How do you find the general form of the line that passes through A(3, -4); perpendicular to the line 2x - 6y = 11?

Dec 11, 2017

$3 x + y - 5 = 0$

Explanation:

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{A x + B y + C = 0} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

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

•color(white)(x)y=mx+b

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

$\text{rearrange "2x-6y=11" into this form}$

$\Rightarrow 6 y = 2 x - 11$

$\Rightarrow y = \frac{1}{3} x - \frac{11}{6} \leftarrow \textcolor{red}{\text{in slope-intercept form}}$

$\Rightarrow \text{slope } = m = \frac{1}{3}$

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

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

$\Rightarrow {m}_{\textcolor{red}{\text{perpendicular}}} = - \frac{1}{\frac{1}{3}} = - 3$

$\Rightarrow y = - 3 x + b \textcolor{b l u e}{\text{ is the partial equation}}$

$\text{to find b substitute "(3,-4)" into the partial equation}$

$- 4 = - 9 + b \Rightarrow b = 5$

$\Rightarrow y = - 3 x + 5 \leftarrow \textcolor{red}{\text{in slope-intercept form}}$

$\Rightarrow 3 x + y - 5 = 0 \leftarrow \textcolor{red}{\text{in general form}}$