Not everyone understands this but if you can help please do mi amor

[tex]D = (x_1,y_1) = (0,-6) \text{ and } E = (x_2, y_2) = (3,1)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(0-3)^2 + (-6-1)^2}\\\\d = \sqrt{(-3)^2 + (-7)^2}\\\\d = \sqrt{9 + 49}\\\\d = \sqrt{58}\\\\d \approx 7.6158\\\\[/tex]

Segment DE is roughly 7.6158 units long.

Let's repeat these steps to find the length of segment EF.

[tex]E = (x_1,y_1) = (3,1) \text{ and } F = (x_2, y_2) = (-2,1)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-(-2))^2 + (1-1)^2}\\\\d = \sqrt{(3+2)^2 + (1-1)^2}\\\\d = \sqrt{(5)^2 + (0)^2}\\\\d = \sqrt{25 + 0}\\\\d = \sqrt{25}\\\\d = 5\\\\[/tex]

Segment EF is exactly 5 units long.

Lastly, we'll compute the length of segment FD.

[tex](x_1,y_1) = (-2,1) \text{ and } (x_2, y_2) = (0,-6)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-2-0)^2 + (1-(-6))^2}\\\\d = \sqrt{(-2-0)^2 + (1+6)^2}\\\\d = \sqrt{(-2)^2 + (7)^2}\\\\d = \sqrt{4 + 49}\\\\d = \sqrt{53}\\\\d \approx 7.2801\\\\[/tex]

Segment FD is about 7.2801 units long.

This is close to the length of DE, but not quite.

To recap, we found these segment lengths:

DE = 7.6158 (approximate)
EF = 5 (exact)
FD = 7.2801 (approximate)

As you can see, all three sides are different lengths. Therefore, triangle DEF is scalene.

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For part (b), follow the same template as part (a).

I'll be skipping the steps.

This is what you should get for the lengths of triangle TRI

segment TR = 6.0828 (approximate)
segment RI = 2 (exact)
segment IT = 6.0828 (approximate)

Segments TR and IT are the same length (both exactly [tex]\sqrt{37}[/tex] units long), which means triangle TRI is isosceles.

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For part (c), I'll also be skipping steps.

Here are the lengths you should find

JK = 4.1231 (approximate)
KL = 8.9443 (approximate)
LJ = 10.6301 (approximate)

All three lengths are different, making this triangle scalene.