Find the Perpendicular Line The line is perpendicular to 3x-y=8 and goes through (-2,7)

The line is perpendicular to $3 x - y = 8$ and goes through $(- 2, 7)$

Solve $3 x - y = 8$ .

Subtract $3 x$ from both sides of the equation.

$- y = 8 - 3 x$

Divide each term in $- y = 8 - 3 x$ by $- 1$ and simplify.

Divide each term in $- y = 8 - 3 x$ by $- 1$ .

$\frac{- y}{- 1} = \frac{8}{- 1} + \frac{- 3 x}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{8}{- 1} + \frac{- 3 x}{- 1}$

Divide $y$ by $1$ .

$y = \frac{8}{- 1} + \frac{- 3 x}{- 1}$

Simplify the right side.

Simplify each term.

Divide $8$ by $- 1$ .

$y = - 8 + \frac{- 3 x}{- 1}$

Move the negative one from the denominator of $\frac{- 3 x}{- 1}$ .

$y = - 8 - 1 \cdot (- 3 x)$

Rewrite $- 1 \cdot (- 3 x)$ as $- (- 3 x)$ .

$y = - 8 - (- 3 x)$

Multiply $- 3$ by $- 1$ .

$y = - 8 + 3 x$

Find the slope when $y = - 8 + 3 x$ .

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.

$y = m x + b$

Reorder $- 8$ and $3 x$ .

$y = 3 x - 8$

Using the slope-intercept form, the slope is $3$ .

$m = 3$

The equation of a perpendicular line must have a slope that is the negative reciprocal of the original slope.

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

Find the equation of the perpendicular line using the point-slope formula.

Use the slope $- \frac{1}{3}$ and a given point $(- 2, 7)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (7) = - \frac{1}{3} \cdot (x - (- 2))$

Simplify the equation and keep it in point-slope form.

$y - 7 = - \frac{1}{3} \cdot (x + 2)$

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

Solve for $y$ .

Simplify $- \frac{1}{3} \cdot (x + 2)$ .

Rewrite.

$y - 7 = 0 + 0 - \frac{1}{3} \cdot (x + 2)$

Simplify by adding zeros.

$y - 7 = - \frac{1}{3} \cdot (x + 2)$

Apply the distributive property.

$y - 7 = - \frac{1}{3} x - \frac{1}{3} \cdot 2$

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

$y - 7 = - \frac{x}{3} - \frac{1}{3} \cdot 2$

Multiply $- \frac{1}{3} \cdot 2$ .

Multiply $2$ by $- 1$ .

$y - 7 = - \frac{x}{3} - 2 (\frac{1}{3})$

Combine $- 2$ and $\frac{1}{3}$ .

$y - 7 = - \frac{x}{3} + \frac{- 2}{3}$

Move the negative in front of the fraction.

$y - 7 = - \frac{x}{3} - \frac{2}{3}$

Move all terms not containing $y$ to the right side of the equation.

Add $7$ to both sides of the equation.

$y = - \frac{x}{3} - \frac{2}{3} + 7$

To write $7$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = - \frac{x}{3} - \frac{2}{3} + 7 \cdot \frac{3}{3}$

Combine $7$ and $\frac{3}{3}$ .

$y = - \frac{x}{3} - \frac{2}{3} + \frac{7 \cdot 3}{3}$

Combine the numerators over the common denominator.

$y = - \frac{x}{3} + \frac{- 2 + 7 \cdot 3}{3}$

Simplify the numerator.

Multiply $7$ by $3$ .

$y = - \frac{x}{3} + \frac{- 2 + 21}{3}$

Add $- 2$ and $21$ .

$y = - \frac{x}{3} + \frac{19}{3}$

Reorder terms.

$y = - (\frac{1}{3} x) + \frac{19}{3}$

Remove parentheses.

$y = - \frac{1}{3} x + \frac{19}{3}$

Find the Perpendicular Line The line is perpendicular to 3x-y=8 and goes through (-2,7)

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