# How do you find the slope that is perpendicular to the line  x = -5y - 5?

Aug 25, 2017

See a solution process below:

#### Explanation:

First, let's put the equation in standard linear form. The standard form of a linear equation is: $\textcolor{red}{A} x + \textcolor{b l u e}{B} y = \textcolor{g r e e n}{C}$

Where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

The slope of an equation in standard form is: $m = - \frac{\textcolor{red}{A}}{\textcolor{b l u e}{B}}$

$x = - 5 y - 5$

$x + \textcolor{red}{5 y} = - 5 y + \textcolor{red}{5 y} - 5$

$x + 5 y = 0 - 5$

$\textcolor{red}{1} x + \textcolor{b l u e}{5} y = \textcolor{g r e e n}{- 5}$

Substituting for $\textcolor{red}{A}$ and $\textcolor{b l u e}{B}$ gives the slope of this line as:

$m = - \frac{\textcolor{red}{1}}{\textcolor{b l u e}{5}}$

Now, let's call the slope for the line perpendicular to this ${m}_{p}$

The rule of perpendicular slopes is:

${m}_{p} = - \frac{1}{m}$

Substituting the slope we calculated gives:

${m}_{p} = \frac{- 1}{- \frac{1}{5}} = 5$

Aug 25, 2017

Slope of line is $5$

#### Explanation:

The slope of the line $x = - 5 y - 5 \mathmr{and} 5 y = - x - 5 \mathmr{and} y = - \frac{x}{5} - 1$ is

${m}_{1} = - \frac{1}{5}$ [Compare with standard slope intercept form

$y = m x + c$] . We know the product of slopes of perpendicular

lines is $m \cdot {m}_{1} = - 1 \therefore m = - \frac{1}{m} _ 1 = - \frac{1}{- \frac{1}{5}} = 5$

Slope of line perpendicular to the line $x = - 5 y - 5$ is $5$ [Ans]