Area of Triangle in Determinant Form/Proof 2

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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

Area-of-Triangle-Determinant.png


Let $A$, $B$ and $C$ be as defined..

Let $O$ be the origin of the Cartesian plane in which $\triangle ABC$ is embedded.

Taking into account the signs of the areas of the various triangles involved:

$\triangle ABC = \triangle OAB + \triangle OBC + \triangle OCA$

as it is seen that $\triangle OBC$ and $\triangle OCA$ are described in clockwise sense.


From proof 2 of Area of Triangle in Determinant Form with Vertex at Origin:

\(\ds \triangle OAB\) \(=\) \(\ds \dfrac 1 2 \paren {x_1 y_2 - x_2 y_1}\)
\(\ds \triangle OBC\) \(=\) \(\ds \dfrac 1 2 \paren {x_2 y_3 - x_3 y_2}\)
\(\ds \triangle OBC\) \(=\) \(\ds \dfrac 1 2 \paren {x_3 y_1 - x_1 y_3}\)
\(\ds \leadsto \ \ \) \(\ds \triangle ABC\) \(=\) \(\ds \dfrac 1 2 \paren {\paren {x_1 y_2 - x_2 y_1} + \paren {x_2 y_3 - x_3 y_2} + \paren {x_3 y_1 - x_1 y_3} }\)
\(\ds \leadsto \ \ \) \(\ds \triangle ABC\) \(=\) \(\ds \frac 1 2 \paren {\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} }\) Determinant of Order 3

The result follows.

$\blacksquare$


Sources

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