Question

Consider the line y = -8x+
Find the equation of the line that is perpendicular to this line and passes through the point (-4, 4).
Find the equation of the line that is parallel to this line and passes through the point (-4, 4).
Equation of perpendicular line:
Equation of parallel line:

Answer 1

Answer:

Part 1) The equation of the perpendicular line is  $y=\frac{1}{8}x+\frac{9}{2}$

Part 2) The equation of the parallel line is $y=-8x-28$

Step-by-step explanation:

Part 1) Find the equation of the line that is perpendicular to the given line and passes through the point (-4, 4)

The slope of the given line is $m=-8$

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of the slopes is equal to -1)

so

The slope of the line perpendicular to the given line is

$m=\frac{1}{8}$

Find the equation of the perpendicular line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{1}{8}$

$point\ (-4,4)$

substitute

$y-4=\frac{1}{8}(x+4)$

Convert to slope intercept form

$y=mx+b$

isolate the variable y

apply distributive property right side

$y-4=\frac{1}{8}x+\frac{1}{2}$

$y=\frac{1}{8}x+\frac{1}{2}+4$

$y=\frac{1}{8}x+\frac{9}{2}$

Part 2) Find the equation of the line that is parallel to the given line and passes through the point (-4, 4)

The slope of the given line is $m=-8$

Remember that

If two lines are parallel, then their slopes are the same

so

The slope of the line parallel to the given line is

$m=-8$

Find the equation of the parallel line in point slope form

$y-y1=m(x-x1)$

we have

$m=-8$

$point\ (-4,4)$

substitute

$y-4=-8(x+4)$

Convert to slope intercept form

$y=mx+b$

isolate the variable y

apply distributive property right side

$y-4=-8x-32$

$y=-8x-32+4$

$y=-8x-28$