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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

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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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    $\begingroup$ Nothing disgusting here. $\endgroup$ – Yves Daoust May 20 '17 at 18:59
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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

$$S=\frac{|AB\times AC|}{2}=\frac{|ad-bc|}{2}$$

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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

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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)

$$AB. H/2=\sqrt {a^2+b^2}.H/2$$ $$=\frac {|bc-ad|}{2} $$

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