Answer:
Perpendicular
Step-by-step explanation:
Given
[tex]A = (-4, 8)[/tex]
[tex]B = (4, 10)[/tex]
[tex]C = (1, 1)[/tex]
[tex]D = (-2, 13)[/tex]
Required
Is AB and CD parallel?
First, we need to calculate the slope (m) of AB and CD
[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
For AB:
[tex]A (x_1,y_1)= (-4, 8)[/tex]
[tex]B(x_2,y_2) = (4, 10)[/tex]
So:
[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
[tex]m = \frac{10 - 8}{4 - (-4)}[/tex]
[tex]m = \frac{10 - 8}{4 +4}[/tex]
[tex]m = \frac{2}{8}[/tex]
[tex]m = \frac{1}{4}[/tex]
For CD:
[tex]C(x_1,y_1) = (1, 1)[/tex]
[tex]D(x_2,y_2) = (-2, 13)[/tex]
So:
[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
[tex]m = \frac{13 - 1}{-2 - 1}[/tex]
[tex]m = \frac{12}{-3}[/tex]
[tex]m = -4[/tex]
For Lines to be parallel, the slope must be equal:
i.e. [tex]m_1 = m_2[/tex]
This condition is not true because:
[tex]\frac{1}{4} \neq -4[/tex]
For Lines to be perpendicular, the slope must be:
[tex]m_1 = -\frac{1}{m_2}[/tex]
This implies:
[tex]\frac{1}{4} = -\frac{1}{-4}[/tex]
[tex]\frac{1}{4} = \frac{-1}{-4}[/tex]
[tex]\frac{1}{4} = \frac{1}{4}[/tex]
Hence, the lines are perpendicular