Class 11 Maths Ncert Solutions Chapter 10 Exercise 10.2

To learn the method of solving problems, students should, at first, get an idea of how the problems given in the exercises are solved. To know the solving process of these problems, understanding the concept is a must. Chapter 10 Straight Lines of Class 11 Maths is categorized under the term – I CBSE Syllabus 2021-22. Exercise 10.2 of NCERT Solutions for Class 11 Maths Chapter 10- Straight Lines is based on the following topics:

1. Various Forms of the Equation of a Line
   1. Horizontal and vertical lines
   2. Point-slope form
   3. Two-point form
   4. Slope-intercept form
   5. Intercept – form
   6. Normal form

The students who practise and prepare for the first term exams with the help of NCERT Solutions for Class 11 Maths would be able to understand the concepts in an effective way. This way, scoring high marks in the term – I examination won't be as difficult as it seems to be.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 10- Straight Lines Exercise 10.2

Access Other Exercise Solutions of Class 11 Maths Chapter 10- Straight Lines

Exercise 10.1 Solutions 14 Questions

Exercise 10.3 Solutions 18 Questions

Miscellaneous Exercise On Chapter 10 Solutions 24 Questions

Access Solutions for Class 11 Maths Chapter 10.2 Exercise

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.

Solution:

The y-coordinate of every point on x-axis is 0.

∴ Equation of x-axis is y = 0.

The x-coordinate of every point on y-axis is 0.

∴ Equation of y-axis is y = 0.

2. Passing through the point (– 4, 3) with slope 1/2

Solution:

Given:

Point (-4, 3) and slope, m = 1/2

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

So, y – 3 = 1/2 (x – (-4))

y – 3 = 1/2 (x + 4)

2(y – 3) = x + 4

2y – 6 = x + 4

x + 4 – (2y – 6) = 0

x + 4 – 2y + 6 = 0

x – 2y + 10 = 0

∴ The equation of the line is x – 2y + 10 = 0.

3. Passing through (0, 0) with slope m.

Solution:

Given:

Point (0, 0) and slope, m = m

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

So, y – 0 = m (x – 0)

y = mx

y – mx = 0

∴ The equation of the line is y – mx = 0.

4. Passing through (2, 2√3) and inclined with the x-axis at an angle of 75^o.

Solution:

Given: point (2, 2√3) and θ = 75°

Equation of line: (y – y₁) = m (x – x₁)

where, m = slope of line = tan θ and (x₁, y₁) are the points through which line passes

∴ m = tan 75°

75° = 45° + 30°

Applying the formula:

We know that the point (x, y) lies on the line with slope m through the fixed point (x₁, y1), if and only if, its coordinates satisfy the equation y – y₁ = m (x – x₁)

Then, y – 2√3 = (2 + √3) (x – 2)

y – 2√3 = 2 x – 4 + √3 x – 2 √3

y = 2 x – 4 + √3 x

(2 + √3) x – y – 4 = 0

∴ The equation of the line is (2 + √3) x – y – 4 = 0.

5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.

Solution:

Given:

Slope, m = -2

We know that if a line L with slope m makes x-intercept d, then equation of L is

y = m(x − d).

If the distance is 3 units to the left of origin then d = -3

So, y = (-2) (x – (-3))

y = (-2) (x + 3)

y = -2x – 6

2x + y + 6 = 0

∴ The equation of the line is 2x + y + 6 = 0.

6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30^o with positive direction of the x-axis.

Solution:

Given: θ = 30°

We know that slope, m = tan θ

m = tan30° = (1/√3)

We know that the point (x, y) on the line with slope m and y-intercept c lies on the line if and only if y = mx + c.

If distance is 2 units above the origin, c = +2

So, y = (1/√3)x + 2

y = (x + 2√3) / √3

√3 y = x + 2√3

x – √3 y + 2√3 = 0

∴ The equation of the line is x – √3 y + 2√3 = 0.

