Question

which coordinate for points a and b would help prove that lines ab and ab are perpendicular? a:(p,m) and b:(z,w) a:(p,m) and b:(z, - w) a:(p, - m) and b:(z,w) a:(p, - m) and b:(z, - w)

Explanation:

Step1: Recall slope - formula

The slope of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Recall perpendicular - line slope relationship

Two non - vertical lines with slopes $m_1$ and $m_2$ are perpendicular if $m_1\times m_2=- 1$.

Step3: Calculate slopes for each option

Let's assume the slope of line $AB$ is $m_{AB}=\frac{z - p}{w + m}$.
For option A: If $A'(p,m)$ and $B'(z,w)$, the slope of $A'B'$ is $m_{A'B'}=\frac{w - m}{z - p}$. Then $m_{AB}\times m_{A'B'}
eq - 1$.
For option B: If $A'(p,m)$ and $B'(z,-w)$, the slope of $A'B'$ is $m_{A'B'}=\frac{-w - m}{z - p}$.
\[m_{AB}\times m_{A'B'}=\frac{z - p}{w + m}\times\frac{-w - m}{z - p}=-1\]
For option C: If $A'(p,-m)$ and $B'(z,w)$, the slope of $A'B'$ is $m_{A'B'}=\frac{w + m}{z - p}$. Then $m_{AB}\times m_{A'B'}
eq - 1$.
For option D: If $A'(p,-m)$ and $B'(z,-w)$, the slope of $A'B'$ is $m_{A'B'}=\frac{-w + m}{z - p}$. Then $m_{AB}\times m_{A'B'}
eq - 1$.

Answer:

A':$(p,m)$ and B':$(z,-w)$