# What is the equation of the line that passes through (-1,1)  and is perpendicular to the line that passes through the following points: (13,1),(-2,3) ?

May 29, 2017

$15 x - 2 y + 17 = 0.$

#### Explanation:

Slope $m '$ of the line through points P(13,1) & Q(-2,3) is,

$m ' = \frac{1 - 3}{13 - \left(- 2\right)} = - \frac{2}{15.}$

So, if the slope of the reqd. line is $m$, then, as the reqd. line is

$\bot$ to the line $P Q , m m ' = - 1 \Rightarrow m = \frac{15}{2.}$

Now, we use the Slope-Point Formula for the reqd. line, known to

be passing through the point $\left(- 1 , 1\right) .$

Thus, the eqn. of the reqd. line, is,

$y - 1 = \frac{15}{2} \left(x - \left(- 1\right)\right) , \mathmr{and} , 2 y - 2 = 15 x + 15.$

$\Rightarrow 15 x - 2 y + 17 = 0.$