# How do you graph the line that passes through (-4,1) perpendicular to the line whose slope is -3/2?

Jul 22, 2017

$2 x - 3 y + 11 = 0.$

#### Explanation:

The product of the slopes of two $\bot$ lines is $- 1.$

Given that, the slope of a line is $- \frac{3}{2} ,$ that of a line $\bot$ to this

reqd. one must be, $\frac{2}{3.}$

The reqd. line passes through the pt. $\left(- 4 , 1\right) .$

Using the Slope-Pt. Form, we find the eqn. of the reqd. line is,

$y - 1 = \frac{2}{3} \left(x - \left(- 4\right)\right) , i . e . , 3 \left(y - 1\right) = 2 \left(x + 4\right) , \mathmr{and} ,$

$2 x - 3 y + 11 = 0.$