7. Passing through the points (–1, 1) and (2, – 4).

Solution:

Given:

Points (-1, 1) and (2, -4)

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

y – 1 = -5/3 (x + 1)

3 (y – 1) = (-5) (x + 1)

3y – 3 = -5x – 5

3y – 3 + 5x + 5 = 0

5x + 3y + 2 = 0

∴ The equation of the line is 5x + 3y + 2 = 0.

8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30^o.

Solution:

Given: p = 5 and ω = 30°

We know that the equation of the line having normal distance p from the origin and angle ω which the normal makes with the positive direction of x-axis is given by x cos ω + y sin ω = p.

Substituting the values in the equation, we get

x cos30° + y sin30° = 5

x(√3 / 2) + y( 1/2 ) = 5

√3 x + y = 5(2) = 10

√3 x + y – 10 = 0

∴ The equation of the line is √3 x + y – 10 = 0.

9. The vertices ofΔPQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.

Solution:

Given:

Vertices of ΔPQR i.e. P (2, 1), Q (-2, 3) and R (4, 5)

Let RL be the median of vertex R.

So, L is a midpoint of PQ.

We know that the midpoint formula is given by
.

∴ L = = (0, 2)

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

y – 5 = -3/-4 (x-4)

(-4) (y – 5) = (-3) (x – 4)

-4y + 20 = -3x + 12

-4y + 20 + 3x – 12 = 0

3x – 4y + 8 = 0

∴ The equation of median through the vertex R is 3x – 4y + 8 = 0.

10. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Solution:

Given:

Points are (2, 5) and (-3, 6).

We know that slope, m = (y₂ – y₁)/(x₂ – x₁)

= (6 – 5)/(-3 – 2)

= 1/-5 = -1/5

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Then, m = (-1/m)

= -1/(-1/5)

= 5

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

Then, y – 5 = 5(x – (-3))

y – 5 = 5x + 15

5x + 15 – y + 5 = 0

5x – y + 20 = 0

∴ The equation of the line is 5x – y + 20 = 0

11. 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.

Solution:

We know that the coordinates of a point dividing the line segment joining the points (x₁, y₁) and (x₂, y₂) internally in the ratio m: n are

We know that slope, m = (y₂ – y₁)/(x₂ – x₁)

= (3 – 0)/(2 – 1)

= 3/1

= 3

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Then, m = (-1/m) = -1/3

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

Here, the point is

3((1 + n) y – 3) = (-(1 + n) x + 2 + n)

3(1 + n) y – 9 = – (1 + n) x + 2 + n

(1 + n) x + 3(1 + n) y – n – 9 – 2 = 0

(1 + n) x + 3(1 + n) y – n – 11 = 0

∴ The equation of the line is (1 + n) x + 3(1 + n) y – n – 11 = 0.

12. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Solution:

Given: the line cuts off equal intercepts on the coordinate axes i.e. a = b.

We know that equation of the line intercepts a and b on x-and y-axis, respectively, which is

x/a + y/b = 1

So, x/a + y/a = 1

x + y = a … (1)

Given: point (2, 3)

2 + 3 = a

a = 5

Substitute value of 'a' in (1), we get

x + y = 5

x + y – 5 = 0

∴ The equation of the line is x + y – 5 = 0.

13. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Solution:

We know that equation of the line making intercepts a and b on x-and y-axis, respectively, is x/a + y/b = 1 . … (1)

Given: sum of intercepts = 9

a + b = 9

b = 9 – a

Now, substitute value of b in the above equation, we get

x/a + y/(9 – a) = 1

Given: the line passes through the point (2, 2),

So, 2/a + 2/(9 – a) = 1

[2(9 – a) + 2a] / a(9 – a) = 1 [18 – 2a + 2a] / a(9 – a) = 1

18/a(9 – a) = 1

18 = a (9 – a)

18 = 9a – a²

a² – 9a + 18 = 0

Upon factorizing, we get

a² – 3a – 6a + 18 = 0

a (a – 3) – 6 (a – 3) = 0

(a – 3) (a – 6) = 0

a = 3 or a = 6

Let us substitute in (1),

Case 1 (a = 3):

Then b = 9 – 3 = 6

x/3 + y/6 = 1

2x + y = 6

2x + y – 6 = 0

Case 2 (a = 6):

Then b = 9 – 6 = 3

x/6 + y/3 = 1

x + 2y = 6

x + 2y – 6 = 0

∴ The equation of the line is 2x + y – 6 = 0 or x + 2y – 6 = 0.

