Find the Slope of a Perpendicular Line x-3y=-6

$x - 3 y = - 6$

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.

$y = m x + b$

Subtract $x$ from both sides of the equation.

$- 3 y = - 6 - x$

Divide each term in $- 3 y = - 6 - x$ by $- 3$ and simplify.

Divide each term in $- 3 y = - 6 - x$ by $- 3$ .

$\frac{- 3 y}{- 3} = \frac{- 6}{- 3} + \frac{- x}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 y}{- 3} = \frac{- 6}{- 3} + \frac{- x}{- 3}$

Divide $y$ by $1$ .

$y = \frac{- 6}{- 3} + \frac{- x}{- 3}$

Simplify the right side.

Simplify each term.

Divide $- 6$ by $- 3$ .

$y = 2 + \frac{- x}{- 3}$

Dividing two negative values results in a positive value.

$y = 2 + \frac{x}{3}$

Write in $y = m x + b$ form.

Reorder $2$ and $\frac{x}{3}$ .

$y = \frac{x}{3} + 2$

Reorder terms.

$y = \frac{1}{3} x + 2$

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

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

The equation of a perpendicular line to $y = \frac{1}{3} x + 2$ 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 $3$ .

$m_{perpendicular} = - 3$