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 (accompanied by a coordinate grid with points m, n, a, b, e, f, j, k and respective lines)

Explanation:

Step1: Recall perpendicular slope rule

For two perpendicular lines, the product of their slopes is \(-1\). Let the slope of the given line be \(m_1 = -\frac{1}{3}\), and the slope of the perpendicular line be \(m_2\). Then \(m_1\times m_2=-1\).

Step2: Calculate the required slope

Substitute \(m_1 = -\frac{1}{3}\) into the formula: \(-\frac{1}{3}\times m_2=-1\). Solve for \(m_2\): \(m_2 = (-1)\div(-\frac{1}{3}) = 3\).

Step3: Analyze each line's slope

• Line MN: Let's find two points on line MN. From the graph, M is \((-1, 4)\) and N is \((3, -4)\). The slope \(m_{MN}=\frac{-4 - 4}{3 - (-1)}=\frac{-8}{4}=-2\), not 3.
• Line AB: Points A \((-2, 2)\) and B \((3, 0)\). Slope \(m_{AB}=\frac{0 - 2}{3 - (-2)}=\frac{-2}{5}\), not 3.
• Line EF: Points E \((0, -4)\) and F \((2, 2)\). Slope \(m_{EF}=\frac{2 - (-4)}{2 - 0}=\frac{6}{2}=3\), which matches the required slope.
• Line JK: Points J \((-3, -3)\) and K \((3, -2)\). Slope \(m_{JK}=\frac{-2 - (-3)}{3 - (-3)}=\frac{1}{6}\), not 3.

Answer:

line EF