14. Find equation of the line through the point (0, 2) making an angle 2π/3 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.

Solution:

Given:

Point (0, 2) and θ = 2π/3

We know that m = tan θ

m = tan (2π/3) = -√3

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

y – 2 = -√3 (x – 0)

y – 2 = -√3 x

√3 x + y – 2 = 0

Given, equation of line parallel to above obtained equation crosses the y-axis at a distance of 2 units below the origin.

So, the point = (0, -2) and m = -√3

From point slope form equation,

y – (-2) = -√3 (x – 0)

y + 2 = -√3 x

√3 x + y + 2 = 0

∴ The equation of line is √3 x + y – 2 = 0 and the line parallel to it is √3 x + y + 2 = 0.

15. The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

Solution:

Given:

Points are origin (0, 0) and (-2, 9).

We know that slope, m = (y₂ – y₁)/(x₂ – x₁)

= (9 – 0)/(-2-0)

= -9/2

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

m = (-1/m) = -1/(-9/2) = 2/9

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

y – 9 = (2/9) (x – (-2))

9(y – 9) = 2(x + 2)

9y – 81 = 2x + 4

2x + 4 – 9y + 81 = 0

2x – 9y + 85 = 0

∴ The equation of line is 2x – 9y + 85 = 0.

16. 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.

Solution:

Let us assume 'L' along X-axis and 'C' along Y-axis, we have two points (124.942, 20) and (125.134, 110) in XY-plane.

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

17. The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs. 14/litre and 1220 litres of milk each week at Rs. 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs. 17/litre?

Solution:

Assuming the relationship between selling price and demand is linear.

Let us assume selling price per litre along X-axis and demand along Y-axis, we have two points (14, 980) and (16, 1220) in XY-plane.

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

y – 980 = 120 (x – 14)

y = 120 (x – 14) + 980

When x = Rs 17/litre,

y = 120 (17 – 14) + 980

y = 120(3) + 980

y = 360 + 980 = 1340

∴ The owner can sell 1340 litres weekly at Rs. 17/litre.

18. P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is x/a + y/b = 2

Solution:

Let AB be a line segment whose midpoint is P (a, b).

Let the coordinates of A and B be (0, y) and (x, 0) respectively.

a (y – 2b) = -bx

ay – 2ab = -bx

bx + ay = 2ab

Divide both the sides with ab, then

Hence proved.

19. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find the equation of the line.

Solution:

Let us consider, AB be the line segment such that r (h, k) divides it in the ratio 1: 2.

So the coordinates of A and B be (0, y) and (x, 0) respectively.

We know that the coordinates of a point dividing the line segment joins the points (x₁, y₁) and (x₂, y₂) internally in the ratio m: n is

h = 2x/3 and k = y/3

x = 3h/2 and y = 3k

∴ A = (0, 3k) and B = (3h/2, 0)

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

3h(y – 3k) = -6kx

3hy – 9hk = -6kx

6kx + 3hy = 9hk

Let us divide both the sides by 9hk, we get,

2x/3h + y/3k = 1

∴ The equation of the line is given by 2x/3h + y/3k = 1

20. By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear.

Solution:

According to the question,

If we have to prove that the given three points (3, 0), (– 2, – 2) and (8, 2) are collinear, then we have to also prove that the line passing through the points (3, 0) and (– 2, – 2) also passes through the point (8, 2).

By using the formula,

The equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

-5y = -2 (x – 3)

-5y = -2x + 6

2x – 5y = 6

If 2x – 5y = 6 passes through (8, 2),

2x – 5y = 2(8) – 5(2)

= 16 – 10

= 6

= RHS

The line passing through the points (3, 0) and (– 2, – 2) also passes through the point (8, 2).

Hence proved. The given three points are collinear.

Class 11 Maths Ncert Solutions Chapter 10 Exercise 10.2

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