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

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Example of Area of Triangle in Determinant Form

Let $A = \tuple {0, 0}, B = \tuple {b, a}, C = \tuple {x, y}$ be points in the Cartesian plane.

Let $T$ the triangle whose vertices are at $A$, $B$ and $C$.

Then the area $\AA$ of $T$ is:

$\map \Area T = \dfrac {\size {b y - a x} } 2$


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

\(\ds B\) \(=\) \(\ds \polar {r_1, \theta_1}\)
\(\ds C\) \(=\) \(\ds \polar {r_2, \theta_2}\)

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

Then we have:

\(\ds \map \Area {\triangle ABC}\) \(=\) \(\ds \dfrac 1 2 AB \cdot AC \sin \theta\) Area of Triangle in Terms of Two Sides and Angle
\(\ds \) \(=\) \(\ds \dfrac 1 2 r_1 r_2 \map \sin {\theta_2 - \theta_1}\) Definition of $\theta$
\(\ds \) \(=\) \(\ds \dfrac 1 2 r_1 r_2 \paren {\sin \theta_2 \cos \theta_1 - \cos \theta_2 \sin \theta_1}\) Sine of Difference
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {r_1 \cos \theta_1 r_2 \sin \theta_2 - r_1 \sin \theta_1 r_2 \cos \theta_2}\) rearranging
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {b y - a x}\) rearranging

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:

$\map \Area {\triangle ABC} = \dfrac {\size {b y - a x} } 2$