NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Straight Lines

The NCERT Solutions for Class 11 Maths Chapter 9, covering Straight Lines, are a helpful resource for students preparing for their Math board exams. The chapter covers important topics like slope, Horizontal and vertical Lines, Point-slope form, Two-point form, Slope-intercept form, Intercept – form, and Normal form. These solutions are designed to help students practice and improve their understanding using straightforward methods and formulas.

NCERT Solutions for Class 11 Maths Chapter 09

EXERCISE 9.2 PAGE NO: 163

In Exercises 1 to 8, find the equation of the line which satisfies the given conditions.

1. Write the equations for the x-and y-axes.

Solution:

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

∴ The equation of the x-axis is y = 0.

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

∴ The equation of the 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) 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) 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), 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 the equation of L is

y = m(x − d).

If the distance is 3 units to the left of the 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 the 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 only if y = mx + c

If the 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. The vertices of ΔPQR are P (2, 1), Q (–2, 3) and R (4, 5). Find the 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

9. 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 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), 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

10. 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 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), 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

11. 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 the x-and the 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

12. Find the 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 the x-and the y-axis, respectively, is x/a + y/b = 1 . … (1)

Given: sum of intercepts = 9

a + b = 9

b = 9 – a

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

x/a + y/(9 – a) = 1

Given: the line passes through 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 factorising, 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

13. Find the equation of the line through the point (0, 2), making an angle 2π/3 with the positive x-axis. Also, find the equation of the 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), 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, the equation of the line parallel to the 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 the line is √3 x + y – 2 = 0, and the line parallel to it is √3 x + y + 2 = 0

14. 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 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) 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 the line is 2x – 9y + 85 = 0

15. The length L (in centimetres) 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

16. 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 the selling price and demand, how many litres could he sell weekly at Rs. 17/litre?

Solution:

Assuming the relationship between the selling price and demand is linear.

Let us assume the 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.

17. P (a, b) is the mid-point of a line segment between axes. Show that the 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 sides with ab, then

Hence, proved.

18. 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 to 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 join 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 sides by 9hk, and we get,

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

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

19. By using the concept of the 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 points (3, 0) and (– 2, – 2) also passes through the point (8, 2).

Hence, proved. The given three points are collinear.