Find the Perpendicular Line y=10x-8 and (-10,-6)

$y = 10 x - 8$ and $(- 10, - 6)$

Use the slope-intercept form to find the slope.

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

$y = m x + b$

Using the slope-intercept form, the slope is $10$ .

$m = 10$

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

$m_{perpendicular} = - \frac{1}{10}$

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

Use the slope $- \frac{1}{10}$ and a given point $(- 10, - 6)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- 6) = - \frac{1}{10} \cdot (x - (- 10))$

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

$y + 6 = - \frac{1}{10} \cdot (x + 10)$

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

Solve for $y$ .

Simplify $- \frac{1}{10} \cdot (x + 10)$ .

Rewrite.

$y + 6 = 0 + 0 - \frac{1}{10} \cdot (x + 10)$

Simplify by adding zeros.

$y + 6 = - \frac{1}{10} \cdot (x + 10)$

Apply the distributive property.

$y + 6 = - \frac{1}{10} x - \frac{1}{10} \cdot 10$

Combine $x$ and $\frac{1}{10}$ .

$y + 6 = - \frac{x}{10} - \frac{1}{10} \cdot 10$

Cancel the common factor of $10$ .

Move the leading negative in $- \frac{1}{10}$ into the numerator.

$y + 6 = - \frac{x}{10} + \frac{- 1}{10} \cdot 10$

Cancel the common factor.

$y + 6 = - \frac{x}{10} + \frac{- 1}{10} \cdot 10$

Rewrite the expression.

$y + 6 = - \frac{x}{10} - 1$

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

Subtract $6$ from both sides of the equation.

$y = - \frac{x}{10} - 1 - 6$

Subtract $6$ from $- 1$ .

$y = - \frac{x}{10} - 7$

Reorder terms.

$y = - (\frac{1}{10} x) - 7$

Remove parentheses.

$y = - \frac{1}{10} x - 7$

Find the Perpendicular Line y=10x-8 and (-10,-6)

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