Area of triangle

[psi20]

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.

[Primarygun]

Why the coordinates of the vertex are totally different?

No triangle is formed?

[psi20]

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.

[Primarygun]

Sorry, just my fault. I am not able to understand it.

[uncool]

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|