# How do you write the standard form of a line given (1/2 , -2/3 ); perpendicular to the line 4x − 8y = 1?

Feb 16, 2018

$6 x + 3 y = 1$

#### Explanation:

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

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

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

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

•color(white)(x)y=mx+b

$\text{where m is the slope and b the y-intercept}$

$\text{rearrange "4x-8y=1" into this form}$

$- 8 y = - 4 x + 1 \Rightarrow y = \frac{1}{2} x - \frac{1}{8} \to m = \frac{1}{2}$

$\text{given a line with slope m then the slope of a line}$
$\text{perpendicular to it is }$

•color(white)(x)m_(color(red)"perpendicular")=-1/m

$\Rightarrow {m}_{\textcolor{red}{\text{perpendicular}}} = - \frac{1}{\frac{1}{2}} = - 2$

$\Rightarrow y = - 2 x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$

$\text{to find b substitute "(1/2,-2/3)" into the partial}$
$\text{equation}$

$- \frac{2}{3} = - 1 + b \Rightarrow b = - \frac{2}{3} + 1 = \frac{1}{3}$

$\Rightarrow y = - 2 x + \frac{1}{3} \leftarrow \textcolor{red}{\text{in slope-intercept form}}$

$\text{multiply through by 3 to eliminate fraction}$

$\Rightarrow 3 y = - 6 x + 1$

$\Rightarrow 6 x + 3 y = 1 \leftarrow \textcolor{red}{\text{in standard form}}$