# 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 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

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

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

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.

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.

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

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

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

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.

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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