Area of Triangle in Determinant Form

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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 1

AreaOfTriangleComplex.png

Let $A$, $B$ and $C$ be defined as complex numbers in the complex plane.

The vectors from $C$ to $A$ and from $C$ to $B$ are given by:

$z_1 = \paren {x_1 - x_3} + i \paren {y_1 - y_3}$
$z_2 = \paren {x_2 - x_3} + i \paren {y_2 - y_3}$

From Area of Triangle in Terms of Side and Altitude, $\AA$ is half that of a parallelogram contained by $z_1$ and $z_2$.

Thus:

\(\ds \AA\) \(=\) \(\ds \frac 1 2 z_1 \times z_2\) Area of Parallelogram in Complex Plane
\(\ds \) \(=\) \(\ds \frac 1 2 \size {\paren {\map \Im {\paren {x_1 - x_3} - i \paren {y_1 - y_3} } \paren {\paren {x_2 - x_3} - i \paren {y_2 - y_3} } } }\) Definition 3 of Vector Cross Product
\(\ds \) \(=\) \(\ds \frac 1 2 \size {\paren {x_1 - x_3} \paren {y_2 - y_3} - \paren {y_1 - y_3} \paren {x_2 - x_3} }\) Definition of Complex Multiplication
\(\ds \) \(=\) \(\ds \frac 1 2 \size {x_1 y_2 - y_1 x_2 + x_2 y_3 - y_2 x_3 + x_3 y_1 - y_3 x_1}\) multiplying out
\(\ds \) \(=\) \(\ds \frac 1 2 \size {\paren {\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } }\) Determinant of Order 3

$\blacksquare$


Proof 2

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.

$\blacksquare$


Proof 3

Area-of-Triangle-Determinant-Proof-3.png

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

$\blacksquare$


Examples

Vertex at Origin

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:

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


Vertices at $\paren {-4 - i}, \paren {1 + 2 i}, \paren {4 - 3 i}$

Let $T$ be a triangle embedded in the complex plane with vertices at $\paren {-4 - i}, \paren {1 + 2 i}, \paren {4 - 3 i}$.

The area of $T$ is given by:

$\map \Area T = 17$


Sources

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