# How do you write an equation in standard form for a line that passes through (-5, -2) and (1, -6)?

May 10, 2018

$2 x + 3 y = - 16$

#### Explanation:

$\text{the equation of a line in "color(blue)"standard form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{A x + B y = C} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where A is a positive integer and B, C are integers}$

$\text{obtain the equation in "color(blue)"point-slope form}$
$\text{and rearrange into standard form}$

•color(white)(x)y-y_1=m(x-x_1)larrcolor(blue)"point-slope form"

$\text{where m is the slope and "(x_1,y_1)" a point on the line}$

$\text{to calculate m use the "color(blue)"gradient formula}$

•color(white)(x)m=(y_2-y_1)/(x_2-x_1)

$\text{let "(x_1,y_1)=(-5,-2)" and } \left({x}_{2} , {y}_{2}\right) = \left(1 , - 6\right)$

$\Rightarrow m = \frac{- 6 - \left(- 2\right)}{1 - \left(- 5\right)} = \frac{- 4}{6} = - \frac{2}{3}$

$\text{using "m=-2/3" and "(x_1,y_1)=(1,-6)" then}$

$y - \left(- 6\right) = - \frac{2}{3} \left(x - 1\right)$

$\Rightarrow y + 6 = - \frac{2}{3} \left(x - 1\right) \leftarrow \textcolor{m a \ge n t a}{\text{in point-slope form}}$

$\Rightarrow y + 6 = - \frac{2}{3} x + \frac{2}{3}$

$\Rightarrow y = - \frac{2}{3} x + \frac{2}{3} - 6$

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

$\text{multiply all terms by 3}$

$\Rightarrow 3 y = - 2 x - 16$

$\Rightarrow 2 x + 3 y = - 16 \leftarrow \textcolor{m a \ge n t a}{\text{in standard form}}$