vertices, is it a right triangle?

Question

not quite sure how to work this, tried graphing it out, and I was able to make some pretty pictures but not sure if that's really how you're supposed to do it. :D

A triangle is defined by the following vertices: (40,85), (15,0), and (-40,35). Is it a right trangle?

Answer 1

A(40,85)
B(15,0)
C(- 40,35)
AB² = (15 - 40)² = (-25)² = 25² = 625
AC² = (80² + 50²) = 6400 + 2500 = 8900
BC² = (55² + 35²) = 3025 + 1225 = 4250

AC² ≠ AB² + BC²
Is NOT right angled triangle.

Answer 2

I think what you're supposed to do is use the distance formula
So, from the first to second:
d = sqr root of ((40-15)^2 +(85)^2) which is sqr root of 7850
second to third:
d = sqr root of ((15 + 40)^2 + (-35)^2) which is sqr root of 4250
third to first:
d = sqr root of ((80)^2 + (85-35)^2) whic his sqr root of 8900
So ithink it is not, because otherwise the squares of both sides would add up to the square of the other, but it doesnt.

Answer 3

Suppose you have one triangle ABC with coordinates
A = (ax, ay) B = (bx, by) C = (cx, cy)
1. Calculate AB^2, BC^2, CA^2 by
AB^2 = (ax-bx)^2 + (ay-by)^2
BC^2 = (bx-cx)^2 + (by-cy)^2
CA^2 = (cx-ax)^2 + (cy-ay)^2
2. Choose maximum from these three results and compare if it is sum of other two.
3. If yes then your triangle is right
4. If no then it is not right.
Please apply with your particular case

Answer 4

Compute the slopes between the three points. If there are two slopes which are negative reciprocals then it is a right triangle; if not then it is not.