Find the Slope of a Perpendicular Line y=1/3x+9

$y = \frac{1}{3} x + 9$

Rewrite in slope-intercept form.

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Simplify the right side.

Combine $\frac{1}{3}$ and $x$ .

$y = \frac{x}{3} + 9$

Reorder terms.

$y = \frac{1}{3} x + 9$

Using the slope-intercept form, the slope is $\frac{1}{3}$ .

$m = \frac{1}{3}$

The equation of a perpendicular line to $y = \frac{1}{3} x + 9$ must have a slope that is the negative reciprocal of the original slope.

$m_{perpendicular} = - \frac{1}{\frac{1}{3}}$

Simplify the result.

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = - (1 \cdot 3)$

Multiply $3$ by $1$ .

$m_{perpendicular} = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$m_{perpendicular} = - 3$