Answer:
Step-by-step explanation:
we can find the equations for both lines and analyze after
[tex]y = mx + b\\where\\m = \frac{y_2-y_1}{x_2-x_1}\\\\[/tex]
AB:
[tex](12, 3), (-3, 9)\\m = \frac{9-3}{-3-12} = \frac{6}{-15} = \frac{-2}{5}[/tex]
[tex]y = \frac{-2}{5}(x) + b\\3= \frac{-2(12)}{5} + b\\3 = \frac{-24}{5} + b\\b = \frac{39}{5} \\y = \frac{-2}{5}(x) + \frac{39}{5}[/tex]
CD:
[tex](-6, 4) (-2, -6)\\m = \frac{4-(-6)}{-6-(-2)} = \frac{10}{-4} = \frac{-5}{2} \\y = \frac{-5}{2}(x) + b\\4 = \frac{-5(-6)}{2} + b\\4 = 15 + b\\b = -11\\y = \frac{-5}{2}(x) - 11[/tex]
Determining whether two lines are parallel or perpendicular involves looking at their slopes. AB has a slope of -2/5, CD has a slope of -5/2. Since these are not equal (not parallel) and not negative reciprocals of each other (not perpendicular), the lines are neither perpendicular or parallel.
