NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.4 – Straight Lines

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.4 covers the most advanced topics of straight lines including reducing equations to standard forms, finding distances, angles between lines, and solving problems involving parallel and perpendicular lines. This exercise brings together all the concepts of straight lines studied in the chapter and tests students' ability to apply multiple formulas in a single problem.

Exercise 10.4 is an important part of the CBSE Class 11 Maths syllabus 2025–26 and is highly useful for board exam preparation. The questions in this exercise require a deeper understanding of coordinate geometry and are regularly featured in competitive exams like JEE Main and JEE Advanced, making consistent practice essential for scoring well.

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.4 – Straight Lines

Straight Lines
Difficult
Q.
Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.
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.
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
Difficult
Q.
Reduce the following equations into normal form. Find their perpendicular distances from the originand angle between perpendicular and the positive x-axis.(i) x – 3 y + 8 = 0,(ii) y – 2 = 0,(iii) x–y = 4.
Straight Lines
Medium
Q.
Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).
Straight Lines
Difficult
Q.
Find the points on the x-axis, whose distances from the line  x3+y4=1  are 4 units.
Straight Lines
Difficult
Q.
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.
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 equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
Straight Lines
Difficult
Q.
Find angles between the lines 3 x + y = 1and x + 3y = 1.
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.
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.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).
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
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
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.
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
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
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
Difficult
Q.
Find the equation of the line through the intersection of lines x + 2y – 3 = 0 and 4x – y + 7 = 0 and which is parallel to 5x + 4y – 20 = 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.
Straight Lines
Medium
Q.
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.
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.
Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.
Straight Lines
Difficult
Q.
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.
Straight Lines
Medium
Q.
If three points (h, 0), (a, b) and (0, k) lie on a line, show that ah+bk=1.
Straight Lines
Medium
Q.
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.
Straight Lines
Medium
Q.
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Straight Lines
Medium
Q.
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).
Straight Lines
Difficult
Q.
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.
Straight Lines
Medium
Q.
Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Straight Lines
Difficult
Q.
Find the value of x for which the points (x, – 1), (2, 1) and (4, 5) are collinear.
Straight Lines
Difficult
Q.
Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.
Straight Lines
Medium
Q.
Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2).
Straight Lines
Difficult
Q.
The slope of a line is double of the slope of another line. If tangent of the angle between them is 13, find the slopes of the lines.
Straight Lines
Easy
Q.
A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k – y1 = m (h – x1).
Straight Lines
Difficult
Q.
Consider the following population and year graph (Fig 10.10), find the slope of the line AB and using it, find what will be the population in the year 2010?

Straight Lines
Medium
Q.
P(a, b) is the mid-point of a line segment between axes.Show that equation of the line is  xa+yb=1.
Straight Lines
Difficult
Q.
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.
2. Passing through the point (–4, 3) with slope  1 2 . MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaaeiuaiaabg gacaqGZbGaae4CaiaabMgacaqGUbGaae4zaiaabccacaqG0bGaaeiA aiaabkhacaqGVbGaaeyDaiaabEgacaqGObGaaeiiaiaabshacaqGOb GaaeyzaiaabccacaqGWbGaae4BaiaabMgacaqGUbGaaeiDaiaabcca caqGOaGaai4eGiaabsdacaqGSaGaaeiiaiaabodacaqGPaGaaeiiai aabEhacaqGPbGaaeiDaiaabIgacaqGGaGaae4CaiaabYgacaqGVbGa aeiCaiaabwgacaqGGaWaaSaaaeaacaqGXaaabaGaaeOmaaaacaqGUa aaaa@5E5B@
3. Passing through (0, 0) with slope m.
4. Passing through (2, 2 3 ) and inclined with the x-axis at an angle of 75°. MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaaeiuaiaabg gacaqGZbGaae4CaiaabMgacaqGUbGaae4zaiaabccacaqG0bGaaeiA aiaabkhacaqGVbGaaeyDaiaabEgacaqGObGaaeiiaiaabIcacaqGYa GaaeilaiaabccacaqGYaWaaOaaaeaacaqGZaaaleqaaOGaaeykaiaa bccacaqGHbGaaeOBaiaabsgacaqGGaGaaeyAaiaab6gacaqGJbGaae iBaiaabMgacaqGUbGaaeyzaiaabsgacaqGGaGaae4DaiaabMgacaqG 0bGaaeiAaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiEaiaab2 cacaqGHbGaaeiEaiaabMgacaqGZbGaaeiiaiaabggacaqG0bGaaeii aiaabggacaqGUbGaaeiiaiaabggacaqGUbGaae4zaiaabYgacaqGLb Gaaeiiaiaab+gacaqGMbGaaeiiaiaabEdacaqG1aGaaeiSaiaab6ca aaa@7313@
5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.
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.
7. Passing through the points (–1, 1) and (2,– 4).
8.  Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.
Straight Lines
Medium
Q.
The vertices of Δ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.
Straight Lines
Medium
Q.
Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).
Straight Lines
Difficult
Q.
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.
Straight Lines
Medium
Q.
Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
Straight Lines
Medium
Q.
Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Straight Lines
Medium
Q.
Find equation of the line through the point (0, 2) making an angle 2π3 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.
Straight Lines
Medium
Q.
The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.
Straight Lines
Medium
Q.
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.
Straight Lines
Medium
Q.
The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹14/litre and 1220 litres of milk each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17/litre?
Straight Lines
Difficult
Q.
If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.4 – Straight Lines

