aimforthehead Posted August 28, 2011 Share Posted August 28, 2011 How do I determine whether a triangle with a given vertices is a right triangle? It gives... (7,-1) (-3,5) (-12,-10) Link to comment Share on other sites More sharing options...
Bignose Posted August 28, 2011 Share Posted August 28, 2011 (edited) you could see if the side lengths obey the Pythagorean Theorem. If it is a right triangle, it has to obey PT, but I am not sure if obeying the PT is necessary and sufficient for the triangle to be right, however... The side lengths can be used in the trig relations, however, which would be necessary and sufficient for finding a 90* angle. Edited August 28, 2011 by Bignose Link to comment Share on other sites More sharing options...
aimforthehead Posted August 28, 2011 Author Share Posted August 28, 2011 I am not sure I understand =/ I've forgotten almost everything about triangles, I have area=1/2b+h and d=sqrt((x2-x1)+(y2-y1)). Not much else. Link to comment Share on other sites More sharing options...
Bignose Posted August 28, 2011 Share Posted August 28, 2011 http://en.wikipedia.org/wiki/Pythagorean_theorem Link to comment Share on other sites More sharing options...
aimforthehead Posted August 29, 2011 Author Share Posted August 29, 2011 Yeah I get that, I don't understand how to get A2+B2=C2 out of 3 points though. Link to comment Share on other sites More sharing options...
DrRocket Posted August 29, 2011 Share Posted August 29, 2011 you could see if the side lengths obey the Pythagorean Theorem. If it is a right triangle, it has to obey PT, but I am not sure if obeying the PT is necessary and sufficient for the triangle to be right, however... . It is both necessary and sufficient. Sufficiency follows from the law of cosines. Link to comment Share on other sites More sharing options...
Arka Posted August 29, 2011 Share Posted August 29, 2011 It is both necessary and sufficient. Sufficiency follows from the law of cosines. This is true that PT is sufficient but in the case when Three vertices r non colinear.... Can we call a line segment as a triangle with area zero?????????? Link to comment Share on other sites More sharing options...
Bignose Posted August 29, 2011 Share Posted August 29, 2011 It is both necessary and sufficient. Sufficiency follows from the law of cosines. thanks, I thought so, but I wasn't sure. Link to comment Share on other sites More sharing options...
DrRocket Posted August 29, 2011 Share Posted August 29, 2011 This is true that PT is sufficient but in the case when Three vertices r non colinear.... Can we call a line segment as a triangle with area zero?????????? That would be the usual convention. Link to comment Share on other sites More sharing options...
Bignose Posted August 29, 2011 Share Posted August 29, 2011 Yeah I get that, I don't understand how to get A2+B2=C2 out of 3 points though. Well, you have 3 points. You have the formula of distance between points, so you have the lengths of three sides. If you can make those 3 sides fit into Pythagoras' Theorem, then you're done. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now