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

## 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:

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

## Proof

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:

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

$\blacksquare$