Answer:
B. Yes. Their slopes have product -1.
Step-by-step explanation:
Given:
Line passing through P(-3, -2) and Q(2, 3)
Line passing through R(10, -1) and S(15, -6)
Required:
To determine if both lines are perpendicular.
SOLUTION:
Two lines are considered perpendicular if the product of their slopes equal -1.
To determine if both lines given in the question are perpendicular, first, calculate their slope using: [tex] m = \frac{y_2 - y_1}{x_2 - x_1} [/tex]
Slope of line passing through P(-3, -2) and Q(2, 3):
Let,
[tex] P(-3, -2) = (x_1, y_1) [/tex]
[tex] Q(2, 3) = (x_2, y_2) [/tex]
[tex] m = \frac{3 -(-2)}{2 -(-3)} [/tex]
[tex] m = \frac{5}{5} = 1 [/tex]
Slope of line passing through R(10, -1) and S(15, -6):
Let,
[tex] R(10, -1) = (x_1, y_1) [/tex]
[tex] S(15, -6) = (x_2, y_2) [/tex]
[tex] m = \frac{-6 -(-1)}{15 - 10} [/tex]
[tex] m = \frac{-5}{5} = -1 [/tex]
The product of their slopes = 1 × -1 = -1
Therefore, the lines are perpendicular.
The answer is: B. "Yes. Their slopes have product -1."
