# How do you write an equation in standard form given point (-1,0) and slope -2/3?

Feb 8, 2017

$2 x + 3 y + 2 = 0$

#### Explanation:

The equation of a line in $\textcolor{b l u e}{\text{standard form}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{a x + b y + c = 0} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where a is a positive integer and b, c are integers.

To begin, express the equation in $\textcolor{b l u e}{\text{point-slope form}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y - {y}_{1} = m \left(x - {x}_{1}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where m represents the slope and $\left({x}_{1} , {y}_{1}\right) \text{ a point on the line}$

$\text{here "m=-2/3" and } \left({x}_{1} , {y}_{1}\right) = \left(- 1 , 0\right)$

substitute values into point-slope equation.

$y - 0 = - \frac{2}{3} \left(x - \left(- 1\right)\right)$

$\Rightarrow y = - \frac{2}{3} \left(x + 1\right) \leftarrow \textcolor{red}{\text{ point-slope form}}$

distributing.

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

add $\frac{2}{3} x \text{ to both sides}$

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

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

$\text{add "2/3" to both sides}$

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

$\Rightarrow \frac{2}{3} x + y + \frac{2}{3} = 0$

multiply ALL terms by 3

$\Rightarrow 2 x + 3 y + 2 = 0 \leftarrow \textcolor{red}{\text{ standard form}}$