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

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2.

• Straight Lines Class 11 Ex 10.1
• Straight Lines Class 11 Ex 10.3

Board	CBSE
Textbook	NCERT
Class	Class 11
Subject	Maths
Chapter	Chapter 10
Chapter Name	Straight Lines
Exercise	Ex 10.2
Number of Questions Solved	20
Category	NCERT Solutions

NCERT Solutions for Class 11 Maths Chapter 10 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 .
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).

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)

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)

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

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°.

Question 9.
The vertices of ∆PQR are P(2, 1), Q(-2, 3) and ff(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.

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.

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.

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 + 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 + 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

Since it passes through point (2, 2), we have
⇒ 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

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 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:
Here, m = tan =
The equation of the line passing through point (0, 2) is y -2 = (x – 0)
⇒ + y – 2 = 0
The slope of line parallel to
+ y – 2 = 0 is .
Since, it passes through (0, -2).
So, the equation of line is
y + 2= (x – 0)
⇒ + y + 2 = 0.

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

Question 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 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

Question 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 L (litres) 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.

Question 18.
P(a, b) is the mid-point of a lone segment between axes. Show that equation of the line is .
Solution:
Let the line AB makes intercepts c and d on the x-axis and y-axis respectively.

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

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.

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

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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