Find a Perpendicular line through a Point Video Demonstration

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.

Video Tutorial

Practice Problems

Problem 1

Write the equation of a line that is perpendicular to $$y = -x$$ at (0, 0).

Step 1

Find negative reciprocal of the slope :

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

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

Step 2

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

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

Step 3

Solve for b

$$0 = 0 + b$$

$$b = 0$$

Step 4

Substitute b = 0 into slope-intercept equation.

$$y = x$$

Problem 2

Write the equation of a line that is drawn from (0, 1) perpendicular to $$y = 5x$$.

Step 1

Find negative reciprocal of the slope :

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

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

Step 2

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

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

Step 3

Solve for b.

$$ 1 = 0 + b$$

$$ b = 1$$

Step 4

Substitute b = 1 into slope-intercept equation.

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

Problem 3

Write the equation of a line that is perpendicular to the line $$y=\frac{1}{4} x-3$$ at the point (-4, -4).

Step 1

Find the negative reciprocal of the slope .

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

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

Step 2

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

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

Step 3

Solve for b.

$$ -4 = 16 + b$$
$$ -16 -16$$
$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$$
$$b = -20$$

Step 4

Substitute b into slope-intercept formula.

$$y = -4x - 20$$

Problem 4

What is the equation of a line that is perpendicular to $$y = -8$$ at (-8, -8)?

Step 1

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

Step 2

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

Step 3

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 equation of a line that is perpendicular to $$y = 4 - 3x$$ at (0, -5).

Step 1

Find the negative reciprocal of the slope .

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

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

Step 2

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

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

Step 3

Solve for b.

$$-5 = 0 + b$$

$$b = -5$$

Step 4

Substitute b into slope-intercept formula.

$$y = 3x - 5$$

Problem 6

Write the equation of a line that is perpendicular to $$x = 32$$ at (-32, 32).

Step 1

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

Step 2

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

$$y = 21$$
$$y = 22$$
$$y = 23$$
$$y = 24$$
$$\dots \dots$$
$$\dots \dots$$
$$\dots \dots$$
$$y = 32$$

Step 3

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

Find the equation of a line that is perpendicular to $$\frac{x+y}{3} =1$$ at (-3, 6).

Step 1

Multiply both sides of the equation by 3.

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

Step 2

Simplify the product on the left.

$$x + y = 3$$

Step 3

Subtract x from each side.

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

Step 4

Cancel out the x' on the left.

$$y = -x + 3$$

Step 5

Find the negative reciprocal of the slope .

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

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

Step 6

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

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

Step 7

Solve for b.

$$6 = -3 + b$$
$$ +3 +3$$
$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$$
$$9 = b$$

Step 8

Substitute b into slope-intercept formula.

$$y = x + 9$$

Problem 8

Find the equation of a line that is perpendicular to $$3x+y-5=0$$ at (0, 5).

Step 1

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

$$3x+y-5-3x+5=-3x+5$$

Step 2

Cancel out -3x and 3x, and -5 and 5 on the left.

$$y=-3x+5$$

Step 3

Find the negative reciprocal of the slope .

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

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

Step 4

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

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

Step 5

Solve for b.

$$5 = 0 + b$$

$$5 = b$$

Step 6

Substitute b into slope-intercept formula.

$$y = 3x + 5$$

Problem 9

Some of the following lines are pairwise perpendicular. Identify two lines that are perpendicular at (-2, 4).

A: $$y = -7x - 10$$

B: $$y=\frac{1}{7} x+12$$

C: $$y=-\frac{1}{7} x+12$$

D: $$y=\frac{1}{7} x+\frac{30}{7} $$

Step 1

The slopes of the lines are as follows:

A: -7

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

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

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

Step 2

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

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

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

Step 3

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

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

$$4 = 14 - 10$$

$$4 = 4$$

Step 4

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

Replace (-2, 4) in B.

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

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

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

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

Step 5

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.

Step 6

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

Problem 10

In order to examine the perpendicularity of two lines, we usually use their slopes; that is, the coefficients of x's in the equations of the lines. If the product of their slopes is -1, then they can be perpendicular. Which pairs of the following lines can be perpendicular?

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

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

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

$$N: y = -4x + 9$$

Step 1

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

Step 2

Equation K: Add 2x to both sides.

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

Step 3

Cancel out the similar terms on the left side.

$$8y = 2x + 11$$

Step 4

Divide both sides by 8.

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

Step 5

Simplify the fractions.

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

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

Step 6

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

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

L: slope is 5.

Step 7

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

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

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

Step 8

Cancel out numbers on the right side.

$$-x - 12 = 4y$$

Step 9

Divide both sides by 4.

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

Step 10

Simplify all the fractions.

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

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

Step 11

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.

Step 12
Now we have four slopes:

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.