PERPENDICULAR LINES WORKSHEET

Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.

Problem 1 :

The slopes of the two lines are 7 and (3k + 2). If the two lines are perpendicular, find the value of 'k'.

Problem 2 :

The equations of the two perpendicular lines are

3x + 2y - 8  =  0

(5k+3) - 3y + 1  =  0

Find the value of 'k'.

Problem 3 :

Find the equation  of a straight line is passing through (2, 3) and perpendicular to the line 2x - y + 7  =  0. 

Problem 4 :

Verify, whether the following two lines  re perpendicular.

3x - 2y - 7  =  0

y  =  -(2x/3) + 4

Problem 5 :

Verify, whether the following two lines are perpendicular.

5x + 7y - 1  =  0

14x - 10y + 5  =  0

Answers

1. Answer :

If the given two lines are perpendicular, then the product of the slopes is equal to  -1.

7(3k + 2)  =  - 1

Use distributive property. 

21k + 14  =  -1

Subtract 14 from each side. 

21k  =  -15

Divide each side by 21. 

k  =  -15/21

k  =  -5/7

2. Answer :

3x + 2y - 8  =  0

(5k+3) - 3y + 1  =  0

If the two lines are perpendicular, then the coefficient 'y' term in the first line is equal to the coefficient of 'x' term in the second line. 

Then,

5k + 3  =  2

Subtract 3 from each side. 

5k  =  -1

Divide each side by 5.

k  =  -1/5

3. Answer :

Required line is perpendicular to 2x - y + 7  =  0.

Then, the equation of the required line is

x  + 2y + k  =  0 -----(1)

The required line is passing through (2, 3).

Substitute x  =  2 and y  = 3  in  (1).

(1)-----> 2 + 2(3) + k  =  0

2 + 6 + k  =  0

8 + k  =  0

Subtract 8 from each side. 

k  =  -8

4. Answer :

3x - 2y - 7  =  0

y  =  -(2x/3) + 4

In the equations of the given two lines, the equation of the second line is not in general form. 

Let us write the equation of the second line in general form. 

y  =  - (2x/3) + 4

Multiply each side by 3.

3y  =  - 2x + 12

2x + 3y - 12  =  0

Compare the equations of two lines, 

3x - 2y - 7 = 0

2x + 3y - 12  =  0

When we look at the general form of equations of the above two lines, we get the following points. 

(i) The sign of 'y' terms are different.

(ii) The coefficient of 'x' term in the first equation is the coefficient of "y" term in the second equation. 

(iii) The coefficient of 'y' term in the first equation is the coefficient of 'x' term in the second equation. 

(iv)  The above equations differ in constant terms. 

Considering the above points, it is clear that the given two lines are perpendicular. 

5. Answer :

5x + 7y - 1  =  0

14x - 10y + 5  =  0

In the equation of the second line 14x - 10y + 5 = 0, the coefficients of 'x' and 'y' have the common divisor 2. 

Divide the second equation by 2.

(14x/2) - (10y/2) + (5/2)  =  (0/2)

7x - 5y + 2.5  =  0

Compare the equations of two lines, 

5x + 7y - 1  =  0

7x - 5y + 2.5  =  0

When we look at the general form of equations of the above two lines, we get the following points. 

(i) The sign of y- terms are different.

(ii) The coefficient of 'x' term in the first equation is the coefficient of "y" term in the second equation. 

(iii) The coefficient of 'y' term in the first equation is the coefficient of 'x' term in the second equation. 

(iv)  The above equations differ in constant terms. 

Considering the above points, it is clear that the given two lines are perpendicular.

Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

About Us  |  Contact Us  |  Privacy Policy

©All rights reserved. onlinemath4all.com

Recent Articles

  1. 10 Hard SAT Math Questions (Part - 40)

    Dec 23, 25 06:12 AM

    digitalsatmath422.png
    10 Hard SAT Math Questions (Part - 40)

    Read More

  2. AP Calculus BC Problems with Solutions

    Dec 20, 25 10:51 AM

    AP Calculus BC Problems with Solutions

    Read More

  3. AP Precalculus Problems and Solutions (Part - 1)

    Dec 20, 25 10:49 AM

    AP Precalculus Problems and Solutions (Part - 1)

    Read More