# What is the equation of the line that passes through (-2,1)  and is perpendicular to the line that passes through the following points: #(-1,11),(13,7)?

Apr 4, 2016

$7 x - 2 y + 16 = 0$

#### Explanation:

Slope of the line that passes through two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by $\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$.

Hence slope of line joining points $\left(- 1 , 11\right)$ and $\left(13 , 7\right)$ is $\frac{7 - 11}{13 - \left(- 1\right)} = - \frac{4}{14} = - \frac{2}{7}$.

Hence line perpendicular to this will have a slope of $- \frac{1}{- \frac{2}{7}} = \frac{7}{2}$

As this line passes through $\left(- 2 , 1\right)$, using point slope equation of line will be

$y - 1 = \frac{7}{2} \left(x - \left(- 2\right)\right)$ or

$2 \left(y - 1\right) = 7 \left(x + 2\right)$ or $2 y - 2 = 7 x + 14$ i.e.

$7 x - 2 y + 16 = 0$