psi20 Posted March 14, 2005 Share Posted March 14, 2005 The area of a triangle with vertices (a,b), (c,d), and (e,f) is (sorry, I don't know how to put it into vertical bars for determinants) +/- .5 determinant of A where A is a b 1 c d 1 e f 1 and the symbol +/- indicates that the appropiate sign should be chosen to yield a positive area. How can one prove this? I think I have a method for proving it, but it seems too tedious and long to do. Link to comment Share on other sites More sharing options...
Primarygun Posted March 14, 2005 Share Posted March 14, 2005 Why the coordinates of the vertex are totally different? No triangle is formed? Link to comment Share on other sites More sharing options...
psi20 Posted March 15, 2005 Author Share Posted March 15, 2005 Was that a question or a statement? This works for a triangle in a rectangular coordinate-system formed by any 3 given points. If the determinant is 0, that means no triangle is formed and the 3 points are co-linear. Link to comment Share on other sites More sharing options...
Primarygun Posted March 15, 2005 Share Posted March 15, 2005 Sorry, just my fault. I am not able to understand it. Link to comment Share on other sites More sharing options...
uncool Posted March 20, 2005 Share Posted March 20, 2005 A = (a, d) B = (b, e) C = (c, f) A(triangle) = length of cevian * width of triangle with respect to cevian - that is, the length of the projection of the opposite side onto the perpendicular to the cevian BC: y = ((f-e)/(c-b))(x-b)+e = ((f-e)/(c-b))x - b((f-e)/(c-b)) + e = ((f-e)/(c-b))x - ((bf - be)/(c-b)) + e = ((f-e)/(c-b))x - ((bf - be + be - ec)/(c-b)) Length of vertical cevian: d - ((f-e)/(c-b))a + ((bf - ec)/(c-b)) Width of triangle = c - b Area = d(c-b) - a(f-e) + bf - ec = d(c-b) + e(a-c) + f(b-a) = |a d 1| |b e 1| |c f 1| Link to comment Share on other sites More sharing options...
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