Question

which line is perpendicular to a line that has a slope of -1/3?
o line mn
o line ab
o line ef
o line jk

Explanation:

Step1: Recall slope - perpendicular relationship

The product of the slopes of two perpendicular lines 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: Solve for $m_2$

Substitute $m_1 =-\frac{1}{3}$ into $m_1\times m_2=-1$. We get $-\frac{1}{3}\times m_2=-1$. Solving for $m_2$ gives $m_2 = 3$.

Step3: Calculate slopes of given lines

For a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$, the slope $m=\frac{y_2 - y_1}{x_2 - x_1}$.
For line $MN$: Let $M(-1,4)$ and $N(1,-5)$. Then $m_{MN}=\frac{-5 - 4}{1+1}=\frac{-9}{2}=-4.5$.
For line $AB$: Let $A(-3,3)$ and $B(5,1)$. Then $m_{AB}=\frac{1 - 3}{5 + 3}=\frac{-2}{8}=-\frac{1}{4}$.
For line $EF$: Let $E(0,-3)$ and $F(2,3)$. Then $m_{EF}=\frac{3+3}{2 - 0}=\frac{6}{2}=3$.
For line $JK$: Let $J(-4,-3)$ and $K(4,-2)$. Then $m_{JK}=\frac{-2 + 3}{4 + 4}=\frac{1}{8}$.

Answer:

line EF