we have
[tex]-x+4y=20[/tex]
Isolate the variable y
[tex]4y=x+20[/tex]
[tex]y=\frac{1}{4}x+5[/tex]
the slope m of the given line is
[tex]m=\frac{1}{4}[/tex]
we know that
If two lines are parallel , then their slopes are the same
so
[tex]m1=m2[/tex]
if two lines are perpendicular, then the product of their slopes is equal to minus one
so
[tex]m1*m2=-1[/tex]
we will proceed to verify each case to determine the solution
case A) [tex]-x+4y=8[/tex]
Isolate the variable y
[tex]4y=x+8[/tex]
[tex]y=\frac{1}{4}x+2[/tex]
the slope is
[tex]m2=\frac{1}{4}[/tex]
Compare the slope of the line of the case A) with the slope of the given line
[tex]m1=\frac{1}{4}[/tex] -----> slope given line
[tex]m2=\frac{1}{4}[/tex] ----> slope line case A)
[tex]m1=m2[/tex] --------> the lines are parallel
case B) [tex]4x+y=-1[/tex]
Isolate the variable y
[tex]y=-4x-1[/tex]
the slope is
[tex]m2=-4[/tex]
Compare the slope of the line of the case B) with the slope of the given line
[tex]m1=\frac{1}{4}[/tex] -----> slope given line
[tex]m2=-4[/tex] ----> slope line case B)
[tex]m1*m2=\frac{1}{4}*-4=-1[/tex] --------> the lines are perpendicular
case C) [tex]y=-\frac{1}{4}x+6[/tex]
the slope is
[tex]m2=-\frac{1}{4}[/tex]
Compare the slope of the line of the case C) with the slope of the given line
[tex]m1=\frac{1}{4}[/tex] -----> slope given line
[tex]m2=-\frac{1}{4}[/tex] ----> slope line case C)
[tex]m1\neq m2[/tex]
[tex]m1*m2\neq-1[/tex]
therefore
the line case C) and the given line are neither parallel nor perpendicular
case D) [tex]y=-4x-3[/tex]
the slope is
[tex]m2=-4[/tex]
Compare the slope of the line of the case D) with the slope of the given line
[tex]m1=\frac{1}{4}[/tex] -----> slope given line
[tex]m2=-4[/tex] ----> slope line case D)
[tex]m1*m2=\frac{1}{4}*-4=-1[/tex] --------> the lines are perpendicular
the answer in the attached figure
