# What is the slope of any line perpendicular to the line passing through (3,13) and (-8,17)?

May 16, 2017

write the equation in the form y=mx + b using the points (3,13) and (-8,17)

Find the slope $\frac{13 - 17}{3 + 8} = - \frac{4}{11}$

Then find the y-intercept, plug in one of the points for (x,y)

$13 = \left(- \frac{4}{11}\right) \cdot \left(3\right) + b$

Simplify

$13 = - \frac{12}{11} + b$

Solve for b, add $\frac{12}{11}$ to both sides to isolate b

$b = 14 \frac{1}{11}$

Then you get the equation

$y = - \frac{4}{11} x + 14 \frac{1}{11}$

To find a PERPENDICULAR equation

The the slope of the perpendicular equation is
Opposite Reciprocal of the original equation

So the original equation had a slope of $- \frac{4}{11}$

Find the opposite reciprocal of that slope to find the slope of the perpendicular equation

The new slope is: $\frac{11}{4}$

Then find b, by plugging in a given point so either (3,13) or (-8,17)

$17 = \left(\frac{11}{4}\right) \cdot \left(- 8\right) + b$

Simplify

$17 = - 22 + b$

Add 22 to both sides to isolate b

$b = 39$

The Perpendicular Equation is: $y = \frac{11}{4} x + 39$