Chapter 10 – Straight Lines (Ex – 10.2)

Straight Lines (Ex – 10.2)

In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:

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

Solution:
We know that the ordinate of each point on the x-axis is 0.
If P(x, y) is any point on the x-axis, then y = 0.
∴ Equation of x-axis is y = 0.
Also, we know that the abscissa of each point on the y-axis is 0. If P(x, y) is any point on the y-axis, then x = 0.
∴ Equation of y-axis is x = 0.

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

Solution:
We know that the equation of a line with slope m and passing through the point (x0, y0) is given by (y – y0) = m (x – x0).

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 1

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

Solution:
We know that the equation of a line with slope m and passing through the point (x0, y0) is given by (y – y0) = m(x – x0)
Here, slope = m, x0 = 0, y0 = 0 Required equation is (y – 0) = m(x – 0)
⇒ y = mx.

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

Solution:
We know that the equation of a line with slope m and passing through the point (X0, y0) is given by (y – y0) = m(x – x0)

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 2

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

Solution:
We know that the equation of a line with slope m and passing through the point
(x0, y0) is given by (y – y0) = m(x – x0).
Here, m = – 2, x0 = – 3, y0 = 0
y-0 = -2(x + 3) ⇒ 2x + y + 6 = 0

Question 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.

Solution:
We know that the equation of line with slope m and passing through the point (x0, y0) is given by (y – y0) = m(x – x0)

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 3

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

Solution:
Let the given points be A(-1, 1) and B(2, -4).
We know that the equation of a line passing through the given points (xx, y,) and (x2, y2) is given by

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 4

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

Solution:
Here, we are given that p = 5 and ⍵ = 30°.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 5

Question 9.
The vertices of ∆PQR are P(2, 1), Q(-2, 3) and f(4, 5). Find equation of the median through the vertex R.

Solution:
The vertices of ∆PQR are P( 2, 1), Q(-2, 3) and R(4, 5).
Let S be the midpoint of PQ.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 6

Question 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:
Let M(2, 5) and N(-3, 6) be the end points of the given line segment.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 7

Question 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:
Let A(1, 0) and B( 2, 3) be the given points and D divides the line segment in the ratio 1 : n.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 8
NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 9

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

Solution:
Let the required line make intercepts a on the x-axis and y-axis.
Then its equation is x/a + y/b = 1
⇒ x + y = a … (i)
Since (i) passes through the point (2, 3), we have
2 + 3 = a ⇒ a = 5
So, required equation of the line is
x/5 + y/5 = 1 
⇒ x + y = 5.

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

Solution:
Let the intercepts made by the line on the x-axis and y-axis be o and 9 – a respectively.
Then its equation is
x/a + y/(9−a) = 1
Since it passes through point (2, 2), we have 2/a + 2/(9−a) = 1
⇒ 2(9 – a) + 2a = a(9 – a)
⇒ 18 – 2a + 2a = 9a – 9a2
⇒ 18 = 9a – a2 v a2 – 9a + 18 = 0
⇒ a2 – 6a – 3a + 18 = 0
⇒ a(a – 6) – 3 (a – 6) = 0 ⇒ a = 3, 6
Now, if a = 3 ⇒ b = 9 – 3 = 6 and if a = 6 ⇒ b = 9 – 6 = 3
So, required equation is
x/3 + y/6 = 1 or x/6 + y/3 = 1
i.e., 2x + y – 6 = 0 or x + 2y – 6 = 0.

Question 14.
Find equation of the line through the point (0, 2) making an angle 2π3 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:

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

Solution:

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 10

Question 16.
The length L (in centimeter) 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 C.

Solution:
Assuming L along x-axis and C along y-axis, we have two points (124.942, 20) and (125.134, 110). By two point form, the point (L, C) satisfies the equation

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 11

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

Solution:
Assuming L (liters) along x-axis and R(rupees) along y-axis, we have two points (980,14) and (1220,16).
By two point form, the point (L, R) satisfies the equation.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 12

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

Solution:
Let the line AB makes intercepts c and d on the x-axis and y-axis respectively.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 13

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

Solution:
Let AB be the given line segment making intercepts a and b on the x-axis & y-axis respectively.
Then, the equation of line AB is x/a + y/b = 2

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 14

So, these points are A(a, 0) and B(0, b).
Now, R(h , k) divides the line segment Ab in the ratio 1 : 2.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 15

Question 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:
Let the given points be A(3, 0), B(-2, -2) and C(8, 2). Then the equation of the line passing through A and B is

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 16

Clearly the point C(8, 2) satisfy the equation 2x – 5y – 6 = 0.
(∵ 2(8) – 5(2) – 6 = 16 – 10 – 6 = 0)
Hence, the given points lie on the same straight line whose equation is 2x – 5y – 6 = 0.

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