### Home > AC > Chapter 11 > Lesson 11.3.1 > Problem11-102

11-102.

Find the equation of the line perpendicular to $5x-2y=13$ that passes through the point $\left(60,-20\right)$.

Solve for $y$ so the equation is in $y = mx + b$ form. $m = \text{ slope (growth)}$

Subtract $5x$ from both sides.

$-2y=-5x+13$

Divide both sides by $-2$.

$y=\frac{-2}{-5}x+\frac{13}{-5}$

$y=\frac{2}{5}x-\frac{13}{5}$

The line perpendicular to the line in the problem has a slope that is the negative reciprocal of the slope of line.

The slope of the line perpendicular to $5x-2y=13$ is $-\frac{5}{2}$

Substitute the point $\left(60,-20\right)$ from the problem into the equation $y=mx+b$ with slope $-\frac{5}{2}$ and solve for $b$.

$y=-\frac{5}{2}x+b$

$-20=-\frac{5}{2}\left(60\right)+b$

$b = 130$

$y=-\frac{5}{2}+130$