Question

Identify the lines that are perpendicular: A. = −5 and = 2 are perpendicularB. + 5 = 2 and − 5 + = 3 are perpendicularC. = 13 + 1 and − 1 = −3( − 5) are perpendicularD. − 5 = + 1 and + = 3 are perpendicularE. + 2 = 13 ( − 6) and = 3 + 4 are perpendicular

Answer 1

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):

$m_1=-\frac{1}{m_2}$

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:

$\begin{gathered} x+\frac{y}{5}=2 \\ 5x+y=10 \\ y=-5x+10\to m=-5 \\ \\ -\frac{x}{5}+y=3 \\ y=\frac{x}{5}+3\to m=\frac{1}{5} \end{gathered}$

These slopes obey the relation stated above, so the lines are perpendicular.

C.

$\begin{gathered} y=\frac{1}{3}x+1\to m=\frac{1}{3} \\ \\ y-1=-3(x-5) \\ y-1=-3x+15 \\ y=-3x+16\to m=-3 \end{gathered}$

These slopes obey the relation stated above, so the lines are perpendicular.

D.

$\begin{gathered} y-5=x+1 \\ y=x+6\to m=1 \\ \\ x+y=3 \\ y=-x+3\to m=-1 \end{gathered}$

These slopes obey the relation stated above, so the lines are perpendicular.

E.

$\begin{gathered} y+2=\frac{1}{3}(x-6) \\ y+2=\frac{1}{3}x-2 \\ y=\frac{1}{3}x-4\to m=\frac{1}{3} \\ \\ y=3x+4\to m=3 \end{gathered}$

These slopes don't obey the relation stated above, so the lines aren't perpendicular.

The correct options are A, B, C and D.