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
 A: In fact, if you are looking for a geometric approach you can get:
$$S=\frac{|AB\times AC|}{2}=\frac{|AB||AC|\sin\alpha}{2}$$
but in another way, you also can't forget that
$$AB\times AC=\begin{vmatrix}
\vec i & \vec j&\vec k\\
a&b&0\\
c&d&0
\end{vmatrix}=\vec k(ad-bc)$$
and then
$$|AB\times AC|=|ad-bc|$$
what then give you
$$S=\frac{|AB\times AC|}{2}=\frac{|ad-bc|}{2}$$
A: 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.
A: Is easy to prove using rotations around the origin(you can try) that $det(v|w)=||v||*||w||*sin(t)$ where $t$ is the angle formed by two vectors $v$ and $w$, so should be easy to conclude for you
A: Wlog, we can take $A $ as origin.
then the line $AB $ has the equation
$$\frac {b}{a}x-y=0$$
the distance from the point $C $ to the line $AB$ (the height) is
$$H=\frac {|\frac {b}{a}c-d|}{\sqrt {1+\frac {b^2}{a^2}   }}$$
the surface is ( base x height /2)
$$AB. H/2=\sqrt {a^2+b^2}.H/2$$
$$=\frac {|bc-ad|}{2} $$
