Find Equations of Lines Perpendicular to an Axis

Lines that are horizontal or vertical are special for two reasons. First, their equations have just one variable.

• Horizontal line $y=b$; because all the $y$-coordinates are the same and the slope equals zero.
• Vertical line $x=a$; because all the $x$-coordinates are the same and the slope is undefined.

Second, they have a special relationship with the x- and y-axes because they are either parallel or perpendicular to them.

Examine the line that passes through $(0,4)$ and $(3,4)$.

What is the rise? The rise is 0.

What is the run? The run is 3.

What is the slope? $m=\frac{\text{rise}}{\text{run}}=\frac{0}{3}=0$

So, the horizontal line passing through $(0,4)$ and $(3,4)$ has a slope of 0. For all of the points on this line, the $y$-values will always equal 4. So, the equation is $y=4$.

Also notice that it is parallel to the $x$-axis and perpendicular to the $y$-axis.

Now consider the vertical line passing through $(3,2)$ and $(3,0)$.

What is the rise? The rise is 2.

What is the run? The run is 0.

What is the slope? $m=\frac{\text{rise}}{\text{run}}=\frac{2}{0}$ (undefined)

So, the vertical line passing through $(3,2)$ and $(3,0)$ has an undefined slope. And, for all of the points on this line, the $x$-values will always equal 3. So, the equation is $x=3$.

Also notice that it is parallel to the $y$-axis and perpendicular to the $x$-axis.

Example 1

Find the equation of the line parallel to the $x$-axis and passing through $(8,1)$.

Step 1 - Identify the point.

$(8,1)$

Step 2 - Identify the slope of the line.

Since the line is parallel to the $x$-axis, the line is horizontal. This means the slope of the line is 0.

$m=0$

Step 3 - Substitute the values into an equation.

From here, you may know the equation, but if you don’t, then we can use the point-slope equation.

$y-y_1=m(x-x_1)$

$y-1=0(x-8)$

Step 4 - Simplify.

$y-1=0$

$y=1$

Thus, the equation of the line parallel to the $x$-axis and passing through $(8,1)$ is $y=1$.

Example 2

Find the equation of the line perpendicular to the $x$-axis and passing through $(8,1)$.

Step 1 - Identify the point.

$(8,1)$

Step 2 - Identify the slope of the line.

Since the line is perpendicular to the $x$-axis, the line is vertical. This means the slope of the line is undefined!

$m=\text{undefined}$

Step 3 - Substitute the values into an equation.

Since the slope is undefined, we cannot use the point-slope equation. But, we know that if $(8,1)$ is on the line, then another point vertically above this point is $(8,2)$. Another point vertically below this point is $(8,0)$. On a vertical line, the $x$-values do not change.

So, $x=x_1$

$x=8$

Example 3

Find an equation of a line perpendicular to $x=5$ that contains the point $(3,-2)$. Write the equation in slope-intercept form.

Again, since we know one point, the point-slope option seems more promising than the slope-intercept option. We need the slope to use this form, and we know the new line will be perpendicular to $x=5$. This line is vertical, so its perpendicular will be horizontal. This tells us the $m_{⟂}=0$.

Step 1 - Identify the point.

$(3,-2)$

Step 2 - Identify the slope of the perpendicular line.

$m_{⟂}=0$

Step 3 - Substitute the values into an equation.

$\begin{array}{ccc}\hfill y-y_1& \hfill =\hfill & m(x-x_1)\hfill \\ \hfill y-(-2)& \hfill =\hfill & 0(x-3)\hfill \end{array}$

Step 4 - Simplify

$\begin{array}{ccc}\hfill y+2& \hfill =\hfill & 0\hfill \\ \hfill y& \hfill =\hfill & -2\hfill \end{array}$

Example 4

Find an equation of a line that is perpendicular to $y=4$ that contains the point $(2,3)$. Write the equation in slope-intercept form.

The line $y=4$ is a horizontal line. Any line perpendicular to it must be vertical, in the form $x=a$. Since the perpendicular line is vertical and passes through $(2,3)$, every point on it has an $x$-coordinate of 2. The equation of the perpendicular line is $x=2$.