
On 10/24/01 11:56:21 PM, Wert Shin wrote: >The Cramer's Rule for area of >a triangle is 
> > > > > > > 
A = (1/2)  ê 

ê 
>but how does it work? 
This formula gives the area of the triangle formed by the vertices
(x_{1},y_{1}), (x_{2},y_{2}), and (x_{3},y_{3})
The determinant of this matrix is
D = x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1}  x_{1}y_{3}  x_{2}y_{1}  x_{3}y_{2}
This can be partially factored as
D = (x_{1}x_{3})(y_{2}y_{3})  (x_{2}x_{3})(y_{1}y_{3})
Now move the vertex (x_{3},y_{3}) to the origin by subtracting it from each of the vertices. So now our triangle becomes
(x_{1}x_{3},y_{1}y_{3}), (x_{2}x_{3},y_{2}y_{3}), and (0,0)
Naturally, this triangle has the same area as the original triangle; All I've done is move it. To simplify the discussion, let
a = (x_{1}x_{3}),
b = (y_{1}y_{3}),
c = (x_{2}x_{3}), and
d = (y_{2}y_{3})
Now the triangle in question has vertices (a,b), (c,d), and (0,0). Substituting these variables into the equation, above, for the determinant of the original matrix, we see it is
D = ad  bc
The area of the triangle, then, using the determinant formula is
A = (1/2)(ad  bc)
Well, not quite. The area can't be negative, so we have to take the absolute value.
A = (1/2)adbc
Let's review. Here is what we know so far
(1) The determinant of the 3x3 matrix, above, is equal to adbc, after we set a, b, c, and d appropriately.
(2) The area of triangle (a,b), (c,d), (0,0) is the same as the area of
triangle
(x_{1},y_{1}), (x_{2},y_{2}), and (x_{3},y_{3})
Now, to finish the explanation of why half the determinant of that 3x3 matrix is equal to the triangle's area, we just need to explain why
A = (1/2)adbc
For this explanation, click here: Triangle Area using Vectors
Polygon Area using Determinant
Heron's formula for the area of a triangle, if all you know is the lengths of its sides.
Algorithm to determine whether a point is inside a triangle.
The webmaster and author of this Math Help site is Graeme McRae.