−1/√3))| = |(−2/√3)/2| = 1/√3

⟹ θ = 30°

Therefore, the angle between the lines is 30° or 150°.

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

m₁ = (1 − 3)/(4 − h) = −2/(4 − h) m₂ = 7/9

Since perpendicular: m₁ × m₂ = −1 (−2/(4 − h)) × (7/9) = −1 ⟹ −14 = −(36 − 9h) ⟹ 14 = 36 − 9h ⟹ 9h = 22

Therefore, h = 22/9.

Question 10. Prove that the line through (x₁, y₁) parallel to Ax + By + C = 0 is A(x − x₁) + B(y − y₁) = 0.

Slope of Ax + By + C = 0 is m = −A/B. Parallel lines have the same slope.

Equation through (x₁, y₁) with slope −A/B: y − y₁ = (−A/B)(x − x₁) B(y − y₁) = −A(x − x₁)

Therefore, A(x − x₁) + B(y − y₁) = 0. (Hence proved)

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

m₁ = 2. Using tan 60° = |(m₁ − m₂)/(1 + m₁m₂)|: √3 = |(2 − m₂)/(1 + 2m₂)|

This gives: m₂ = (2 − √3)/(2√3 + 1)   or   m₂ = −(2 + √3)/(2√3 − 1)

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

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

Midpoint of AB = ((3 − 1)/2, (4 + 2)/2) = (1, 3) Slope of AB = (2 − 4)/(−1 − 3) = −2/−4 = 1/2 Slope of perpendicular bisector = −2

Equation through (1, 3) with slope −2: y − 3 = −2(x − 1) ⟹ y − 3 = −2x + 2 ⟹ 2x + y = 5

Therefore, the equation of the right bisector is 2x + y = 5.

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

Let (a, b) be the foot. m₁ = (b − 3)/(a + 1), m₂ = 3/4.

m₁ × m₂ = −1 ⟹ (b − 3)/(a + 1) × (3/4) = −1 ⟹ 3b − 9 = −4a − 4 ⟹ 4a + 3b = 5 …(1)

Since (a, b) lies on 3x − 4y = 16: 3a − 4b = 16 …(2)

Solving (1) and (2): a = 68/25, b = −49/25

Therefore, the foot of perpendicular = (68/25, −49/25).

Question 14. The perpendicular from the origin to y = mx + c meets it at (−1, 2). Find m and c.

Slope of line joining (0, 0) and (−1, 2) = 2/(−1) = −2 Since perpendicular to y = mx + c: m × (−2) = −1 ⟹ m = 1/2

Since (−1, 2) lies on y = mx + c: 2 = (1/2)(−1) + c ⟹ c = 5/2

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

Question 15. Prove p² + 4q² = k² for perpendiculars from origin to x cosθ − y sinθ = k cos2θ and x secθ + y cosecθ = k.

For line (1): p = |−k cos2θ| / √(cos²θ + sin²θ) = |k cos2θ|

For line (2): q = |−k| / √(sec²θ + cosec²θ) = k sinθ cosθ

p² + 4q² = k²cos²2θ + 4k²sin²θcos²θ = k²cos²2θ + k²(2sinθcosθ)² = k²cos²2θ + k²sin²2θ = k²(cos²2θ + sin²2θ) = k²

Hence proved that p² + 4q² = k².

Question 16. In △ABC with A(2, 3), B(4, −1) and C(1, 2), find the equation and length of altitude from vertex A.

Slope of BC = (2 − (−1))/(1 − 4) = 3/(−3) = −1 Slope of altitude AD = −1/(−1) = 1

Equation through A(2, 3): y − 3 = 1(x − 2) ⟹ y − x = 1

Equation of BC: x + y − 3 = 0

Length of AD = |1(2) + 1(3) − 3| / √(1² + 1²) = |2| / √2 = √2

Therefore, equation of altitude: y − x = 1 and Length = √2 units.

Question 17. If p is the length of perpendicular from origin to a line with intercepts a and b on axes, show that 1/p² = 1/a² + 1/b².

Equation of line: x/a + y/b = 1 ⟹ bx + ay − ab = 0

p = |b(0) + a(0) − ab| / √(b² + a²) = ab / √(a² + b²)

p² = (ab)² / (a² + b²)

⟹ (a² + b²) / (a²b²) = 1/p²

⟹ 1/a² + 1/b² = 1/p²

Therefore, 1/p² = 1/a² + 1/b². (Hence proved)

Frequently Asked Questions (FAQs)

Q1. What is the slope-intercept form of a line? The slope-intercept form is y = mx + c, where m is the slope and c is the y-intercept. It directly gives the steepness and the point where the line crosses the y-axis.

Q2. How do you find the distance between two parallel lines? For parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0, the distance is d = |C₁ − C₂| / √(A² + B²).

Q3. What is the intercept form of a line? The intercept form is x/a + y/b = 1, where a is the x-intercept and b is the y-intercept.

Q4. How do you find the foot of perpendicular from a point to a line? Let (a, b) be the foot. Apply two conditions: the perpendicularity condition (product of slopes = −1) and the fact that (a, b) satisfies the equation of the given line. Solve the two equations simultaneously.

Q5. How many questions are in NCERT Class 11 Maths Chapter 10 Exercise 10.3? Exercise 10.3 contains 17 questions covering slope-intercept form, intercept form, distance of a point from a line, distance between parallel lines, and equations of lines under geometric conditions like parallelism, perpendicularity, and right bisectors.