**RD Sharma Solutions Class 11 Maths Chapter 23 Exercise 23.7: **The solutions are created by Kopykitab experts with the grasping ability of students in mind. Students can use **RD Sharma Solutions Class 11 Maths Chapter 23** Exercise 23.7, which are available in pdf format, to improve their conceptual knowledge. The pdf can be downloaded by students using the links provided below.

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## Download RD Sharma Class 11 Solutions Chapter 23 Exercise 23.7 Free PDF

RD Sharma Solutions Class 11 Maths Chapter 23 Exercise 23.7

In this exercise, we’ll look at one of the several types of equations for a straight line, such as the normal or perpendicular form of a line, using examples to help students understand the concepts. Students can utilise the **RD Sharma Class 11 Solutions** pdf to learn how to solve problems more effectively.

### Access answers to RD Sharma Solutions Class 11 Maths Chapter 23 Exercise 23.7- Important Question with Answers

**1. Find the equation of a line for which**

**(i) p = 5, α = 60°**

**(ii) p = 4, α = 150°**

**Solution:**

**(i) **p = 5, α = 60°

Given:

p = 5, α = 60°

The equation of the line in normal form is given by

Using the formula,

x cos α + y sin α = p

Now, substitute the values, we get

x cos 60° + y sin 60° = 5

x/2 + **√**3y/2 = 5

x + √3y = 10

∴ The equation of line in normal form is x + √3y = 10.

**(ii) **p = 4, α = 150°

Given:

p = 4, α = 150°

The equation of the line in normal form is given by

Using the formula,

x cos α + y sin α = p

Now, substitute the values, we get

x cos 150° + y sin 150° = 4

cos (180° – θ) = – cos θ , sin (180° – θ) = sin θ

x cos(180° – 30°) + y sin(180° – 30°) = 4

– x cos 30° + y sin 30° = 4

–**√**3x/2 + y/2 = 4

**-√**3x + y = 8

∴ The equation of a line in normal form is **-√**3x + y = 8.

**2. Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of the x-axis is 30°.**

**Solution:**

Given:

p = 4, α = 30°

The equation of the line in normal form is given by

Using the formula,

x cos α + y sin α = p

Now, substitute the values, we get

x cos 30° + y sin 30° = 4

x**√**3/2 + y1/2 = 4

√3x + y = 8

∴ The equation of line in normal form is √3x + y = 8.

**3. Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of the x-axis are 15°.**

**Solution:**

**Straight lines Ex 23.7 Q4****Straight lines Ex 23.7 Q5****Straight lines Ex 23.7 Q6****Straight lines Ex 23.7 Q7****Straight lines Ex 23.7 Q8****Straight lines Ex 23.7 Q9****Straight lines Ex 23.7 Q10**

We have provided complete details of RD Sharma Solutions Class 11 Maths Chapter 23 Exercise 23.7. 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 23.7

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The RD Sharma Solutions Class 11 Maths Chapter 23 Exercise 23.7 available on Kopykitab’s website has been created by highly qualified experts to assist students in achieving high scores on the board exam. The solutions are well-organized and logical, giving pupils a clear picture of the most important questions.

### Is RD Sharma Solutions for Class 12 Maths Chapter 23 Exercise 23.7 for free?

Yes, You can get RD Sharma Solutions Class 12 Maths Chapter 23 Exercise 23.7 for free.