Finding a Perpendicular Line Given an Equation and Point

In this section, you will use what you know about writing the equation of a line given the slope and a point to write the equation of perpendicular lines. The graph shows the equation of the line $y=2x-3$. The point $P(-2,1)$ is also plotted. You can use the fact that perpendicular lines have slopes that are opposite reciprocals to find a line that is perpendicular to line $l$ and goes through point $P(-2,1)$.

From the equation, you know the slope of the line is 2. The second line will pass through $P(-2,1)$ and have slope $-\frac{1}{2}$. To graph the line, start at $(-2,1)$ and count out the rise and run. The slope is $\frac{rise}{run}=\frac{-1}{2}$. Count out the rise, $-1$, and run, 2, and plot the point. You can graph the line as shown. Line $n$ is perpendicular to line $l$ and goes through point $(-2,1)$.

To find the equation of a line perpendicular to a line through a given point algebraically, you can use what you know about finding the equation of a line given the slope and a point.

Example

Write an equation of a line perpendicular to $y=2x-3$ that contains the point $(-2,1)$. Write the equation in slope-intercept form.

Step 1 - Find the slope of the given line.

The line is in slope-intercept form $y=2x-3$

$m=2$

Step 2 - Find the slope of the perpendicular line perpendicular lines have the inverse slope

$m_1=-\frac{1}{2}$

Step 3 - Identify the point

The given point is $(-2,1)$

$(x_1,y_1)=(-2,1)$

Step 4 - Substitute values into the point-slope form $y-y_1=m(x-x_1)$

Simplify

$\begin{array}{ccc}\hfill y-y_1& \hfill =\hfill & m(x-x_1)\hfill \\ \hfill y-1& \hfill =\hfill & -\frac{1}{2}(x-(-2))\hfill \\ \hfill y-1& \hfill =\hfill & -\frac{1}{2}(x+2)\hfill \\ \hfill y-1& \hfill =\hfill & -\frac{1}{2}x-1\hfill \end{array}$

Step 5 - Write the equation in slope-intercept form

$y=-\frac{1}{2}x$

Use this table for reference to write an equation of a line perpendicular to a given line: