Question

Is each line parallel, perpendicular, or neither parallel nor perpendicular to the line −x+4y=20?

Answer 1

we have

$-x+4y=20$

Isolate the variable y

$4y=x+20$

$y=\frac{1}{4}x+5$

the slope m of the given line is

$m=\frac{1}{4}$

we know that

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

so

$m1=m2$

if two lines are perpendicular, then the product of their slopes is equal to minus one

so

$m1*m2=-1$

we will proceed to verify each case to determine the solution

case A) $-x+4y=8$

Isolate the variable y

$4y=x+8$

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

the slope  is

$m2=\frac{1}{4}$

Compare the slope of the line of the case A) with the slope of the given line

$m1=\frac{1}{4}$  -----> slope given line

$m2=\frac{1}{4}$ ----> slope line case A)

$m1=m2$ --------> the lines are parallel

case B) $4x+y=-1$

Isolate the variable y

$y=-4x-1$

the slope  is

$m2=-4$

Compare the slope of the line of the case B) with the slope of the given line

$m1=\frac{1}{4}$  -----> slope given line

$m2=-4$ ----> slope line case B)

$m1*m2=\frac{1}{4}*-4=-1$ --------> the lines are perpendicular

case C) $y=-\frac{1}{4}x+6$

the slope  is

$m2=-\frac{1}{4}$

Compare the slope of the line of the case C) with the slope of the given line

$m1=\frac{1}{4}$  -----> slope given line

$m2=-\frac{1}{4}$ ----> slope line case C)

$m1\neq m2$

$m1*m2\neq-1$

therefore

the line case C) and the given line are neither parallel nor perpendicular

case D) $y=-4x-3$

the slope  is

$m2=-4$

Compare the slope of the line of the case D) with the slope of the given line

$m1=\frac{1}{4}$  -----> slope given line

$m2=-4$ ----> slope line case D)

$m1*m2=\frac{1}{4}*-4=-1$ --------> the lines are perpendicular

the answer in the attached figure

Answer 2

-x+4y=8: Parallel
4x+y=-1: Perpendicular
y=-1/4x+6: neither
y=-4x-3: Perpendicular