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In the given triangle , Coordinates of A (-1,3) , B (-5,-1), C (3,-1).

Let's find the slope of AB and AC .

Formula of slope is

[tex] m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}} [/tex]

For AB, slope is,

[tex] m = \frac{-1-3}{-5+1} = \frac{-4}{-4} =1 [/tex]

And slope of AC is

[tex] \frac{-1-3}{3+1} = \frac{-4}{4} = -1 [/tex]

Product of the slopes of AB and AC is

[tex] = 1* (-1) =-1 [/tex]

Since the product of the two slopes is -1, so AB and AC are perpendicular.

And if they are perpendicular, angle BAC is 90 degree, so the triangle is right triangle .

Now we check the length of the sides of the triangle, and for that we use distance formula , which is

[tex] d = \sqrt{ (x_{2} - x_{1} )^2 + (y_{2} - y_{1} )^2} [/tex]

So for AB,

[tex] AB = \sqrt{ (-3-1)^2 + (-5+1)^2} = \sqrt{32} = 4 \sqrt 2 [/tex]

For AC,

[tex] AC = \sqrt{ (-1-3)^2 + (3+1)^2 } =\sqrt{32} = 4 \sqrt 2 [/tex]

For BC,

[tex] \sqrt{(-1+1)^2 + (3+5)^2 } = 8 [/tex]

And since two sides are equal, so the triangle is isosceles triangle .



Answer:

Fx13

Step-by-step explanation:

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