Question

Consider the line y=-8x+7

Answer 1

If we have an equation of the form

$y=mx+b$

then the equation of the perpendicular line is given by

$y=-\frac{1}{m}x+c$

Now, keeping that in mind, the equation of the perpendicular line to y =-8x+7 is

$y=\frac{1}{8}x+c$

We find c by using the fact that this perpendicular line passes through (-6, -5)

Putting in x = -6 and y = -5 gives

$-5=\frac{1}{8}(-6)+c$$-5-\frac{3}{4}=c$$c=-\frac{23}{4}$

Hence, the equation of the perpendicular line is

$y=-\frac{1}{8}x-\frac{23}{4}$

The equation of the parallel line is

$y=-8x+b$

We find b using the point (-6, -5). Putting tn x = -6 and y = -5 gives

$-5=-8(-6)+b$$b=43$

Hence, the equation of the parallel line is

$y=-8x+43$