Question

VOn a coordinate plane, 4 lines are shown. Line A B has points (negative 3, 2) and (3, 0). E F has points (0, negative 3) and (2, 3). Line J K has points (negative 3, negative 4) and (3, negative 2). Line M N has points (negative 1, 4) and (2, negative 5).

Which line is perpendicular to a line that has a slope
of Negative one-third?

line MN
line AB
line EF
line JK

Answer 1

line EF

Explanation:

The slopes of perpendicular lines are negative reciprocals. Their product is -1. The given line has slope -1/3, so the line we are looking for has slope 3.

Slope AB (-3, 2), (3, 0)

m = (2 - 0)/(-3 - 3) = 2/(-6) = -1/3

Slope EF (0, -3), (2, 3)

m = (-3 - 3)/(0 - 2) = -6/(-2) = 3

Slope JK (-3, -4), (3, -2)

m = (-4 - (-2))/(-3 - 3) = -2/(-6) = 1/3

Slope MN (-1, 4), (2, -5)

m = (-5 - 4)/(2 - (-1)) = -9/3 = -3

The only line with slope = 3 is line EF.

Answer: line EF