How to use slopes (3 points are given) to prove that they form a right triangle?

Question

Question: Use slopes to show that $A(-3, -1)$, $B(3, 3)$ and $C(-9, 8)$ are vertices of a right triangle.

My try at the problem: I know that we can find the slopes of $AB$, $BC$ and $CA$ and then prove that the slope of one line is the negative reciprocal of another and hence, two lines are perpendicular. However, it can be the case that two of the lines are perpendicular, but the other one is not perpendicular/parallel to either of them as follows:

In these case, we do see two lines perpendicular to each other, but no right triangle. How to we actually solve the problem them?

Answer 1

There are actually two process where we can show its a right triangle

1. By showing that any two of those line's slope are perpendicular. Actually Its a right triangle so basically multiplying the slope of the hypotenuse wont get the result with either of the lines. We have to find out all the slopes of those lines and multiply all of them simultaneously.
   Given, A(−3,−1), B(3,3) and C(−9,8) Slope=(difference of y/difference of x) Slope of AB=2/3, Slope of AC=-3/2, Slope of BC=-5/12 Now we have to multiply any two of them two find out whether they are perpendicular or not By using my intuitive, I could spot it and it should be AB x AC=-1 If you are still unable to reckon that you can try all of them You can do all of them AB x BC, BC x AC.
2. Another method is calculating the distance between those points and find out the length then use the Pythagorean theorem Thanks for asking

Answer 2

There are three angles and three pairs of adjacent lines in a triangle.

Calculate slope product pairs $(m_1m_2, m_2m_3, m_3m_1)$ and one among them should be equal to $-1$... that is sufficient to locate the vertex making the right angle.