Perpendicular Lines and Slopes

Example :

Write the equation of a line that passes through the point $\left(1,3\right)$ and is perpendicular to the line $y=3x+2$ .

Perpendicular lines are lines that intersect at right angles.

The slope of the line with equation $y=3x+2$ is $3$ . If you multiply the slopes of two perpendicular lines, you get $-1$ .

$3\cdot \left(-\frac{1}{3}\right)=-1$

So, the line perpendicular to $y=3x+2$ has the slope $-\frac{1}{3}$ .

Now use the point-slope form to find the equation.

$y-y_1=m\left(x-x_1\right)$

We have to find the equation of the line which has the slope $-\frac{1}{3}$ and passes through the point $\left(1,3\right)$ . So, replace $m$ with $-\frac{1}{3}$ , $x_1$ with $1$ , and $y_1$ with $3$ .

$y-3=-\frac{1}{3}\left(x-1\right)$

Use the distributive property .

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

Add $3$ to each side.

$\begin{array}{l}y-3+3=-\frac{1}{3}x+\frac{1}{3}+3\\ y=-\frac{1}{3}x+\frac{10}{3}\end{array}$

Therefore, the line $y=-\frac{1}{3}x+\frac{10}{3}$ is perpendicular to the line $y=3x+2$ and passes through the point $\left(1,3\right)$ .