# Area of Triangle in Determinant Form

## 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 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 $\ds$ $=$ $\ds \frac 1 2 \size {\paren {\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } }$ Definition of Determinant of Order 3

$\blacksquare$

## Proof 2 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.

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

The result follows.

$\blacksquare$

## Proof 3 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 $\ds$ $=$ $\ds \dfrac 1 2 \size {\paren {\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } }$ Definition of 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:

$\map \Area T = 17$