Students are often asked to find the equation of a line that is perpendicular to another line and that passes through a point. Watch the video tutorial below to understand how to do these problems and, if you want, download this free worksheet if you want some extra practice.

**Practice** Problems

##### Problem 1

Find negative reciprocal of the slope:

Slope = -1 or $$-\frac{1}{1} $$

Negative reciprocal = $$\frac{1}{1} = 1$$

Plug the x and y given in the equation into the slope interecept formula.

$$0 = 1\left(0\right) + b$$

Solve for b.

$$0 = 0 + b$$

$$b = 0$$

Substitute b = 0 into slope-intercept equation.

$$y = x$$

##### Problem 2

Find negative reciprocal of the slope:

Slope = 5 or $$\frac{5}{1} $$

Negative reciprocal = $$-\frac{1}{5} $$

Plug the x and y given in the equation into the slope interecept formula.

$$ 1= -\frac{1}{5} \left(0\right) + b$$

Solve for b.

$$ 1 = 0 + b$$

$$ b = 1$$

Substitute b = 1 into slope-intercept equation.

$$y = -\frac{1}{5} x + 1$$

##### Problem 3

Find the negative reciprocal of the slope.

Slope = $$\frac{1}{4} $$

Negative reciprocal = $$-\frac{4}{1} = -4$$

Plug the x and y given in the question into the slope interecept formula.

$$ -4 = -4(-4) + b$$

Solve for b.

$$ -4 = 16 + b$$

$$ -16 -16$$

$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$$

$$b = -20$$

Substitute b into slope-intercept formula.

$$y = -4x - 20$$

##### Problem 4

Line $$y = -8$$ is parallel to the x-axis. So, the perpendicular line to this line must be parallel to the y-axis.

The equation of the perpendicular line is in the form $$x = n$$.

This line is perpendicular to the first line at $$y = -8$$; therefore, the second line must be perpendicular at a point with the x-coordinate of -8. That is, $$x = -8$$.

##### Problem 5

Find the negative reciprocal of the slope.

Slope = -3 or $$-\frac{3}{1} $$

Negative reciprocal = $$\frac{3}{1} = 3$$

Plug the x and y given in the question into the slope interecept formula.

$$ -5 = 3(0) + b$$

Solve for b.

$$-5 = 0 + b$$

$$b = -5$$

Substitute b into slope-intercept formula.

$$y = 3x - 5$$

### Practice Problems II

##### Problem 6

Line $$x = 32$$ is a vertical line. Any line to be perpendicular to this line must be horizontal.

We can draw infinite number of perpendicular lines to this line. Some of them are, for example.

$$y = 21$$

$$y = 22$$

$$y = 23$$

$$y = 24$$

$$\dots \dots$$

$$\dots \dots$$

$$\dots \dots$$

$$y = 32$$

Out of all the infinite number of perpendicular lines only one passes through (-32, 32).

Therefore, $$y = -32$$ is the only perpendicular line the problem asked for.

##### Problem 7

Multiply both sides of the equation by 3.

$$3\left(\frac{x+y}{3} \right) = 3 $$

Simplify the product on the left.

$$x + y = 3$$

Subtract x from each side.

$$x + y - x = 3 - x$$

Cancel out the x' on the left.

$$y = -x + 3$$

Find the negative reciprocal of the slope.

Slope = -1 or $$-\frac{1}{1} $$

Negative reciprocal = $$\frac{1}{1} =1$$

Plug the x and y given in the question into the slope interecept formula.

$$ 6 = 1(-3) + b$$

Solve for b.

$$6 = -3 + b$$

$$ +3 +3$$

$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$$

$$9 = b$$

Substitute b into slope-intercept formula.

$$y = x + 9$$

##### Problem 8

Find the slope of the given line first. The line is not in slope-Intercept Form. Add -3x + 5 to both sides.

Find the negative reciprocal of the slope.

Slope = -3 or $$-\frac{3}{1} $$

Negative reciprocal = $$\frac{3}{1} =3$$

Plug the x and y given in the question into the slope interecept formula.

$$5 = 3(0) + b$$

Solve for b.

$$5 = 0 + b$$

$$5 = b$$

Substitute b into slope-intercept formula.

$$y = 3x + 5$$

##### Problem 9

The slopes of the lines are as follows:

A: -7

B: $$\frac{1}{7} $$

C: $$-\frac{1}{7} $$

D: $$\frac{1}{7} $$

Only the products of the slopes of the following pairs give -1.

$$(Slope of A)(Slope of B) = -1$$

$$(Slope of A)(Slope of D) = -1$$

Check the given point to see which lines it fits. Replace (-2, 4) in A.

$$4 = -7(-2) - 10$$

$$4 = 14 - 10$$

$$4 = 4$$

So, A is acceptable as one of the perpendicular lines.

Replace (-2, 4) in B.

$$4=\frac{1}{7} (-2)+12$$

$$4=\frac{-2}{7} +12$$

$$4=\frac{-2+84}{7}$$

$$4=\frac{82}{7}$$

Both sides are not equal. Thus, line B is not accepted.

Replace (4, -2) in D.

$$4=\frac{1}{7} \left(-2\right)+\frac{30}{7}$$

$$4=\frac{-2}{7} +\frac{30}{7}$$

$$4=\frac{-2+30}{7}$$

$$4=\frac{28}{7}$$

$$4=4$$

Both sides are equal. So, the given point is on D.

Point (4, -2) fits two perpendicular lines A and D. Therefore, A and D only are perpendicular lines at (2, -4).

##### Problem 10

All the equations are not in the form of $$y = mx + b$$. So, we must change their forms to y = mx + b first.

Equation K: Add 2x to both sides.

$$-2x + 8y + 2x = 11 + 2x$$

Cancel out the similar terms on the left side.

$$8y = 2x + 11$$

Divide both sides by 8.

$$\frac{8y}{8} =\frac{2x}{8} +\frac{11}{8}$$

Simplify the fractions.

$$y=\frac{1}{4} x+\frac{11}{8}$$

Thus, the slope of K is $$\frac{1}{4} $$.

L: $$y = 5x - 6$$

This line is already in Slope-Intercept form. The coefficient of x is 5.

L: slope is 5.

M: $$-x = 4y + 12$$

This line is not in slope-intercept form. Subtract 12 from each side.

$$-x - 12 = 4y + 12 - 12$$

Cancel out numbers on the right side.

$$-x - 12 = 4y$$

Divide both sides by 4.

$$\frac{x}{4} -\frac{12}{4} =\frac{4y}{4}$$

Simplify all the fractions.

$$y=-\frac{1}{4} x-3$$

That is, the slope of M is $$-\frac{1}{4} $$.

N: $$y = -4x + 9$$ is in slope -- intercept form.

The coefficient of x is the slope of the line. Thus, -4 is the slope of N.

Slope of $$K = \frac{1}{4} $$

Slope of $$L = 5$$

Slope of $$M = -\frac{1}{4} $$

Slope of $$N = -4$$

The product of the slopes of K and N is $$\left(\frac{1}{4} \right)\left(-4\right)=-1$$. So K and N are perpendicular.