Answer to Question #167734 in Algebra for Augustine

Question

How to find the equation of a line which passes through the point of intersection of the of the lines x - 3y =4 and 3x + y =0 and are relatively parallel and perpendicular to the line 3x + 4y =0

Answer 1

Find the point of intersection of the of the lines x - 3y =4 and 3x + y =0 

$y=\dfrac{1}{3}x-\dfrac{4}{3}\ and\ y=-3x$

$\dfrac{1}{3}x-\dfrac{4}{3}=-3x$

$x=\dfrac{2}{5}$

$y=-\dfrac{6}{5}$

$Point\ \big(\dfrac{2}{5}, -\dfrac{6}{5}\big)$

Parallel to the line 3x + 4y =0

$slope=m_1=-\dfrac{3}{4}$

$y-(-\dfrac{6}{5})=-\dfrac{3}{4}(x-\dfrac{2}{5})$

$3x+4y=-\dfrac{18}{5}$

Perpendicular to the line 3x + 4y =0

$slope=m_2=\dfrac{4}{3}$

$y-(-\dfrac{6}{5})=\dfrac{4}{3}(x-\dfrac{2}{5})$

$4x-3y=\dfrac{26}{5}$