Answer: Choice C
No, because the product of the slopes is not -1.
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Explanation:
Let's find the slope of line DE.
[tex]D = (x_1,y_1) = (1,-2) \text{ and } E = (x_2,y_2) = (3,4)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{4 - (-2)}{3 - 1}\\\\m = \frac{4 + 2}{3 - 1}\\\\m = \frac{6}{2}\\\\m = 3\\\\[/tex]
The slope of line DE is 3.
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Use similar steps to find the slope of line FG.
[tex]F = (x_1,y_1) = (-1,2) \text{ and } G = (x_2,y_2) = (4,0)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{0 - 2}{4 - (-1)}\\\\m = \frac{0 - 2}{4 + 1}\\\\m = -\frac{2}{5}\\\\[/tex]
Line FG has a slope of -2/5
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Now multiply the two slopes together
3*(-2/5) = -6/5 = -1.2
The product of the slopes is not -1, so the lines are not perpendicular.
This confirms why choice C is the answer.
