prove that the triangle with vertices (3,5), (-2,6), and (1,3) is a right triangle.?

Question

please someone show me how to do this? i think i have to find the slop am i right? but how would i do that with 3 vertices? i have to pair them up and find all the slopes?

Answer 1

Nope. The pythagorean theorem states that if a and b are the lengths of the two legs and c the length of the hypotenuse, then a²+b²=c². The converse of this is true as well, so all you have to do is find the lengths of the legs, show that they satisfy this relation, and you have thus proven the triangle is right.

a² = (3-1)²+(5-3)² = 4+4 = 8
b² = (-2-1)² + (6-3)² = 9+9 = 18
c² = (3-(-2))² + (5-6)² = 25+1 = 26

a²+b² = 8 + 18 = 26 = c²

So the triangle is a right triangle.

Edit: actually, LaxPlayer's method is also valid. There's certainly nothing saying you have to use the converse of the pythagorean theorem, any more than you have to find the slope. This is a good example of a fundamental truth in mathematics: most of the time, there is more than one valid method of doing a problem. You needn't concern yourself with doing it in a particular way, as long as you do it in a way that is based on mathematically correct principles, you should get the same answer no matter how you do it.

Answer 2

Correct.
Let A = (3,5) B = (-2,6) C = (1,3)

Slope AB = (5-6)/(3--2) = -1/5 = -1/5
Slope BC = (6-3)/(-2-1) = 3/-3 = -1
Slope AC = (5-3)/(3-1) = 2/2 = 1

Therefore, since no slopes are equal, there are no parallel lines thus a triangle would be formed between these three vertices. Also, Since BC and AC are perpendicular (perpendicular if slope1 = -1/slope2), this would form a right angle thus a right triangle.

Answer 3

plug the points into the graph
find the distances
use a^2+b^2=c^2