Respuesta :
We'll need the slope formula which is
[tex]m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\[/tex]
We subtract the y values together, and divide that over the difference in the x values when subtracted in the same order.
Let's find the slope of line DE
[tex]D = (x_1,y_1) = (6,1) \text{ and } E = (x_2,y_2) = (2,3)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{3 - 1}{2 - 6}\\\\m = \frac{2}{-4}\\\\m = -\frac{1}{2}\\\\[/tex]
The slope of line DE is -1/2.
Next, compute the slope of line EF
[tex]E = (x_1,y_1) = (2,3) \text{ and } F = (x_2,y_2) = (-1,-3)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{-3 - 3}{-1 - 2}\\\\m = \frac{-6}{-3}\\\\m = 2\\\\[/tex]
The slope of line EF is 2.
Lastly, compute the slope of line FD
[tex]F = (x_1,y_1) = (-1,-3) \text{ and } D = (x_2,y_2) = (6,1)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{1 - (-3)}{6 - (-1)}\\\\m = \frac{1 + 3}{6 + 1}\\\\m = \frac{4}{7}\\\\[/tex]
The slope of line FD is 4/7.
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To recap everything so far, we found the following:
- slope of DE = -1/2
- slope of EF = 2
- slope of FD = 4/7
The product of the first two slopes gets us (-1/2)*(2) = -1 showing that DE is perpendicular to EF.
Perpendicular slopes multiply to -1 as long as neither line is vertical nor horizontal.
Since DE is perpendicular to EF, this proves we have a 90 degree angle at point E.
Therefore triangle DEF is a right triangle.
