Find the Perpendicular Line What is an equation of the line that passes through the point (-6,-1) and is perpendicular to the line 6x-y=7 ?

What is an equation of the line that passes through the point $(- 6, - 1)$ and is perpendicular to the line $6 x - y = 7$ ?

Step 1

Solve $6 x - y = 7$ .

Subtract $6 x$ from both sides of the equation.

$- y = 7 - 6 x$

Divide each term in $- y = 7 - 6 x$ by $- 1$ and simplify.

Divide each term in $- y = 7 - 6 x$ by $- 1$ .

$\frac{- y}{- 1} = \frac{7}{- 1} + \frac{- 6 x}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{7}{- 1} + \frac{- 6 x}{- 1}$

Divide $y$ by $1$ .

$y = \frac{7}{- 1} + \frac{- 6 x}{- 1}$

Simplify the right side.

Simplify each term.

Divide $7$ by $- 1$ .

$y = - 7 + \frac{- 6 x}{- 1}$

Move the negative one from the denominator of $\frac{- 6 x}{- 1}$ .

$y = - 7 - 1 \cdot (- 6 x)$

Rewrite $- 1 \cdot (- 6 x)$ as $- (- 6 x)$ .

$y = - 7 - (- 6 x)$

Multiply $- 6$ by $- 1$ .

$y = - 7 + 6 x$

Step 2

Find the slope when $y = - 7 + 6 x$ .

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 $- 7$ and $6 x$ .

$y = 6 x - 7$

Using the slope-intercept form, the slope is $6$ .

$m = 6$

Step 3

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

$m_{perpendicular} = - \frac{1}{6}$

Step 4

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

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

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

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

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

Step 5

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

Solve for $y$ .

Simplify $- \frac{1}{6} \cdot (x + 6)$ .

Rewrite.

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

Simplify by adding zeros.

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

Apply the distributive property.

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

Combine $x$ and $\frac{1}{6}$ .

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

Cancel the common factor of $6$ .

Move the leading negative in $- \frac{1}{6}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Subtract $1$ from both sides of the equation.

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

Subtract $1$ from $- 1$ .

$y = - \frac{x}{6} - 2$

Reorder terms.

$y = - (\frac{1}{6} x) - 2$

Remove parentheses.

$y = - \frac{1}{6} x - 2$

Step 6

Find the Perpendicular Line What is an equation of the line that passes through the point (-6,-1) and is perpendicular to the line 6x-y=7 ?

Related Questions

Support LCMGCF.com by disabling your adblocker.