Question

which line is perpendicular to a line that has a slope of $-\frac{1}{3}$? \bigcirc line mn \bigcirc line ab \bigcirc line ef \bigcirc line jk

Explanation:

Step1: Recall perpendicular slope rule

The slope of a line perpendicular to a line with slope \( m \) is the negative reciprocal, i.e., \( -\frac{1}{m} \) (or \( \frac{-1}{m} \), equivalent to \( \frac{1}{-m} \)). For a slope of \( -\frac{1}{3} \), the perpendicular slope \( m_{\perp} \) is \( -\frac{1}{-\frac{1}{3}} = 3 \) (since negative reciprocal of \( -\frac{1}{3} \) is \( 3 \), as \( \frac{-1}{-\frac{1}{3}} = 3 \)).

Step2: Analyze each line's slope

• Line MN: Let's find its slope. Points M(-1, 4) and N(2, -5). Slope \( m_{MN} = \frac{-5 - 4}{2 - (-1)} = \frac{-9}{3} = -3 \). Not 3.
• Line AB: Points A(-3, 2) and B(3, 0). Slope \( m_{AB} = \frac{0 - 2}{3 - (-3)} = \frac{-2}{6} = -\frac{1}{3} \). Same slope, parallel.
• Line EF: Points E(0, -3) and F(2, 3). Slope \( m_{EF} = \frac{3 - (-3)}{2 - 0} = \frac{6}{2} = 3 \). This matches the perpendicular slope.
• Line JK: Points J(-4, -4) and K(3, -2). Slope \( m_{JK} = \frac{-2 - (-4)}{3 - (-4)} = \frac{2}{7} \). Not 3.

Answer:

line EF