which line is perpendicular to a line that has a slope of $\frac{1}{2}$?\nline ab\nline cd\nline fg\nline hj
which line is perpendicular to a line that has a slope of $\frac{1}{2}$?\nline ab\nline cd\nline fg\nline hj
Answer
Explanation:
Step1: Recall slope - perpendicularity rule
The product of slopes of two perpendicular lines is - 1. If one line has slope $m_1=\frac{1}{2}$, then the slope of the perpendicular line $m_2$ satisfies $m_1\times m_2=-1$. So, $m_2=-2$.
Step2: 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 $AB$: Let $A=( - 4,-2)$ and $B=(3,2)$. Then $m_{AB}=\frac{2+2}{3 + 4}=\frac{4}{7}$. For line $CD$: Let $C=( - 4,1)$ and $D=(3,-4)$. Then $m_{CD}=\frac{-4 - 1}{3+4}=-\frac{5}{7}$. For line $FG$: Let $F=( - 2,-3)$ and $G=(0,4)$. Then $m_{FG}=\frac{4 + 3}{0+2}=\frac{7}{2}$. For line $HJ$: Let $H=( - 3,3)$ and $J=( - 2,1)$. Then $m_{HJ}=\frac{1 - 3}{-2+3}=-2$.
Answer:
line HJ