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

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

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

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