# How do you find two ways to determine if the triangle with vertices A(-7,3), B(4,0), and C(-4,4) is a right triangle?

Nov 13, 2017

Check the Problem.

#### Explanation:

We are given the points (pts.)

$A = A \left(- 7 , 3\right) , B = B \left(4 , 0\right) , C = C \left(- 4 , 4\right) .$

Method 1 :

Using the Distance Formula, we have,

$A {B}^{2} = {\left(- 7 - 4\right)}^{2} + {\left(3 - 0\right)}^{2} = 121 + 9 = 130.$

$B {C}^{2} = {\left(4 - \left(- 4\right)\right)}^{2} + {\left(0 - 4\right)}^{2} = 64 + 16 = 80.$

$A {C}^{2} = {\left(- 7 - \left(- 4\right)\right)}^{2} + {\left(3 - 4\right)}^{2} = 9 + 1 = 10.$

Since, $A {B}^{2} \ne B {C}^{2} + A {C}^{2} , \angle C \ne {90}^{\circ} .$