NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines Ex 9.2

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines Ex 9.2

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines Ex 9.2 are the part of NCERT Solutions for Class 11 Maths. Here you can find the NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines Ex 9.2.


NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines Ex 9.2


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

Ex 9.2 Class 11 Maths Question 1.

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

Solution:
We know that the y-coordinate 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 x-coordinate 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.

Ex 9.2 Class 11 Maths Question 2.

Passing through the point (-4, 3) with slope ½.

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).


Ex 9.2 Class 11 Maths 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.

Ex 9.2 Class 11 Maths 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)

Ex 9.2 Class 11 Maths 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
Required equation is  y – 0 = -2(x + 3)

 2x + y + 6 = 0

Ex 9.2 Class 11 Maths 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).


Ex 9.2 Class 11 Maths 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 two given points (x1, y1) and (x2, y2) is given by:

Ex 9.2 Class 11 Maths Question 8.

The vertices of ∆PQR are P(2, 1), Q(-2, 3) and R(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 mid-point of PQ.

Ex 9.2 Class 11 Maths Question 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:
Let M(2, 5) and N(-3, 6) be the end points of the given line segment.


Ex 9.2 Class 11 Maths Question 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:
Let A(1, 0) and B(2, 3) be the given points and the point D divides the line segment in the ratio 1 : n.

Ex 9.2 Class 11 Maths Question 11.

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/y/1 
 x + y = 5.

 

Ex 9.2 Class 11 Maths Question 12.

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 a and 9 – a respectively.
Then, its equation is
x/a y/(9a) 1
Since it passes through point (2, 2), we have 2/2/(9a) 1
 2(9 – a) + 2a = a(9 – a)
 18 – 2a + 2a = 9a – a2
 18 = 9a – a2 

⇒ a2 – 9a + 18 = 0
 a2 – 6a – 3a + 18 = 0
 a(a – 6) – 3 (a – 6) = 0

 (a – 3) (a – 6) = 0

 a = 3, 6
Now, if a = 3 
 b = 9 – 3 = 6 and if a = 6  b = 9 – 6 = 3
So, the required equation is:
x/y/or x/y/1
i.e., 2x + y – 6 = 0 or x + 2y – 6 = 0.

 

Ex 9.2 Class 11 Maths Question 13.

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 the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Solution:
Here, m = tan 2π/3 = 3
The equation of the line passing through point (0, 2) is y – 2 = 3(x – 0)
 3x + y – 2 = 0
The slope of line parallel to 3x + y – 2 = 0 is 3.
Since, it passes through (0, -2).
So, the equation of line is y + 2= 3(x – 0)
 3x + y + 2 = 0.

 

Ex 9.2 Class 11 Maths Question 14.

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

Solution:


Ex 9.2 Class 11 Maths Question 15.

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

Ex 9.2 Class 11 Maths Question 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 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.


Ex 9.2 Class 11 Maths Question 17.

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 the line AB makes intercepts c and d on the x-axis and y-axis, respectively.

 

Ex 9.2 Class 11 Maths Question 18.

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 and y-axis, respectively.
Then, the equation of the line AB is x/a y/b 2.

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.

Ex 9.2 Class 11 Maths Question 19.

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.


Related Links:

NCERT Solutions for Maths Class 9

NCERT Solutions for Maths Class 10

NCERT Solutions for Maths Class 12

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