Respuesta :
Answer:
They are not perpendicular because their slopes are not negative reciprocals.
Step-by-step explanation:
Well first we need to find slope.
[tex]\frac{y^2-y^1}{x^2-x^1}[/tex]
Line HJ)
(-4,-2) , (0,4)
y2 is 4 y1 is -2, so 4 - -2 = 6
0 - -4 = 4
6/4 -> 3/2
Due to the point (0,4) having no x value 4 is the y intercept.
Hence, y = 3/2x + 4 is the slope of line HJ
Line FG)
(-4,1) , (0,-2)
y2 is -2 y1 is 1, so -2 - 1 = -3
0- -4 = 4
Because (0,-2) is missing an x value -2 is the y intercept,
Equation: y = -3/4x - 2
They are not perpendicular because their slopes are not negative reciprocals.

The slope of HJ (3/2) and the slope of FG (-3/4) are not negative reciprocal, so, they are not perpendicular. (Option D).
Recall:
- Lines that are parallel will have the same slope.
- Lines that are perpendicular to each other will have slope values that are negative reciprocal of each other.
- Slope (m) = [tex]\frac{y_2- y_1}{x_2 - x_1}[/tex]
Given that lines HJ (blue line) and FG (red line) are on a coordinate plane as shown in the diagram attached below, let's find their slope:
Slope of line HJ:
[tex]Slope (m) = \frac{-2 - 4}{-4 -0} = \frac{-6}{-4} = \frac{3}{2}[/tex]
- Slope of HJ is 3/2
Slope of line FG:
[tex]Slope (m) = \frac{-2 - 1}{0-(-4)} = \frac{-3}{4} = -\frac{3}{4}[/tex]
- Slope of FG is -3/4
Therefore, the slope of HJ (3/2) and the slope of FG (-3/4) are not negative reciprocal, so, they are not perpendicular. (Option D).
Learn more here:
https://brainly.com/question/18975049
