NCERT Solutions for Class 11 Maths Chapter 10- Straight Lines Exercise 10.3

NCERT Solutions are provided to help the students in understanding the steps to solve mathematical problems that are provided in the textbook. The Exercise 10.3 of NCERT Solutions for Class 11 Maths Chapter 10- Straight Lines is based on the following topics:

  1. General Equation of a Line
    1. Different forms of Ax + By + C = 0
  2. The distance of a Point From a Line
    1. Distance between two parallel lines

The subject experts at BYJU’S stick to the syllabus while preparing the solutions. The problem-solving method provided in the examples is followed while preparing the NCERT Solutions for class 11 as well.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 10- Straight Lines Exercise 10.3

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Access Other Exercise Solutions of Class 11 Maths Chapter 10

Exercise 10.1 Solutions 14 Questions

Exercise 10.2 Solutions 20 Questions

Miscellaneous Exercise On Chapter 10 Solutions 24 Questions

Access Solutions for Class 11 Maths Chapter 10 Exercise 10.3

1. 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

Solution:

(i) x + 7y = 0

Given:

The equation is x + 7y = 0

Slope – intercept form is represented in the form ‘y = mx + c’, where m is the slope and c is the y intercept

So, the above equation can be expressed as

y = -1/7x + 0

∴ The above equation is of the form y = mx + c, where m = -1/7 and c = 0.

(ii) 6x + 3y – 5 = 0

Given:

The equation is 6x + 3y – 5 = 0

Slope – intercept form is represented in the form ‘y = mx + c’, where m is the slope and c is the y intercept

So, the above equation can be expressed as

3y = -6x + 5

y = -6/3x + 5/3

= -2x + 5/3

∴ The above equation is of the form y = mx + c, where m = -2 and c = 5/3.

(iii) y = 0

Given:

The equation is y = 0

Slope – intercept form is given by ‘y = mx + c’, where m is the slope and c is the y intercept

y = 0 × x + 0

∴ The above equation is of the form y = mx + c, where m = 0 and c = 0.

2. 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

Solution:

(i) 3x + 2y – 12 = 0

Given:

The equation is 3x + 2y – 12 = 0

Equation of line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepts on x axis and y – axis respectively.

So, 3x + 2y = 12

now let us divide both sides by 12, we get

3x/12 + 2y/12 = 12/12

x/4 + y/6 = 1

∴ The above equation is of the form x/a + y/b = 1, where a = 4, b = 6

Intercept on x – axis is 4

Intercept on y – axis is 6

(ii) 4x – 3y = 6

Given:

The equation is 4x – 3y = 6

Equation of line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepts on x axis and y – axis respectively.

So, 4x – 3y = 6

Now let us divide both sides by 6, we get

4x/6 – 3y/6 = 6/6

2x/3 – y/2 = 1

x/(3/2) + y/(-2) = 1

∴ The above equation is of the form x/a + y/b = 1, where a = 3/2, b = -2

Intercept on x – axis is 3/2

Intercept on y – axis is -2

(iii) 3y + 2 = 0

Given:

The equation is 3y + 2 = 0

Equation of line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepts on x axis and y – axis respectively.

So, 3y = -2

Now, let us divide both sides by -2, we get

3y/-2 = -2/-2

3y/-2 = 1

y/(-2/3) = 1

∴ The above equation is of the form x/a + y/b = 1, where a = 0, b = -2/3

Intercept on x – axis is 0

Intercept on y – axis is -2/3

3. Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

(i) x – √3y + 8 = 0

(ii) y – 2 = 0

(iii) x – y = 4

Solution:

(i) x – √3y + 8 = 0

Given:

The equation is x – √3y + 8 = 0

Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.

So now, x – √3y + 8 = 0

x – √3y = -8

Divide both the sides by √(12 + (√3)2) = √(1 + 3) = √4 = 2

x/2 – √3y/2 = -8/2

(-1/2)x + √3/2y = 4

This is in the form of: x cos 120o + y sin 120o = 4

∴ The above equation is of the form x cos θ + y sin θ = p, where θ = 120° and p = 4.

Perpendicular distance of line from origin = 4

Angle between perpendicular and positive x – axis = 120°

(ii) y – 2 = 0

Given:

The equation is y – 2 = 0

Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.

So now, 0 × x + 1 × y = 2

Divide both sides by √(02 + 12) = √1 = 1

0 (x) + 1 (y) = 2

This is in the form of: x cos 90o + y sin 90o = 2

∴ The above equation is of the form x cos θ + y sin θ = p, where θ = 90° and p = 2.

Perpendicular distance of line from origin = 2

Angle between perpendicular and positive x – axis = 90°

(iii) x – y = 4

Given:

The equation is x – y + 4 = 0

Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.

