Area of Triangle in Determinant Form/Proof 3
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Theorem
Let $A = \tuple {x_1, y_1}, B = \tuple {x_2, y_2}, C = \tuple {x_3, y_3}$ be points in the Cartesian plane.
The area $\AA$ of the triangle whose vertices are at $A$, $B$ and $C$ is given by:
- $\AA = \dfrac 1 2 \size {\paren {\begin {vmatrix}
x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } }$
Proof
Let $A$, $B$ and $C$ be defined as $\tuple {x_1, y_1}$, $\tuple {x_2, y_2}$ and $\tuple {x_3, y_3}$ respectively.
From the figure, we see that:
| \(\ds \map \Area {ABC}\) | \(=\) | \(\ds \map \Area {PACR} + \map \Area {RCBQ} - \map \Area {PABQ}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {\paren {x_3 - x_1} \paren {y_3 + y_1} } 2 + \dfrac {\paren {x_2 - x_3} \paren {y_2 + y_3} } 2 - \dfrac {\paren {x_2 - x_1} \paren {y_1 + y_2} } 2\) | Area of Trapezoid | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {x_1 y_2 - x_2 y_1 + x_2 y_3 - x_3 y_2 + x_3 y_1 - x_1 y_3} 2\) | simplification | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 2 \size {\paren {\begin {vmatrix}
x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } }\) |
Determinant of Order 3 |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {III}$. Analytical Geometry: The Straight Line: The area of the triangle $ABC$