Given:
The vertices of ΔWXY are W(-10, 4), X(-3, -1), and Y(-5, 11).
To find:
Which type of triangle is ΔWXY by its sides.
Solution:
Distance formula:
![d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
Using distance formula, we get
![WX=\sqrt{(-3-(-10))^2+(-1-4)^2}](https://tex.z-dn.net/?f=WX%3D%5Csqrt%7B%28-3-%28-10%29%29%5E2%2B%28-1-4%29%5E2%7D)
![WX=\sqrt{(-3+10)^2+(-5)^2}](https://tex.z-dn.net/?f=WX%3D%5Csqrt%7B%28-3%2B10%29%5E2%2B%28-5%29%5E2%7D)
![WX=\sqrt{(7)^2+(-5)^2}](https://tex.z-dn.net/?f=WX%3D%5Csqrt%7B%287%29%5E2%2B%28-5%29%5E2%7D)
![WX=\sqrt49+25}](https://tex.z-dn.net/?f=WX%3D%5Csqrt49%2B25%7D)
![WX=\sqrt{74}](https://tex.z-dn.net/?f=WX%3D%5Csqrt%7B74%7D)
Similarly,
![XY=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(11-\left(-1\right)\right)^2}=2\sqrt{37}](https://tex.z-dn.net/?f=XY%3D%5Csqrt%7B%5Cleft%28-5-%5Cleft%28-3%5Cright%29%5Cright%29%5E2%2B%5Cleft%2811-%5Cleft%28-1%5Cright%29%5Cright%29%5E2%7D%3D2%5Csqrt%7B37%7D)
![WY=\sqrt{\left(-5-\left(-10\right)\right)^2+\left(11-4\right)^2}=\sqrt{74}](https://tex.z-dn.net/?f=WY%3D%5Csqrt%7B%5Cleft%28-5-%5Cleft%28-10%5Cright%29%5Cright%29%5E2%2B%5Cleft%2811-4%5Cright%29%5E2%7D%3D%5Csqrt%7B74%7D)
Now,
![WX=WY](https://tex.z-dn.net/?f=WX%3DWY)
So, triangle is an isosceles triangles.
and,
![WX^2+WY^2=(\sqrt{74})^2+(\sqrt{74})^2](https://tex.z-dn.net/?f=WX%5E2%2BWY%5E2%3D%28%5Csqrt%7B74%7D%29%5E2%2B%28%5Csqrt%7B74%7D%29%5E2)
![WX^2+WY^2=74+74](https://tex.z-dn.net/?f=WX%5E2%2BWY%5E2%3D74%2B74)
![WX^2+WY^2=148](https://tex.z-dn.net/?f=WX%5E2%2BWY%5E2%3D148)
![WX^2+WY^2=(2\sqrt{37})^2](https://tex.z-dn.net/?f=WX%5E2%2BWY%5E2%3D%282%5Csqrt%7B37%7D%29%5E2)
![WX^2+WY^2=WY^2](https://tex.z-dn.net/?f=WX%5E2%2BWY%5E2%3DWY%5E2)
So, triangle is right angled triangle.
Therefore, the ΔWXY is an isosceles right angle triangle.