So now, x – y = 4

Divide both the sides by √(12 + 12) = √(1+1) = √2

x/√2 – y/√2 = 4/√2

1/√2x + (-1/√2)y = 2√2

This is in the form: x cos 315o + y sin 315o = 2√2

∴ The above equation is of the form x cos θ + y sin θ = p, where θ = 315° and p = 2√2.

Perpendicular distance of line from origin = 2√2

Angle between perpendicular and positive x – axis = 315°

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

Solution:

Given:

The equation of the line is 12(x + 6) = 5(y – 2).

12x + 72 = 5y – 10

12x – 5y + 82 = 0 … (1)

Now, compare equation (1) with general equation of line Ax + By + C = 0, where A = 12, B = –5, and C = 82

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 33

∴ The distance is 5units.

5. Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.

Solution:

Given:

The equation of line is x/3 + y/4 = 1

4x + 3y = 12

4x + 3y – 12 = 0 …. (1)

Now, compare equation (1) with general equation of line Ax + By + C = 0, where A = 4, B = 3, and C = -12

Let (a, 0) be the point on the x-axis, whose distance from the given line is 4 units.

So, the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 34

|4a – 12| = 4 × 5

± (4a – 12) = 20

4a – 12 = 20 or – (4a – 12) = 20

4a = 20 + 12 or 4a = -20 + 12

a = 32/4 or a = -8/4

a = 8 or a = -2

∴ The required points on the x – axis are (-2, 0) and (8, 0)

6. 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

Solution:

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

Given:

The parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.

By using the formula,

The distance (d) between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 35

∴ The distance between parallel lines is 65/17

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

Given:

The parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0.

lx + ly + p = 0 and lx + ly – r = 0

by using the formula,

The distance (d) between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 36

∴ The distance between parallel lines is |p+r|/l2

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

Solution:

Given:

The line is 3x – 4y + 2 = 0

So, y = 3x/4 + 2/4

= 3x/4 + ½

Which is of the form y = mx + c, where m is the slope of the given line.

The slope of the given line is 3/4

We know that parallel line have same slope.

∴ Slope of other line = m = 3/4

Equation of line having slope m and passing through (x1, y1) is given by

y – y1 = m (x – x1)

∴ Equation of line having slope 3/4 and passing through (-2, 3) is

y – 3 = ¾ (x – (-2))

4y – 3 × 4 = 3x + 3 × 2

3x – 4y = 18

∴ The equation is 3x – 4y = 18

8. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Solution:

Given:

The equation of line is x – 7y + 5 = 0

So, y = 1/7x + 5/7 [which is of the form y = mx + c, where m is the slope of the given line.]

Slope of the given line is 1/7

Slope of the line perpendicular to the line having slope m is -1/m

Slope of the line perpendicular to the line having a slope of 1/7 is -1/(1/7) = -7

So, the equation of line with slope -7 and x intercept 3 is given by y = m(x – d)

y = -7 (x – 3)

y = -7x + 21

7x + y = 21

∴ The equation is 7x + y = 21

9. Find angles between the lines √3x + y = 1 and x + √3y = 1.

Solution:

Given:

The lines are √3x + y = 1 and x + √3y = 1

So, y = -√3x + 1 … (1) and

y = -1/√3x + 1/√3 …. (2)

Slope of line (1) is m1 = -√3, while the slope of line (2) is m2 = -1/√3

Let θ be the angle between two lines

So,

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 37

θ = 30°

∴ The angle between the given lines is either 30° or 180°- 30° = 150°

10. 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.

Solution:

Let the slope of the line passing through (h, 3) and (4, 1) be m1

Then, m1 = (1-3)/(4-h) = -2/(4-h)

Let the slope of line 7x – 9y – 19 = 0 be m2

7x – 9y – 19 = 0

So, y = 7/9x – 19/9

m2 = 7/9

Since, the given lines are perpendicular

m1 × m2 = -1

-2/(4-h) × 7/9 = -1

-14/(36-9h) = -1

-14 = -1 × (36 – 9h)

36 – 9h = 14

9h = 36 – 14

h = 22/9

∴ The value of h is 22/9

11. 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.

Solution:

Let the slope of line Ax + By + C = 0 be m

Ax + By + C = 0

So, y = -A/Bx – C/B

m = -A/B

By using the formula,

Equation of the line passing through point (x1, y1) and having slope m = -A/B is

y – y1 = m (x – x1)

y – y1= -A/B (x – x1)

B (y – y1) = -A (x – x1)

∴ A(x – x1) + B(y – y1) = 0

So, the line through point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x – x1) + B (y – y1) = 0

Hence proved.

