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

$x + 3 y = 6$

Step 1

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 $3$ .

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

Move the negative in front of the fraction.

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

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

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

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

Reorder terms.

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

Remove parentheses.

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

Step 2

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

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

Step 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.

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

Step 4

Simplify the result.

Cancel the common factor of $1$ and $- 1$ .

Rewrite $1$ as $- 1 (- 1)$ .

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

Move the negative in front of the fraction.

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

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = 1 \cdot 3$

Multiply $- - (1 \cdot 3)$ .

Multiply $3$ by $1$ .

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

Multiply $- 1$ by $3$ .

$m_{perpendicular} = 3$

Multiply $- 1$ by $- 3$ .

$m_{perpendicular} = 3$

Step 5

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

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