# How do you write the standard form of the equation through: (3,1), perpendicular to y=-2/3x+4?

Nov 6, 2016

$3 x - 2 y = 7$

#### Explanation:

Write the standard form of the line that goes through $\left(3 , 1\right)$ and is perpendicular to $y = - \frac{2}{3} x + 4$.

The equation $y = \textcolor{red}{- \frac{2}{3}} x + 4$ is in slope intercept form
$y = \textcolor{red}{m} x + b$ where $\textcolor{red}{m}$= slope and $b$ = the $y$ intercept.

The slope of this line is then $m = \textcolor{red}{- \frac{2}{3}}$

A perpendicular slope is the opposite sign reciprocal. So, we change the sign of $\textcolor{red}{- \frac{2}{3}}$ and switch the numerator and denominator.

Perpendicular slope $\textcolor{b l u e}{m} = \textcolor{b l u e}{\frac{3}{2}}$

To find the equation of the new line, use the point slope equation
$y - {y}_{1} = m \left(x - {x}_{1}\right)$ where $m =$ slope and $\left({x}_{1} , {y}_{1}\right)$ is a point.

The slope is $\textcolor{b l u e}{\frac{3}{2}}$ and the point is the given point $\left(3 , 1\right)$.

$y - 1 = \textcolor{b l u e}{\frac{3}{2}} \left(x - 3\right) \textcolor{w h i t e}{a a a}$Distribute

$y - 1 = \frac{3}{2} x - \frac{9}{2}$

Standard form is $a x + b y = c$ where $a , b \mathmr{and} c$ are integers and $a$ is positive.

$\textcolor{w h i t e}{a a} 2 \left(y - 1 = \frac{3}{2} x - \frac{9}{2}\right) \textcolor{w h i t e}{a a a}$Multiply the equation by $2$

$\textcolor{w h i t e}{a a a a a} 2 y - 2 = 3 x - 9$
$- 3 x \textcolor{w h i t e}{a a a a a a a} - 3 x \textcolor{w h i t e}{a a a}$Subtract $3 x$ from both sides

$- 3 x + 2 y - 2 = - 9$
$\textcolor{w h i t e}{a a a a a a a a} + 2 \textcolor{w h i t e}{a a a} + 2 \textcolor{w h i t e}{a a a}$Add $2$ to both sides

$- 3 x + 2 y = - 7$

$- 1 \left(- 3 x + 2 y = - 7\right) \textcolor{w h i t e}{a a a}$Multiply the equation by $- 1$

$3 x - 2 y = 7$