# What is the equation of the line that passes through (-1,4) and is perpendicular to the line that passes through the following points: (-2,2),(5,-6) ?

Mar 11, 2016

$8 y = 7 x + 39$

#### Explanation:

The slope m, of the line passing through (x1,y1) & (x2,y2) is

$m = \frac{y 2 - y 1}{x 2 - x 1}$
Thus the slope of the line passing through (-2,2) & (5, -6) is
m = (-6 - 2) / ((5 - (-2)) = $- \frac{8}{7}$
Now if the slope of two lines which are perpendicular to each other are m and m', we have the relationship
$m \cdot m ' = - 1$
So, in our problem, the slope, m2, of the first line = $- \frac{1}{- \frac{8}{7}}$
= $\frac{7}{8}$
Let the equation of the line be $y = m 2 x + c$
Here, $m 2 = \frac{7}{8}$
So the equation is $y = \frac{7}{8} x + c$
It passes through the points, $\left(- 1 , 4\right)$
Substituting the x and y values,
$4 = \frac{7}{8} \cdot \left(- 1\right) + c$
or $c = 4 + \frac{7}{8} = \frac{39}{8}$
So the equation is
$y = \frac{7}{8} x + \frac{39}{8}$
or $8 y = 7 x + 39$