# How do you write an equation of a line passing through (5, 8), perpendicular to  y = 2/3 x - 3?

May 17, 2017

$y = - \frac{3}{2} x + \frac{31}{2}$

#### Explanation:

First find the slope of the perpendicular line:

The slope of a perpendicular line is defined as $m \bot = - \frac{1}{m}$

We know that the slope of the given line is $\frac{2}{3}$ so the slope of the perpendicular line is then:

$- \frac{1}{\frac{2}{3}}$ or $- 1 \cdot \frac{3}{2} = - \frac{3}{2}$

Next, we can use the point slope formula: $y - {y}_{1} = m \left(x - {x}_{1}\right)$ to find the equation of the line by substituting the perpendicular slope and the point $\left(5 , 8\right)$ for $\left({x}_{1} , {y}_{1}\right)$ Thus:

$y - 8 = - \frac{3}{2} \left(x - 5\right)$

We can express this in $y = m x + b$ form is desired by simply solving for $y$:

$y - 8 = - \frac{3}{2} x + \frac{15}{2}$

$y \cancel{- 8 + 8} = - \frac{3}{2} x + \frac{15}{2} + 8$

*Find the LCD for $\frac{15}{2}$ and $8$ which is $2$

$y = - \frac{3}{2} x + \frac{15}{2} + \frac{8}{1} \left(\frac{2}{2}\right)$

$y = - \frac{3}{2} x + \frac{15}{2} + \frac{16}{2}$

$y = - \frac{3}{2} x + \frac{31}{2}$

The graph of the two lines will then look like this: graph{(y-2/3x+3)(y+3/2x-31/2)=0 [-1.92, 18.08, -0.32, 9.68]}