Question

Triangle ABC has vertices at A(–2, 3), B(–3, –6), and C(2, –1). Is triangle ABC a right triangle? If so, which angle is the right angle?

Answer 1

We'll use rise over run to find the slopes of two of the sides of the triangle:

$\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

Because C is the rightmost point, we'll have it represent (x1,y1), and have A and B represent (x2,y2):

A to C
(-2,3) to (2,-1)

$\frac{-1 - 3}{2 + 2} = \frac{-4}{4} = -1$

The slope of the line from A to C is -1.

B to C
(-3,-6) to (2,-1)

$\frac{-1 + 6}{2+3} = \frac{5}{5} = 1$

The slope of the line from B to C is 1.

These lines intersect at (2,-1), and they have negative reciprocal slopes of each other, which creates a perpendicular angle. This means that triangle ABC is a right triangle, and angle ACB is the right angle.