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
![m = (y_(2) - y_(1) )/(x_(2) - x_(1))](https://img.qammunity.org/2019/formulas/mathematics/middle-school/dx67684z5suem39010lsk8vvi7sydhkh08.png)
For AB, slope is,
![m = (-1-3)/(-5+1) = (-4)/(-4) =1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/fjq0nxe80vdqocqaobo4ckodj1vzbh38jk.png)
And slope of AC is
![(-1-3)/(3+1) = (-4)/(4) = -1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/gp5mn2ppxvhhlyh0uqpdlmqv8xts2ljz4z.png)
Product of the slopes of AB and AC is
![= 1* (-1) =-1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/gel9pup4qn6ccljtmn22xc02j3hdgdcnf7.png)
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
![d = \sqrt{ (x_(2) - x_(1) )^2 + (y_(2) - y_(1) )^2}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/hm8fgpqr1yhx3adabm1g8o62u1a3lfdb0x.png)
So for AB,
![AB = √( (-3-1)^2 + (-5+1)^2) = √(32) = 4 \sqrt 2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/mxcn2svl6rk5ypbaejjuhekaz586fraou3.png)
For AC,
![AC = √( (-1-3)^2 + (3+1)^2 ) =√(32) = 4 \sqrt 2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/shcpikibzi3tjtx977nx0eff0texng7ske.png)
For BC,
![√((-1+1)^2 + (3+5)^2 ) = 8](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ljtjnj4r723wwc0xwcizhkqksrwkddq0w0.png)
And since two sides are equal, so the triangle is isosceles triangle .