# How do you show that a triangle with vertices (13,-2), (9,-8), (5,-2) is isosceles?

Jan 6, 2016

Find the length of the triangle's segments and prove that two are equal but the third is different. In case, $P 1 P 1 = P 2 P 3 = \sqrt{50}$ and $P 1 P 3 = 8$ => ${\triangle}_{P 1 P 2 P 3}$ is isosceles

#### Explanation:

Length of the triangle's segments:
$P 1 P 2 = \sqrt{{\left(9 - 13\right)}^{2} + {\left(- 8 + 2\right)}^{2}} = \sqrt{16 + 36} = \sqrt{50}$
P1P3=sqrt((5-13^2+(-2+2)^2)=sqrt(64+0)=8
P2P3=sqrt((5-9)^2+(-2+8)^2)=sqrt(16+36)=sqrt(50

As we can see two sides of the triangle are equal ($P 1 P 2 = P 2 P 3 = \sqrt{50}$) but the third one is different ($P 1 P 3 = 8$): so ${\triangle}_{P 1 P 2 P 3}$ is isosceles.