# Which equation is the equation of a line that passes through (-10. 3) and is perpendicular to y=5x-7?

Jan 8, 2017

$y = - \frac{1}{5} x + 1$

#### Explanation:

I assume there is a typo and the problem should be:

write the equation of a line that passes through $\left(- 10 , 3\right)$ and is perpendicular to $y = 5 x - 7$.

The line $y = 5 x - 7$ is in slope-intercept form $y = m x + b$ where $m$ is the slope. The slope of this line is thus $m = 5$.

Perpendicular lines have slopes which are negative reciprocals. In other words, take the reciprocal of the slope and change the sign.

The negative reciprocal of $5$ is $- \frac{1}{5}$.

To find the equation of a line which passes through $\left(\textcolor{red}{- 10} , \textcolor{red}{3}\right)$ and with a slope of $\textcolor{b l u e}{m} = \textcolor{b l u e}{- \frac{1}{5}}$, use the point- slope formula:

$y - \textcolor{red}{{y}_{1}} = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$ where $\left(\textcolor{red}{{x}_{1}} , \textcolor{red}{{y}_{1}}\right)$ is a point and $\textcolor{b l u e}{m}$ is the slope.

$y - \textcolor{red}{3} = \textcolor{b l u e}{- \frac{1}{5}} \left(x - \textcolor{red}{- 10}\right)$

$y - 3 = - \frac{1}{5} \left(x + 10\right) \textcolor{w h i t e}{a a a}$ Equation in point-slope form

To put the equation in slope-intercept form, distribute the $- \frac{1}{5}$.

$y - 3 = - \frac{1}{5} x - 2$

$y - 3 = - \frac{1}{5} x - 2$
$\textcolor{w h i t e}{a} + 3 \textcolor{w h i t e}{a a a a a a a a} + 3$
$y = - \frac{1}{5} x + 1$