This exercise discusses the concepts related to intercepts of a line on the axes and equations of lines parallel to the coordinate axes, with suitable examples for better understanding. The problems present in the RD Sharma textbook are solved by experts at BYJU’S in a simple manner, depending upon the students’ grasping abilities. It is very important for the students to get well-versed in these solutions to get a good score in the Class 11 examination. These solutions will help you understand and master different types of questions on straight lines. The PDF of RD Sharma Class 11 Maths Solutions is provided here. Students can refer to and download the PDF from the given links.
RD Sharma Solutions for Class 11 Maths Exercise 23.2 Chapter 23 – The Straight Lines
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Access answers to RD Sharma Solutions for Class 11 Maths Exercise 23.2 Chapter 23 – The Straight Lines
EXERCISE 23.2 PAGE NO: 23.17
1. Find the equation of the parallel to x–axis and passing through (3, –5).
Solution:
Given: A line which is parallel to the x-axis and passing through (3, –5)
By using the formula,
The equation of line: [y – y1 = m(x – x1)]
We know that the parallel lines have equal slopes
And the slope of the x-axis is always 0
Then
The slope of the line, m = 0
Coordinates of line are (x1, y1) = (3, –5)
The equation of line = y – y1 = m(x – x1)
Now, substitute the values, and we get
y – (– 5) = 0(x – 3)
y + 5 = 0
∴ The equation of the line is y + 5 = 0
2. Find the equation of the line perpendicular to x–axis and having intercept – 2 on x–axis.
Solution:
Given: A line which is perpendicular to the x-axis and has intercept –2
By using the formula,
The equation of line: [y – y1 = m(x – x1)]
We know that the line is perpendicular to the x–axis, then x is 0 and y is –1.
The slope of the line is m = y/x
= -1/0
It is given that x–intercept is –2, so y is 0.
Coordinates of line are (x1, y1) = (–2, 0)
The equation of line = y – y1 = m(x – x1)
Now, substitute the values, and we get
y – 0 = (-1/0) (x – (– 2))
x + 2 = 0
∴ The equation of line is x + 2 = 0
3. Find the equation of the line parallel to x–axis and having intercept – 2 on y – axis.
Solution:
Given: A line which is parallel to the x-axis and has intercept –2 on the y-axis
By using the formula,
The equation of line: [y – y1 = m(x – x1)]
The parallel lines have equal slopes,
And the slope of the x-axis is always 0
Then
The slope of the line, m = 0
It is given that intercept is –2, on y-axis then
Coordinates of line are (x1, y1) = (0, – 2)
The equation of line is y – y1 = m(x – x1)
Now, substitute the values, and we get
y – (– 2) = 0 (x – 0)
y + 2 = 0
∴ The equation of the line is y + 2 = 0
4. Draw the lines x = –3, x = 2, y = –2, y = 3 and write the coordinates of the vertices of the square so formed.
Solution:
Given: x = –3, x = 2, y = –2 and y = 3
∴ The Coordinates of the square are: A(2, 3), B(2, –2), C(–3, 3), and D(–3, –2).
5. Find the equations of the straight lines which pass through (4, 3) and are respectively parallel and perpendicular to the x–axis.
Solution:
Given: A line which is perpendicular and parallel to the x-axis respectively and passing through (4, 3)
By using the formula,
The equation of line: [y – y1 = m(x – x1)]
Let us consider,
Case 1: When the line is parallel to the x-axis
The parallel lines have equal slopes,
And, the slope of the x-axis is always 0, then
The slope of the line, m = 0
Coordinates of line are (x1, y1) = (4, 3)
The equation of line is y – y1 = m(x – x1)
Now substitute the values, and we get
y – (3) = 0(x – 4)
y – 3 = 0
Case 2: When the line is perpendicular to x–axis
The line is perpendicular to the x–axis, then x is 0 and y is – 1.
The slope of the line is m = y/x
= -1/0
Coordinates of line are (x1, y1) = (4, 3)
The equation of line = y – y1 = m(x – x1)
Now substitute the values, and we get
y – 3 = (-1/0) (x – 4)
x = 4
∴ The equation of the line when it is parallel to the x-axis is y = 3, and it is perpendicular is x = 4.
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