NCERT Solutions Class 11th Maths Chapter – 10 Straight Lines Exercise 10.2

NCERT Solutions Class 11th Maths Chapter – 10 Straight Lines Exercise 10.2

NCERT Solutions Class 11th Maths Chapter – 10 Straight Lines Exercise 10.2

?Chapter – 10?

✍Straight Lines✍

?Exercise 10.2?

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 of ΔPQR 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

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 obtaine 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.