Find the Perpendicular Line Through (-2,-6) and perpendicular to y=1/3x+13

Through $(- 2, - 6)$ and perpendicular to $y = \frac{1}{3} x + 13$

Find the slope when $y = \frac{1}{3} x + 13$ .

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 $x$ .

$y = \frac{x}{3} + 13$

Reorder terms.

$y = \frac{1}{3} x + 13$

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

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

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

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

Simplify $- \frac{1}{\frac{1}{3}}$ to find the slope of the perpendicular line.

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 $3$ .

$m_{perpendicular} = - 3$

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

Use the slope $- 3$ and a given point $(- 2, - 6)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- 6) = - 3 \cdot (x - (- 2))$

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

$y + 6 = - 3 \cdot (x + 2)$

Solve for $y$ .

Simplify $- 3 \cdot (x + 2)$ .

Rewrite.

$y + 6 = 0 + 0 - 3 \cdot (x + 2)$

Simplify by adding zeros.

$y + 6 = - 3 \cdot (x + 2)$

Apply the distributive property.

$y + 6 = - 3 x - 3 \cdot 2$

Multiply $- 3$ by $2$ .

$y + 6 = - 3 x - 6$

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

Subtract $6$ from both sides of the equation.

$y = - 3 x - 6 - 6$

Subtract $6$ from $- 6$ .

$y = - 3 x - 12$

Find the Perpendicular Line Through (-2,-6) and perpendicular to y=1/3x+13

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