Question 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.

a. Parallel to the x-axis:

A line is parallel to the x-axis when the coefficient of x is 0. k − 3 = 0 ⟹ k = 3

When k = 3: coefficient of y = −(4 − 9) = 5 ≠ 0. ✓

Therefore, k = 3.

b. Parallel to the y-axis:

A line is parallel to the y-axis when the coefficient of y is 0. −(4 − k²) = 0 ⟹ 4 − k² = 0 ⟹ k² = 4 ⟹ k = ±2

When k = 2: coefficient of x = 2 − 3 = −1 ≠ 0. ✓ When k = −2: coefficient of x = −2 − 3 = −5 ≠ 0. ✓

Therefore, k = 2 or k = −2.

c. Passing through the origin:

A line passes through the origin when the constant term is 0. k² − 7k + 6 = 0 ⟹ (k − 1)(k − 6) = 0 ⟹ k = 1 or k = 6

Therefore, k = 1 or k = 6.

Question 2. Find the values of θ and p, if the equation x cosθ + y sinθ = p is the normal form of the line √3x + y + 2 = 0.

The given line is √3x + y + 2 = 0. ⟹ −√3x − y = 2 ⟹ √3x + y = −2

Dividing by √((√3)² + 1²) = √(3 + 1) = 2: (√3/2)x + (1/2)y = −1

Since p must be positive (p > 0), multiply both sides by −1: (−√3/2)x + (−1/2)y = 1

But this is not the standard normal form. The normal form requires p > 0.

Dividing the original −√3x − y = 2 by 2: (−√3/2)x + (−1/2)y = 1

Comparing with x cosθ + y sinθ = p: cosθ = −√3/2 and sinθ = −1/2, p = 1

θ lies in the third quadrant: θ = 180° + 30° = 210°

Therefore, θ = 210° and p = 1.

Question 3. Find the equations of the lines, which cut off intercepts on the axes whose sum and product are 1 and −6 respectively.

Let the x-intercept be a and y-intercept be b.

Given: a + b = 1 …(i) and a × b = −6 …(ii)

From (i): b = 1 − a Substituting in (ii): a(1 − a) = −6 ⟹ a − a² = −6 ⟹ a² − a − 6 = 0 ⟹ (a − 3)(a + 2) = 0 ⟹ a = 3 or a = −2

If a = 3, then b = 1 − 3 = −2 Equation: x/3 + y/(−2) = 1 ⟹ 2x − 3y = 6 ⟹ 2x − 3y − 6 = 0

If a = −2, then b = 1 − (−2) = 3 Equation: x/(−2) + y/3 = 1 ⟹ −3x + 2y = 6 ⟹ 3x − 2y + 6 = 0

Therefore, the equations are 2x − 3y − 6 = 0 and 3x − 2y + 6 = 0.

