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# NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.3 Straight Lines - PDF Download NCERT solutions for class 11 maths chapter 9 exercise 9.3 Straight Lines explore the role of algebraic equations in the understanding of the trajectory of a straight line. You can have a straight line in a lot of different ways. For example, you can have it parallel to the x or y axis, or it can pass through the origin. There are questions that can be used to understand the relationship between a set of lines. Using the equation of lines, we can figure out whether the lines are parallel to each other or perpendicular to each other. If the lines intersect, what is the point of intersection between them.

## Topics Covered in Exercise 9.3 Class 11 Mathematics Questions

Ex 9.3 class 11 maths ch 9 is based on distance of a point from a line and distance between two parallel lines. These topics are described here by our subject experts of eSaral.

 1 Distance of a Point From a Line 2 Distance between two parallel lines
1. Distance of a Point From a Line

The shortest distance from a point to a line is the distance between two points. It defines the minimum distance (or length) that a point can travel on a line.

We know that the height of a triangle can be found by constructing an altitude. Therefore, we must draw a perpendicular in order to make a right triangle. The longest side of a triangle is called the “hypotenuse”. For example, if we draw a right triangle by drawing a foot of the perpendicular between the point and the line, and any other segment connecting the point to the perpendicular, we will always have a right triangle. The second line segment drawn in this manner will always be the right triangle's hypotenuse.

Perpendicular Distance of a Point From a Line

In this part, you will find out how to mathematically derive the formula for distance of a point from a line.

Here are the steps on how to find the shortest distance between a point and a line.

Step 1- Let’s say we have a line L = ax + by + c = 0, where d is the distance from point p (x1 y1).

Step 2- A perpendicular PM is drawn from point P to point L.

Step 3- Q and R are the points where the line meets with the x-axis and y-axis respectively.

Step 4- You can write the coordinates of the points like Q(-C/A, 0) or R(0,-C/B).

By solving the equations in a detailed manner as described in class 11 maths NCERT textbook chapter 9.

The perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1 , y1 ) is given by

$\mathrm{d}=\frac{\left|A_{X 1}+B_{\gamma_1}+C\right|}{\sqrt{A^2+B^2}}$

1. Distance between two parallel lines

Parallel lines are lines that do not intersect with each other. If the distance between two lines is equal throughout, they can be referred to as parallel lines. The term ‘parallel’ is indicated by the letter "||".

Distance between parallel lines is calculated using the following steps:

Step 1- Make sure the parallel lines you're dealing with have their equations in the slope-intercept form (y =mx + c).

Step 2- The intercepts (C1 and C2) and slope values, which are common for both lines, have to be determined.

Step 3- Once you get the above values, use them in the equation to find y in slope-intercept.

Step 4- Finally, add all these values to the distance formula above to figure out the distance between two parallel lines.

two parallel lines can be taken in the form,

y = mx + C1

y = mx + C2

Line (1) will intersect x-axis at the point  A $\left(-\frac{C_1}{m}, 0\right)$

Distance between two lines is equal to the length of the perpendicular from point A to line (2). . Therefore, distance between the lines (1) and (2) is

$\frac{\left|(-m)\left(-\frac{\varepsilon_1}{m}\right)+\left(-c_2\right)\right|}{\sqrt{1+m^2}}$ or $\mathrm{d}=\left|\frac{c_1-c_2}{\sqrt{1+m^2}}\right|$

Thus, the distance d between two parallel lines y=mx + C1 and y= mx+C2 is given by

$\mathrm{d}=\left|\frac{c_1-c_2}{\sqrt{1+m^2}}\right|$

If lines are given in general form, i.e., Ax + By + C1 = 0 and Ax + By + C2 = 0, then above formula will take the form

$d=\left|\frac{c_1-c_2}{\sqrt{A^2+B^2}}\right|$

## Tips for Solving Exercise 9.3 Class 11 Chapter 9 Straight Lines

NCERT solutions class 11 maths chapter 9 ex 9.3 Straight Lines focuses on the use of the various forms of the slope and the finding of certain aspects associated with them. In order to solve the questions of ex 9.3, you need to follow the tips given by the subject experts of eSaral.

1. To solve ex 9.3, you need to remember the formula for the general form of slopes and know how to use it to find different parameters for a straight line.

2. To solve NCERT solution class 11 maths chapter 9 ex 9.3, you should read the problem and visualize how it would be arranged on the axis.

3. In addition, it can also help with the study process by visualizing what it would look like for various values of these variables while still keeping the emphasis on the unique conditions.

## Importance of Solving Ex 9.3 Class 11 Maths Chapter 9 Straight Lines

NCERT solutions class 11 maths chapter 9 ex 9.3 has combined numerous benefits of solving questions. Here, we have provided some significant benefits to make your learning simple by step by step solutions.

1. Most of the questions of ex 9.3 class 11 maths chapter 9 is based on distance of a point from a line which provides precise and clear understanding of concepts that helps you solve the questions with ease.

2. The questions included in ex 9.3 are a bit lengthy, students must solve NCERT solutions for class 11 maths ex 9.3 provided by expert teachers of eSaral.

3. Ex 9.3 class 11 maths chapter 9 has some useful formulas to solve exercise questions that will help you to solve the problems in a stepwise manner.

4. NCERT solution PDF for ex 9.3 class 11 maths will also help you in finding the accurate answers for all the questions.

$\mathrm{d}=\left|\frac{c_1-c_2}{\sqrt{A^2+B^2}}\right|$
$\mathrm{d}=\frac{\left|A_{X 1}+B_{\gamma 1}+C\right|}{\sqrt{A^2+B^2}}$