Question

Explain in words how to write an equation that is part one:parallel and then also part two: perpendicular to the equation y= 2/3x-4 passing through the point (-2,-5) write your answer in standard form

Answer 1

You are given the line l with equation $y=\dfrac{2}{3}x-4.$

1. The equation of line that passes through the point (-2,-5) and is parallel to the line l.

Parallel lines have the same slope. So the slope of unknown line is $\dfrac{2}{3}.$

Then the equation is

$y=\dfrac{2}{3}x+a.$

This line passes through point (-2,-5), this means that coordinates of this point satisfy the equation, substitute x=-2 and y=-5 into equation:

$-5=\dfrac{2}{3}\cdot (-2)+a,\\ \\a=-5+\dfrac{4}{3}=\dfrac{-15+4}{3}=-\dfrac{11}{3}.$

Thus, the equation of parallel line is

$y=\dfrac{2}{3}x-\dfrac{11}{3}.$

2. The equation of line that passes through the point (-2,-5) and is perpendicular to the line l.

Perpendicular lines have slopes that satisfy the condition

$m_1\cdot m_2=-1.$

Therefore, the slope of perpendicular line is

$\dfrac{2}{3}\cdot m_2=-1,\\ \\m_2=-\dfrac{3}{2}.$

Then the equation is

$y=-\dfrac{3}{2}x+b.$

This line passes through point (-2,-5), this means that coordinates of this point satisfy the equation, substitute x=-2 and y=-5 into equation:

$-5=-\dfrac{3}{2}\cdot (-2)+b,\\ \\b=-5-3=-8.$

Thus, the equation of perpendicular line is

$y=-\dfrac{3}{2}x-8.$