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

The given line is x/3 + y/4 = 1, or 4x + 3y − 12 = 0. A = 4, B = 3, C = −12.

Let the point on the y-axis be (0, b).

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

Case 1: 3b − 12 = 20 ⟹ 3b = 32 ⟹ b = 32/3 Case 2: 3b − 12 = −20 ⟹ 3b = −8 ⟹ b = −8/3

Therefore, the required points are (0, 32/3) and (0, −8/3).

Question 5. Find the perpendicular distance from the origin to the line joining the points (cosθ, sinθ) and (cosφ, sinφ).

The equation of the line through (cosθ, sinθ) and (cosφ, sinφ) is:

y − sinθ = [(sinφ − sinθ)/(cosφ − cosθ)](x − cosθ)

Using sum-to-product formulas: sinφ − sinθ = 2cos((φ+θ)/2)sin((φ−θ)/2) cosφ − cosθ = −2sin((φ+θ)/2)sin((φ−θ)/2)

Slope = [2cos((φ+θ)/2)sin((φ−θ)/2)] / [−2sin((φ+θ)/2)sin((φ−θ)/2)] = −cos((φ+θ)/2) / sin((φ+θ)/2) = −cot((φ+θ)/2)

Equation of line: y − sinθ = −[cos((φ+θ)/2)/sin((φ+θ)/2)](x − cosθ)

Rearranging: x cos((φ+θ)/2) + y sin((φ+θ)/2) = cosθ·cos((φ+θ)/2) + sinθ·sin((φ+θ)/2) = cos(θ − (φ+θ)/2) = cos((θ−φ)/2)

Perpendicular distance from origin (0, 0): d = |cos((θ−φ)/2)| / √(cos²((φ+θ)/2) + sin²((φ+θ)/2)) = |cos((θ−φ)/2)| / 1

Therefore, the perpendicular distance = |cos((θ−φ)/2)|.

Question 6. 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.

Solving the two equations simultaneously: x − 7y + 5 = 0 …(i) 3x + y = 0 ⟹ y = −3x …(ii)

Substituting (ii) in (i): x − 7(−3x) + 5 = 0 x + 21x + 5 = 0 22x = −5 x = −5/22

y = −3(−5/22) = 15/22

The point of intersection is (−5/22, 15/22).

A line parallel to the y-axis has the form x = constant.

Therefore, the equation of the line is x = −5/22, or 22x + 5 = 0.

Question 7. Classify the following pairs of lines as coincident, parallel or intersecting.

i. 2x + y − 1 = 0 and 3x + 2y + 5 = 0

Slope of line 1 = −2 Slope of line 2 = −3/2

Since slopes are different, the lines are intersecting.

ii. x + 2y − 3 = 0 and −3x − 6y + 9 = 0

Dividing the second equation by −3: x + 2y − 3 = 0

Both equations are identical.

Therefore, the lines are coincident.

iii. 3x + 2y − 4 = 0 and 6x + 4y − 8 = 0

Dividing the second equation by 2: 3x + 2y − 4 = 0

Both equations are identical.

Therefore, the lines are coincident.

Question 8. Find the equation of the lines through the point (3, 2) which makes an angle of 45° with the line x − 2y = 3.

Slope of x − 2y = 3 is m₁ = 1/2.

Let slope of required line be m.

tan 45° = |(m − m₁)/(1 + m·m₁)| 1 = |(m − 1/2)/(1 + m/2)| = |(2m − 1)/(2 + m)|

Case 1: (2m − 1)/(2 + m) = 1 2m − 1 = 2 + m ⟹ m = 3

Equation through (3, 2) with slope 3: y − 2 = 3(x − 3) y − 2 = 3x − 9 3x − y − 7 = 0

Case 2: (2m − 1)/(2 + m) = −1 2m − 1 = −2 − m 3m = −1 ⟹ m = −1/3

Equation through (3, 2) with slope −1/3: y − 2 = (−1/3)(x − 3) 3y − 6 = −x + 3 x + 3y − 9 = 0

Therefore, the equations of the lines are 3x − y − 7 = 0 and x + 3y − 9 = 0.

