Question

1. given p(12, -2), q(5, -10), r(-4, 10), and s(4, y), find the value of y so that \\(\overline{pq} \perp \overline{rs}\\).

Explanation:

Step1: Find slope of \( \overline{PQ} \)

The slope formula is \( m = \frac{y_2 - y_1}{x_2 - x_1} \). For \( P(12, -2) \) and \( Q(5, -10) \),
\( m_{PQ} = \frac{-10 - (-2)}{5 - 12} = \frac{-8}{-7} = \frac{8}{7} \).

Step2: Find slope of \( \overline{RS} \)

For \( R(-4, 10) \) and \( S(4, y) \),
\( m_{RS} = \frac{y - 10}{4 - (-4)} = \frac{y - 10}{8} \).

Step3: Use perpendicular slope property

If two lines are perpendicular, their slopes multiply to \( -1 \):
\( m_{PQ} \cdot m_{RS} = -1 \)
\( \frac{8}{7} \cdot \frac{y - 10}{8} = -1 \).

Step4: Solve for \( y \)

Simplify: \( \frac{y - 10}{7} = -1 \)
Multiply both sides by 7: \( y - 10 = -7 \)
Add 10: \( y = 3 \).

Answer:

\( y = 3 \)