# How do you write an equation of the line in standard form that is Perpendicular to 3y-2x=6 and through (-1,2)?

##### 1 Answer
May 19, 2018

$y = - \frac{3}{2} x + \frac{1}{2}$

#### Explanation:

The first step to solve this is to transform the original equation into $y = m x + b$ form.

$3 y - 2 x = 6$
→Add $2 x$ to both sides

$3 y = 2 x + 6$
→Divide $3$ on both sides

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

A perpendicular line to this equation means that the slope of the new line will be the opposite of the original equation's slope.

For example if the original equation was $y = 2 x + 3$, the slope of the perpendicular line would be $- \frac{1}{2}$.

So for this problem, the slope of perpendicular line will be $- \frac{3}{2}$.

Now that you have the slope, you have to solve for the $b$-value:

$y = - \frac{3}{2} x + b$
$2 = - \frac{3}{2} \left(- 1\right) + b$
$2 = \frac{3}{2} + b$
→Subtract $\frac{3}{2}$ to both sides
$\frac{1}{2} = b$

Now that you have both the $b$ and the $m$-value, you plug them into the $y = m x + b$ form to get your answer.

$y = - \frac{3}{2} x + \frac{1}{2}$