RD Sharma Class 11 Solutions Chapter 23 Exercise 23.12 (Updated for 2021-22)

RD Sharma Solutions Class 11 Maths Chapter 23 Exercise 23.12: Practicing textbook questions will assist you in determining your level of preparation and understanding of the concept. From the links provided below, you can quickly get the pdf of RD Sharma Solutions Class 11 Maths Chapter 23 Exercise 23.12.

Download RD Sharma Class 11 Solutions Chapter 23 Exercise 23.12 Free PDF

RD Sharma Solutions Class 11 Maths Chapter 23 Exercise 23.12

We will cover the notions of lines parallel and perpendicular to a given line in this part, with examples to help students understand the concepts. Students who want to do well in their board exams should refer to RD Sharma Class 11 Solutions on a daily basis to practise the solutions.

Access answers to RD Sharma Solutions Class 11 Maths Chapter 23 Exercise 23.12- Important Question with Answers

1. Find the equation of a line passing through the point (2, 3) and parallel to the line 3x– 4y + 5 = 0.

Solution:

Given: 

The equation is parallel to 3x − 4y + 5 = 0 and pass-through (2, 3)

The equation of the line parallel to 3x − 4y + 5 = 0 is

3x – 4y + λ = 0,

Where, λ is a constant.

It passes through (2, 3).

Substitute the values in above equation, we get

3 (2) – 4 (3) + λ = 0

6 – 12 + λ = 0

λ = 6

Now, substitute the value of λ = 6 in 3x – 4y + λ = 0, we get

3x − 4y + 6

∴ The required line is 3x − 4y + 6 = 0.

2. Find the equation of a line passing through (3, -2) and perpendicular to the line x – 3y + 5 = 0.

Solution:

Given: 

The equation is perpendicular to x – 3y + 5 = 0 and passes through (3,-2)

The equation of the line perpendicular to x − 3y + 5 = 0 is

3x + y + λ = 0,

Where, λ is a constant.

It passes through (3, − 2).

Substitute the values in above equation, we get

3 (3) + (-2) + λ = 0

9 – 2 + λ = 0

λ = – 7

Now, substitute the value of λ = − 7 in 3x + y + λ = 0, we get

3x + y – 7 = 0

∴ The required line is 3x + y – 7 = 0.

3. Find the equation of the perpendicular bisector of the line joining the points (1, 3) and (3, 1).

Solution:

Given: 

A (1, 3) and B (3, 1) be the points joining the perpendicular bisector

Let C be the midpoint of AB.

So, coordinates of C = [(1+3)/2, (3+1)/2] 

= (2, 2)

Slope of AB = [(1-3) / (3-1)] 

= -1

The slope of the perpendicular bisector of AB = 1

Thus, the equation of the perpendicular bisector of AB is given as,

y – 2 = 1(x – 2)

y = x

x – y = 0

∴ The required equation is y = x.

4. Find the equations of the altitudes of a ΔABC whose vertices are A (1, 4), B (-3, 2), and C (-5, -3).

Solution:

Given: 

The vertices of ∆ABC are A (1, 4), B (− 3, 2), and C (− 5, − 3).

Now let us find the slopes of ∆ABC.

Slope of AB = [(2 – 4) / (-3-1)] 

= ½

Slope of BC = [(-3 – 2) / (-5+3)] 

= 5/2

Slope of CA = [(4 + 3) / (1 + 5)] 

= 7/6 

Thus, we have:

Slope of CF = -2

Slope of AD = -2/5 

Slope of BE = -6/7 

Hence,

Equation of CF is:

y + 3 = -2(x + 5)

y + 3 = -2x – 10

2x + y + 13 = 0

Equation of AD is:

y – 4 = (-2/5) (x – 1)

5y – 20 = -2x + 2

2x + 5y – 22 = 0

Equation of BE is:

y – 2 = (-6/7) (x + 3)

7y – 14 = -6x – 18

6x + 7y + 4 = 0

∴ The required equations are 2x + y + 13 = 0, 2x + 5y – 22 = 0, 6x + 7y + 4 = 0.

5. Find the equation of a line that is perpendicular to the line√3x – y + 5 = 0 and which cuts off an intercept of 4 units with the negative direction of the y-axis.

Solution:

Given: 

The equation is perpendicular to √the 3x – y + 5 = 0 equation and cuts off an intercept of 4 units with the negative direction of the y-axis.

The line perpendicular to √3x – y + 5 = 0 is x + √3y + λ = 0

It is given that the line x + √3y + λ = 0 cuts off an intercept of 4 units with the negative direction of the y-axis.

This means that the line passes through (0,-4).

So,

Let us substitute the values in the equation x + √3y + λ = 0, we get

0 – √3 (4) + λ = 0 

λ = 4√3

Now, substitute the value of λ back, we get

x + √3y + 4√3 = 0

∴ The required equation of line is x + √3y + 4√3 = 0.

Straight lines Ex 23.12 Q6

Straight lines Ex 23.12 Q7

Straight lines Ex 23.12 Q8

Straight lines Ex 23.12 Q9

Straight lines Ex 23.12 Q10

Straight lines Ex 23.12 Q11

Straight lines Ex 23.12 Q12

Straight lines Ex 23.12 Q13

Straight lines Ex 23.12 Q14

Straight lines Ex 23.12 Q15

Straight lines Ex 23.12 Q16

Straight lines Ex 23.12 Q17

Straight lines Ex 23.12 Q18

Straight lines Ex 23.12 Q19

Straight lines Ex 23.12 Q20

Straight lines Ex 23.12 Q21

Straight lines Ex 23.12 Q22

Straight lines Ex 23.12 Q23

Straight lines Ex 23.12 Q24

Straight lines Ex 23.12 Q25

Straight lines Ex 23.12 Q26


Straight lines Ex 23.12 Q27

We have provided complete details of RD Sharma Solutions Class 11 Maths Chapter 23 Exercise 23.12. If you have any queries related to CBSE Class 11, feel free to ask us in the comment section below.

FAQs on RD Sharma Class 11 Solutions Chapter 23 Exercise 23.12