RD Sharma Class 11 Solutions Chapter 23 - The Straight Lines (Ex 23.4) Exercise 23.4

The focus of RD Sharma Class 11 chapter 23 is on lines and straight lines. It is explained in a simplified way to ensure students understand it all with ease. We have covered these topics here and ensured the explanations are detailed and simple for students to grasp. With the help of the solutions provided by Vedantu, students will be fully equipped to handle all questions from this chapter. 

About the Topic

In this exercise, we will be looking at the following topics: 

The slope of a line, Horizontal and vertical lines, Point-slope form, Two-point form, Slope-intercept form, Intercept-form, and Normal form. 

Let us give you a brief introduction to a few. 

What is the slope of a line?

A line is said to be sloped if its slope, the ratio of the change in y to the change in x when x and y are varied separately, is nonzero. The slope of a line may be written as either a fraction, a decimal, or a fractional equivalent.

In the first case, the fraction is the change in y divided by the change in x. The expression a/b means that a = number of changes in y, and b = change in x. If the change in y is negative, then a is negative, and the negative is converted to a fraction by multiplying by -1.

The slope of a line may be defined as the ratio of the change in y to the change in x when x and y are varied separately.

To determine whether a line of slope m passes through the origin of the coordinate system is horizontal, vertical, or neither, we need to know the value of m. This is because the x and y-intercepts are expressed as functions of m, and a line is horizontal when its x-axis intercept is zero and its y-axis intercept is not, and a line is vertical when its x-axis intercept is not zero and its y-axis intercept is zero, and it is neither horizontal nor vertical when its x-axis intercept is not zero and its y-axis intercept is zero.

Here is an Example from the exercise : 

Find the equation of the straight line which passes through the point (1, 2) and makes such an angle with the positive direction of the x – axis whose sine is \[\frac{3}{5}\].

Solution:

A line which is passing through (1, 2)

To Find: The equation of a straight line.

By using the formula,

The equation of line is [y – y₁ = m(x – x₁)]

Here, sin θ = \[\frac{3}{5}\]

We know, sin θ = perpendicular/hypotenuse = \[\frac{3}{5}\]

So, according to Pythagoras theorem,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2

(5)2 = (Base)2 + (3)2

(Base) = \[\sqrt{(25-9)}\]

(Base)2 = \[\sqrt{16}\]

Base = 4

Hence, tan θ = \[\frac{Perpendicular}{Base}\]

= \[\frac{3}{4}\]

The slope of the line, m = tan θ

= \[\frac{3}{4}\]

The line passing through (x1,y1) = (1,2)

The required equation of line is y – y₁= m(x – x1)

Now, substitute the values, we get

y – 2 = \[\frac{3}{4}\](x – 1)

4y – 8 = 3x – 3

3x – 4y + 5 = 0

∴ The equation of line is 3x – 4y + 5 = 0

We at Vedantu hope that you make the best of this book and the resources that we provide you. All the best!