# Could the points (-4,3), (-1,1) and (1,3) form the vertices of a right triangle?

Dec 11, 2016

No

#### Explanation:

Using distance formula $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$, the distance between (-4,3) and (-1,1) would be $\sqrt{{3}^{2} + {\left(- 2\right)}^{2}} = \sqrt{13}$, distance between (-1,1) and (1,3) would be $\sqrt{{2}^{2} + {2}^{2}} = \sqrt{8}$, and the distance between (-4,3) and (1,3) would be $\sqrt{{5}^{2} + {0}^{2}} = \sqrt{25}$

Now the sum of the squares of any of these two distances does not equal to the square of the the third distance. Hence the triangle is not a right triangle.