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

Find the slopes of the sides of the triangles:
Slope of AB = 9
Slope of BC = 1
Slope of CA = -1

The triangle is a right triangle as BC and CA are perpendicular* to each other, which means that angle ACB = 90.
* If a slope is the negative reciprocal of another slope, they are perpendicular 

Answer 2

OK. At the beginning, we calculate the length of the sides of the triangle.
We will use the formula for the length of the segment

$A(x_A;\ y_A);\ B(x_B;\ y_B)\\\\|AB|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$
$A(-2;\ 3);\ B(-3;\ -6)\\\\|AB|=\sqrt{(-3-(-2))^2+(-6-3)^2}=\sqrt{(-1)^2+(-9)^2}\\\\=\sqrt{1+81}=\sqrt{82}\\\\A(-2;\ 3);\ C(2;\ -1)\\|AC|=\sqrt{(2-(-2))^2+(-1-3)^2}=\sqrt{4^2+(-4)^2}\\\\=\sqrt{16+16}=\sqrt{32}\\\\B(-3;\ -6);\ C(2;\ -1)\\|BC|=\sqrt{(2-(-3))^2+(-1-(-6))^2}=\sqrt{5^2+5^2}\\\\=\sqrt{25+25}=\sqrt{50}$
AB is the longest side.
If the equation
$|AB|^2=|AC|^2+|BC|^2$
is true, then the triangle is right triangle.
check:
$L_s=(\sqrt{82})^2=82\\\\R_s=(\sqrt{32})^2+(\sqrt{50})^2=32+50=82\\\\L_s=R_s$
Therefore your answer is:
The triangle ABC is a right triangle and angle ACB is the right angle.