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\\). \\(\circ\\) a. \\(p - q = 7\\) \\(\circ\\) b. \\(q - p = 7\\) \\(\circ\\) c. \\(p + q = 7\\) \\(\circ\\) d. \\(-q - p = 7\\)

Explanation:

Step1: Calculate slope of $\overleftrightarrow{AB}$

Slope formula: $m_{AB}=\frac{y_2-y_1}{x_2-x_1}$
$m_{AB}=\frac{1-4}{6-p}=\frac{-3}{6-p}$

Step2: Calculate slope of $\overleftrightarrow{BC}$

Slope formula: $m_{BC}=\frac{y_3-y_2}{x_3-x_2}$
$m_{BC}=\frac{q-1}{9-6}=\frac{q-1}{3}$

Step3: Multiply slopes, set to -1

Perpendicular slopes product = -1
$\frac{-3}{6-p} \times \frac{q-1}{3} = -1$

Step4: Simplify the equation

Cancel 3, solve for $p,q$:
$\frac{-(q-1)}{6-p} = -1$
$-(q-1) = -1(6-p)$
$-q+1 = -6+p$
$1+6 = p+q$
$p+q=7$

Answer:

C. $p+q=7$