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:
- $\map \Area T = \dfrac {\size {b y - a x} } 2$
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:
- $\map \Area {\triangle ABC} = \dfrac {\size {b y - a x} } 2$
$\blacksquare$
Sources
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text I$. Coordinates: $6$. Area of a triangle