Ulearngo

Using Slopes to Identify Perpendicular Lines

Using Slopes to Identify Perpendicular Lines

Let’s look at the lines whose equations are \(y=\frac{1}{4}x-1\) and \(y=-4x+2\), shown in the figure below.

The figure shows two lines graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. One line is labeled with the equation y equals negative 4x plus 2 and goes through the points (0,2) and (1, negative 2). The other line is labeled with the equation y equals one fourth x minus 1 and goes through the points (0, negative 1) and (4,0).

These lines lie in the same plane and intersect in right angles. We call these lines perpendicular.

What do you notice about the slopes of these two lines? As we read from left to right, the line \(y=\frac{1}{4}x-1\) rises, so its slope is positive. The line \(y=-4x+2\) drops from left to right, so it has a negative slope. Does it make sense to you that the slopes of two perpendicular lines will have opposite signs?

If we look at the slope of the first line, \({m}_{1}=\frac{1}{4}\), and the slope of the second line, \({m}_{2}=-4\), we can see that they are negative reciprocals of each other. If we multiply them, their product is \(-1.\)

\(\begin{array}{c}{m}_{1}·{m}_{2}\hfill \\ \frac{1}{4}\left(-4\right)\hfill \\ -1\hfill \end{array}\)

This is always true for perpendicular lines and leads us to this definition.

Perpendicular Lines

Perpendicular lines are lines in the same plane that form a right angle.

If \({m}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{m}_{2}\) are the slopes of two perpendicular lines, then:

\({m}_{1}·{m}_{2}=-1\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}{m}_{1}=\frac{-1}{{m}_{2}}\)

Vertical lines and horizontal lines are always perpendicular to each other.

We were able to look at the slope–intercept form of linear equations and determine whether or not the lines were parallel. We can do the same thing for perpendicular lines.

We find the slope–intercept form of the equation, and then see if the slopes are negative reciprocals. If the product of the slopes is \(-1\), the lines are perpendicular. Perpendicular lines may have the same y-intercepts.

Example

Use slopes to determine if the lines, \(y=-5x-4\) and \(x-5y=5\) are perpendicular.

Solution

\(\begin{array}{ccccccccccc}\text{The first equation is in slope–intercept form.}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & -5x-4\hfill & & & & & \\ \text{Solve the second equation for}\phantom{\rule{0.2em}{0ex}}y.\hfill & & & \hfill \phantom{\rule{2em}{0ex}}x-5y& =\hfill & 5\hfill & & & & & \\ & & & \hfill \phantom{\rule{2em}{0ex}}-5y& =\hfill & \text{−}x+5\hfill & & & & & \\ & & & \hfill \phantom{\rule{2em}{0ex}}\frac{-5y}{-5}& =\hfill & \frac{\text{−}x+5}{-5}\hfill & & & & & \\ & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & \frac{1}{5}x-1\hfill & & & & & \\ \text{Identify the slope of each line.}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & -5x-4\hfill & & & & & \hfill y& =\hfill & \frac{1}{5}x-1\hfill \\ & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & mx+b\hfill & & & & & \hfill y& =\hfill & mx+b\hfill \\ & & & \hfill \phantom{\rule{2em}{0ex}}{m}_{1}& =\hfill & -5\hfill & & & & & \hfill {m}_{2}& =\hfill & \frac{1}{5}\hfill \end{array}\)

The slopes are negative reciprocals of each other, so the lines are perpendicular. We check by multiplying the slopes,

\(\begin{array}{c}{m}_{1}·{m}_{2}\hfill \\ \\ -5\left(\frac{1}{5}\right)\hfill \\ -1✓\hfill \end{array}\)

Example

Use slopes to determine if the lines, \(7x+2y=3\) and \(2x+7y=5\) are perpendicular.

Solution

\(\begin{array}{ccccccccccccccc}\text{Solve the equations for}\phantom{\rule{0.2em}{0ex}}y.\hfill & & & & & \hfill \phantom{\rule{2em}{0ex}}7x+2y& =\hfill & 3\hfill & & & & & \hfill 2x+7y& =\hfill & 5\hfill \\ & & & & & \hfill \phantom{\rule{2em}{0ex}}2y& =\hfill & -7x+3\hfill & & & & & \hfill 7y& =\hfill & -2x+5\hfill \\ & & & & & \hfill \phantom{\rule{2em}{0ex}}\frac{2y}{2}& =\hfill & \frac{-7x+3}{2}\hfill & & & & & \hfill \frac{7y}{7}& =\hfill & \frac{-2x+5}{7}\hfill \\ & & & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & -\frac{7}{2}x+\frac{3}{2}\hfill & & & & & \hfill y& =\hfill & -\frac{2}{7}x+\frac{5}{7}\hfill \\ \text{Identify the slope of each line.}\hfill & & & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & mx+b\hfill & & & & & \hfill y& =\hfill & mx+b\hfill \\ & & & & & \hfill \phantom{\rule{2em}{0ex}}{m}_{1}& =\hfill & -\frac{7}{2}\hfill & & & & & \hfill {m}_{2}& =\hfill & -\frac{2}{7}\hfill \end{array}\)

The slopes are reciprocals of each other, but they have the same sign. Since they are not negative reciprocals, the lines are not perpendicular.

Helpful Resource:

Access this online resource for additional instruction and practice with graphs.

This lesson is part of:

Graphs and Equations

View Full Tutorial

Track Your Learning Progress

Sign in to unlock unlimited practice exams, tutorial practice quizzes, personalized weak area practice, AI study assistance with Lexi, and detailed performance analytics.