PROPERTIES OF PARALLEL AND PERPENDICULAR LINES WORKSHEET

Problem 1 :

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

Problem 2 :

If the following equations of two lines are parallel, then find the value of k.

3x + 2y - 8 = 0

(5k + 3)x + 2y + 1 = 0

Problem 3 :

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

Problem 4 :

Verify, whether the following equations of two lines are parallel.

3x + 2y - 7 = 0

y = -1.5x + 4

Problem 5 :

Verify, whether the following equations of two lines are parallel.

5x + 7y - 1 = 0

10x + 14y + 5 = 0

Problem 6 :

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

Problem 7 :

The equations of the two perpendicular lines are

3x + 2y - 8 = 0

(5k + 3) - 3y + 1 = 0

Find the value of k.

Problem 8 :

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

Problem 9 :

Verify, whether the following two lines  re perpendicular.

3x - 2y - 7 = 0

y = -(2x/3) + 4

Problem 10 :

Verify, whether the following two lines are perpendicular.

5x + 7y - 1 = 0

14x - 10y + 5 = 0

1. Answer :

If two lines are parallel, then their slopes are equal.

3k + 2 = 7

Subtract 2 from both sides.

3k = 5

Divide both sides by 5.

k = 5/3

2. Answer :

3x + 2y - 8 = 0

(5k + 3)x + 2y + 1 = 0

If the two lines are parallel, then their general forms of equations will differ only in the constant term and they will have the same coefficients of x and y.

To find the value of k, equate the coefficients of x.

5k + 3 = 3

Subtract 3 from both sides.

5k  =  0

Divide both sides by 5.

k = 0

3. Answer :

Because the required line is parallel to 2x - y + 7 = 0, the equation of the required line and the equation of the given line 2x - y + 7 = 0 will differ only in the constant term.

Then, the equation of the required line is

2x - y + k = 0 ----(1)

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

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

2(2) - 3 + k = 0

4 - 3 + k = 0

1 + k = 0

k = -1

So, the equation of the required line is

(1)----> 2x - y - 1 = 0

4. Answer :

3x + 2y - 7 = 0

y = -1.5x + 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 = -1.5x + 4

1.5x + y - 4 = 0

Multiply by 2 on both sides,

3x + 2y - 8 = 0

Now, let us compare the equations of two lines,

3x + 2y - 7 = 0

3x + 2y - 8 = 0

The above two equations differ only in the constant term. 

So, the equations of the given two lines are parallel.

5. Answer :

5x + 7y - 1 = 0

10x + 14y + 5 = 0

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

So, divide the second equation by 2.

5x + 7y + 2.5 = 0

Now, let us compare the equations of two lines,

5x + 7y - 1 = 0

5x + 7y + 2.5 = 0

The above two equations differ only in the constant term.

So, the equations of the given two lines are parallel.

6. 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

7. Answer :

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.

5k + 3 = 2

Subtract 3 from both sides.

5k = -1

Divide both sides by 5.

k = -1/5

8. 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 both sides. 

k = -8

9. 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.

10. 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.

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.

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