Area of Triangle in Determinant Form with Vertex at Origin/Proof 2

Proof
Let the polar coordinates of $B$ and $C$ be:

Let $\theta$ be the angle between $AB$ and $AC$.

Then we have:

We can define the area of $\triangle ABC$ as being positive or negative according to the sign of $\dfrac 1 2 \paren {b y - a x}$.

However, if we are interested only in the absolute value of $\triangle ABC$, as in this context, we can report:
 * $\map \Area {\triangle ABC} = \dfrac {\size {b y - a x} } 2$