we know that
If two lines are perpendicular then, the product of their slopes is equal to minus one, that means that their slopes are negative reciprocals
so
[tex]m1*m2=-1[/tex]
Step [tex]1[/tex]
Find the slope of the line HJ
[tex]H(-4,-2)\\J(0,4)[/tex]
we know that
the slope between two points is equal to
[tex]m=\frac{(y2-y1)}{(x2-x1)}[/tex]
substitute the values
[tex]m1=\frac{(4+2)}{(0+4)}[/tex]
[tex]m1=\frac{(6)}{(4)}[/tex]
[tex]m1=\frac{3}{2}[/tex]
Step [tex]2[/tex]
Find the slope of the line HJ
[tex]F(-4,1)\\G(0,-2)[/tex]
we know that
the slope between two points is equal to
[tex]m=\frac{(y2-y1)}{(x2-x1)}[/tex]
substitute the values
[tex]m2=\frac{(-2-1)}{(0+4)}[/tex]
[tex]m2=\frac{(-3)}{(4)}[/tex]
[tex]m2=-\frac{3}{4}[/tex]
Step [tex]3[/tex]
Verify if the two lines are perpendicular
[tex]m1=\frac{3}{2}[/tex]
[tex]m2=-\frac{3}{4}[/tex]
Find the product m1 by m2
[tex]\frac{3}{2}*-\frac{3}{4}=-\frac{9}{8}[/tex]
so
[tex]-\frac{9}{8}\neq -1[/tex] --------> the slopes are not negative reciprocals
therefore
the answer is the option
D.They are not perpendicular because their slopes are not negative reciprocals