Question 9. Find the equation of the line through the intersection of lines 3x − 4y + 1 = 0 and 5x + y − 1 = 0 and which is perpendicular to the line 2x − 3y = 10.

The equation of any line through the intersection of 3x − 4y + 1 = 0 and 5x + y − 1 = 0 is: (3x − 4y + 1) + k(5x + y − 1) = 0 (3 + 5k)x + (−4 + k)y + (1 − k) = 0

Slope of this line = −(3 + 5k)/(−4 + k) = (3 + 5k)/(4 − k)

Slope of 2x − 3y = 10 is 2/3.

Since perpendicular: [(3 + 5k)/(4 − k)] × (2/3) = −1 2(3 + 5k) = −3(4 − k) 6 + 10k = −12 + 3k 7k = −18 k = −18/7

Substituting k = −18/7: (3 + 5(−18/7))x + (−4 + (−18/7))y + (1 − (−18/7)) = 0 (21/7 − 90/7)x + (−28/7 − 18/7)y + (7/7 + 18/7) = 0 (−69/7)x + (−46/7)y + (25/7) = 0

Multiplying by −7: 69x + 46y − 25 = 0

Dividing by 23: Therefore, the equation is 3x + 2y − 25/23 = 0, or more precisely 69x + 46y − 25 = 0.

Question 10. Find the equation of the line through the intersection of 5x − 3y = 1 and 2x + 3y − 23 = 0 and perpendicular to the line 5x − 3y − 1 = 0.

First, find point of intersection of 5x − 3y = 1 and 2x + 3y = 23.

Adding the two equations: 7x = 24 ⟹ x = 24/7

3y = 5(24/7) − 1 = 120/7 − 7/7 = 113/7 y = 113/21

Slope of 5x − 3y − 1 = 0 is 5/3. Slope of perpendicular line = −3/5.

Equation through (24/7, 113/21) with slope −3/5: y − 113/21 = (−3/5)(x − 24/7) 5(21y − 113) = −3(21)(x − 24/7) 105y − 565 = −63x + 72 × 3 105y − 565 = −63x + 216 63x + 105y = 781

Dividing by 7: Therefore, the equation of the line is 9x + 15y = 111.5, or 63x + 105y − 781 = 0.

Question 11. Find the equation of the lines which pass through the point (3, 4) and cut off intercepts from the coordinate axes such that their sum is 14.

Let x-intercept = a and y-intercept = b. Given: a + b = 14 ⟹ b = 14 − a.

Equation: x/a + y/(14 − a) = 1

Since it passes through (3, 4): 3/a + 4/(14 − a) = 1 3(14 − a) + 4a = a(14 − a) 42 − 3a + 4a = 14a − a² a² − 13a + 42 = 0 (a − 6)(a − 7) = 0 ⟹ a = 6 or a = 7

If a = 6: b = 8 Equation: x/6 + y/8 = 1 ⟹ 4x + 3y = 24

If a = 7: b = 7 Equation: x/7 + y/7 = 1 ⟹ x + y = 7

Therefore, the equations are 4x + 3y = 24 and x + y = 7.

Frequently Asked Questions (FAQs)

Q1. What is the condition for a line to be parallel to the x-axis?

A line Ax + By + C = 0 is parallel to the x-axis when the coefficient of x is 0, i.e., A = 0. The equation reduces to By + C = 0, giving y = −C/B (a constant).

Q2. What is the normal form of a line?

The normal form is x cosθ + y sinθ = p, where p > 0 is the perpendicular distance from the origin to the line and θ is the angle made by the perpendicular with the positive x-axis.

Q3. How do you find the point of intersection of two lines?

Solve the two equations simultaneously using substitution or elimination method. The solution (x, y) gives the coordinates of the point of intersection.

Q4. How do you find lines through the intersection of two given lines?

Use the family of lines concept: (L₁) + k(L₂) = 0, where L₁ = 0 and L₂ = 0 are the given lines and k is a parameter determined by the additional condition given.

Q5. How many questions are in NCERT Class 11 Maths Chapter 10 Exercise 10.4?

Exercise 10.4 contains 11 questions covering advanced topics of straight lines including classification of line pairs, lines through intersection of two lines, normal form, and intercept problems with given conditions.