Question

the coordinates of a, b, and c in the diagram are a(p,4), b(6,1), and c(9,q). which equation correctly relates p and q? hint: since $overleftrightarrow{ab}$ is perpendicular to $overleftrightarrow{bc}$, the slope of $overleftrightarrow{ab}$×the slope of $overleftrightarrow{bc}=-1$. a. -q - p = 7 b. p - q = 7 c. q - p = 7 d. p + q = 7

Explanation:

Step1: Calculate the slope of $\overrightarrow{AB}$

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For $\overrightarrow{AB}$ with $A(p,4)$ and $B(6,1)$, the slope $m_{AB}=\frac{1 - 4}{6 - p}=\frac{-3}{6 - p}$.

Step2: Calculate the slope of $\overrightarrow{BC}$

For $\overrightarrow{BC}$ with $B(6,1)$ and $C(9,q)$, the slope $m_{BC}=\frac{q - 1}{9 - 6}=\frac{q - 1}{3}$.

Step3: Use the perpendicular - slope relationship

Since the slope of $\overrightarrow{AB}\times$ the slope of $\overrightarrow{BC}=-1$, we have $\frac{-3}{6 - p}\times\frac{q - 1}{3}=-1$.
Simplify the left - hand side: $\frac{-(q - 1)}{6 - p}=-1$.
Cross - multiply to get $-(q - 1)=-(6 - p)$.
Expand: $-q + 1=-6 + p$.
Rearrange the terms: $p - q=7$.

Answer:

B. $p - q = 7$