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Chapter 10 – Straight Lines (Ex – 10.3)

Straight Lines (Ex – 10.3)

Question 1.
Reduce the following equations into slope- intercept form and find their slopes and the y-intercepts.
(i) x + 7y = 0,
(ii) 6x + 3y-5 = 0,
(iii) y=0

Solution:

Question 2.
Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0,
(ii) 4x – 3y = 6,
(iii) 3y + 2 = 0

Solution:

Question 3.
Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

Solution:

Question 4.
Find the distance of the point (-1, 1) from the line 12 (x+ 6) = 5(y — 2).

Solution:

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 2

Question 5.
Find the points on the x-axis, whose distances from the line x/3 + y/4 =1 are 4 units.

Solution:

Question 6.
Find the distance between parallel lines
(i) 15x+8y-34 = 0and 15x + 8y+31 =0
(ii) |(x + y) + p = 0 and |(x + y) – r = 0.

Solution:
If lines are Ax + By + Q = 0
and Ax + By + C2 = 0, then distance between

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 3

Question 7.
Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (-2, 3).

Solution:
We have given an equation of line 3x – 4y + 2 = 0
Slop of the line(i) = 3/4
Thus, slope of any line parallel to the given line (i) is 3/4 and passes through (-2, 3), then its equation is

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 4

Question 8.
Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Solution:
Given equation is x – 7y + 5 = 0 … (i)
Slope of this line = 1/7
∴ Slope of any line perpendicular to the line (i) is -7 and passes through (3, 0) then
y – 0 = -7(x – 3)
[∵ Product of slope of perpendicular lines is -1]
⇒ y = -7x + 21
⇒ 7x + y – 21 = 0, is the required equation of line.

Question 9.

Solution:

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 5
NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 6

Question 10.
The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.

Solution:
Given points are (h, 3) and (4,1).
∴ Slope of the line joining (h, 3) & (4,1)

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 7

Question 11.
Prove that the line through the point (x1 y1) and parallel to the line Ax + By + C = 0 is A(x-x1) + B(y-y1) = 0.

Solution:

Question 12.
Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.

Solution:
We have given a point (2, 3), through which two lines are passing and intersects at an angle of 60°.
Let m be the slope of the other line

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 8
NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 9
NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 10

Question 13.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).

Solution:
suppose the given points are A and B.
Let M be the mid point of AB.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 11
NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 12

Question 14.
Find the coordinates of the foot of perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.

Solution:
We have, 3x – 4y – 16 = 0
Slope of the kine(i) = 3/4
Then equation of any line ⊥ from (-1, 3) to the given line(i) is

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 13

Question 15.
The perpendicular from the origin to the line y = mx + c meets it at the point (-1,2). Find the values of m and c.

Solution:

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 14

Given, the perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2)
∴ 2 = m (-1) + c … (i)
⇒ c – m = 2

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 15

Question 16.
If p and q are the lengths of perpendiculars from the origin to the lines x cosθ – y sinθ = k cos 2θ and x secθ + y cosecθ = k, respectively, prove that p2 + 4q2 = k2.

Solution:
Given p and q are the lengths of perpendiculars from the origin to the lines x cos θ – ysinθ=k cos 2θ and xsecθ+y cosec θ = k.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 16

Question 17.
In the triangle ABC with vertices A(2, 3), 8(4, -1) and C( 1, 2), find the equation and length of altitude from the vertex A.

Solution:
We have given a AABC with the vertices, A (2, 3), B (4, -1) and C (1, 2)

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 17
NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 18

Question 18.

Solution:
Given, p be the length of perpendicular from the origin to the line whose intercepts

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 19

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