12. Two lines passing through the point (2, 3) intersects each other at an angle of 60o. If slope of one line is 2, find equation of the other line.

Solution:

Given: m1 = 2

Let the slope of the first line be m1

And let the slope of the other line be m2.

Angle between the two lines is 60°.

So,

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 38

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13. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

Solution:

Given:

The right bisector of a line segment bisects the line segment at 90°.

End-points of the line segment AB are given as A (3, 4) and B (–1, 2).

Let mid-point of AB be (x, y)

x = (3-1)/2= 2/2 = 1

y = (4+2)/2= 6/2 = 3

(x, y) = (1, 3)

Let the slope of line AB be m1

m1 = (2 – 4)/(-1 – 3)

= -2/(-4)

= 1/2

And let the slope of the line perpendicular to AB be m2

m2 = -1/(1/2)

= -2

The equation of the line passing through (1, 3) and having a slope of –2 is

(y – 3) = -2 (x – 1)

y – 3 = – 2x + 2

2x + y = 5

∴ The required equation of the line is 2x + y = 5

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

Solution:

Let us consider the co-ordinates of the foot of the perpendicular from (-1, 3) to the line 3x – 4y – 16 = 0 be (a, b)

So, let the slope of the line joining (-1, 3) and (a, b) be m1

m1 = (b-3)/(a+1)

And let the slope of the line 3x – 4y – 16 = 0 be m2

y = 3/4x – 4

m2 = 3/4

Since these two lines are perpendicular, m1 × m2 = -1

(b-3)/(a+1) × (3/4) = -1

(3b-9)/(4a+4) = -1

3b – 9 = -4a – 4

4a + 3b = 5 …….(1)

Point (a, b) lies on the line 3x – 4y = 16

3a – 4b = 16 ……..(2)

Solving equations (1) and (2), we get

a = 68/25 and b = -49/25

∴ The co-ordinates of the foot of perpendicular is (68/25, -49/25)

15. 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.

Solution:

Given:

The perpendicular from the origin meets the given line at (–1, 2).

The equation of line is y = mx + c

The line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.

So, the slope of the line joining (0, 0) and (–1, 2) = 2/(-1) = -2

Slope of the given line is m.

m × (-2) = -1

m = 1/2

Since, point (-1, 2) lies on the given line,

y = mx + c

2 = 1/2 × (-1) + c

c = 2 + 1/2 = 5/2

∴ The values of m and c are 1/2 and 5/2 respectively.

16. If p and q are the lengths of perpendiculars from the origin to the lines x cos θ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p2 + 4q2 = k2

Solution:

Given:

The equations of given lines are

x cos θ – y sin θ = k cos 2θ …………………… (1)

x sec θ + y cosec θ = k ……………….… (2)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 41

q = k cos θ sin θ

Multiply both sides by 2, we get

2q = 2k cos θ sin θ = k × 2sin θ cos θ

2q = k sin 2θ

Squaring both sides, we get

4q2 = k2 sin22θ …………………(4)

Now add (3) and (4) we get

p2 + 4q2 = k2 cos2 2θ + k2 sin2 2θ

p2 + 4q2 = k2 (cos2 2θ + sin2 2θ) [Since, cos2 2θ + sin2 2θ = 1]

∴ p2 + 4q2 = k2

Hence proved.

17. 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.

Solution:

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 42

Let AD be the altitude of triangle ABC from vertex A.

So, AD is perpendicular to BC

Given:

Vertices A (2, 3), B (4, –1) and C (1, 2)

Let slope of line BC = m1

m1 = (- 1 – 2)/(4 – 1)

m1 = -1

Let slope of line AD be m2

AD is perpendicular to BC

m1 × m2 = -1

-1 × m2 = -1

m2 = 1

The equation of the line passing through point (2, 3) and having a slope of 1 is

y – 3 = 1 × (x – 2)

y – 3 = x – 2

y – x = 1

Equation of the altitude from vertex A = y – x = 1

Length of AD = Length of the perpendicular from A (2, 3) to BC

Equation of BC is

y + 1 = -1 × (x – 4)

y + 1 = -x + 4

x + y – 3 = 0 …………………(1)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 43

Now compare equation (1) to the general equation of line i.e., Ax + By + C = 0, we get

Length of AD =
NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 44

[where, A = 1, B = 1 and C = -3]

∴ The equation and the length of the altitude from vertex A are y – x = 1 and

√2 units respectively.

18. 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 1/p2 = 1/a2 + 1/b2 

Solution:

Equation of a line whose intercepts on the axes are a and b is x/a + y/b = 1

bx + ay = ab

bx + ay – ab = 0 ………………..(1)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 45

Now square on both the sides we get

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 46

∴ 1/p2 = 1/a2 + 1/b2

Hence proved.

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