In order to check if the lines are perpendicular, we need to check if their slopes have the following relation (to find the slope we can use the slope-intercept form y = mx + b):
[tex]m_1=-\frac{1}{m_2}[/tex]A.
In this option, y = -5 is an horizontal line and x = 2 is a vertical line, therefore they are perpendicular.
B.
First let's find the slope of each line:
[tex]\begin{gathered} x+\frac{y}{5}=2 \\ 5x+y=10 \\ y=-5x+10\to m=-5 \\ \\ -\frac{x}{5}+y=3 \\ y=\frac{x}{5}+3\to m=\frac{1}{5} \end{gathered}[/tex]These slopes obey the relation stated above, so the lines are perpendicular.
C.
[tex]\begin{gathered} y=\frac{1}{3}x+1\to m=\frac{1}{3} \\ \\ y-1=-3(x-5) \\ y-1=-3x+15 \\ y=-3x+16\to m=-3 \end{gathered}[/tex]These slopes obey the relation stated above, so the lines are perpendicular.
D.
[tex]\begin{gathered} y-5=x+1 \\ y=x+6\to m=1 \\ \\ x+y=3 \\ y=-x+3\to m=-1 \end{gathered}[/tex]These slopes obey the relation stated above, so the lines are perpendicular.
E.
[tex]\begin{gathered} y+2=\frac{1}{3}(x-6) \\ y+2=\frac{1}{3}x-2 \\ y=\frac{1}{3}x-4\to m=\frac{1}{3} \\ \\ y=3x+4\to m=3 \end{gathered}[/tex]These slopes don't obey the relation stated above, so the lines aren't perpendicular.
The correct options are A, B, C and D.
