<s>[INST] Proof the area of a given triangle with coordinates is half determinant I was given a problem, tried to solved it but couldn't get to a solution. 
It goes like that:
There's a triangle ABC with area S.
$$ \vec{AB} = (a,b) $$
$$ \vec{AC} = (c,d) $$
Prove that 
$$ S = \frac{\lvert ad - bc \rvert}{2} $$
I tried to solve it that way:
Express $AB \cdot AC$ as $ac+bd$ and as $\sqrt{a^2+b^2}\sqrt{c^2+d^2}\cos(\alpha)$
then I expressed with $a,b,c,d \sin(\alpha)$ (by squaring both sides and using $\cos^2\alpha=1-\sin^2\alpha$)
then I expressed S as $|AC||AB|\sin(\alpha)/2$, put $\sin(\alpha)$, $|AC|$ and $|AB|$ expressed with $a,b,c,d$ and got a disgusting expression which is probably not equal $ad-bc$...
Would be happy to get your help,
Thanks
 [/INST] Your approach:
From
$$\cos\alpha=\frac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}},$$
you draw
$$1-\cos^2\alpha=\frac{(a^2+b^2)(c^2+d^2)-(ac+bd)^2}{(a^2+b^2)(c^2+d^2)}.$$
You transform the numerator with
$$a^2c^2+a^2d^2+b^2c^2+b^2d^2-a^2c^2-2abcd-b^2d^2=a^2d^2+b^2c^2-2abcd=(ad-bc)^2$$
hence the claim.
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