Area of quadrilateral as a determinant

The Endeavour 2024-02-25

I’ve written several posts about how determinants come up in geometry. These determinants often look similar, having columns related to coordinates and a column of ones. You can find several examples here along with an explanation for this pattern.

If you have three points z₁, z₂, and z₃ in the complex plane, you can find the area of a triangle with these points as vertices

$A(z_1, z_2, z_3) = \frac{i}{4} \, \left| \begin{matrix} z_1 & \bar{z}_1 & 1 \\ z_2 & \bar{z}_2 & 1 \\ z_3 & \bar{z}_3 & 1 \\ \end{matrix} \right|$

You can read more about this here.

If you add the second column to the first, and subtract the first column from the second, you can get the equation for the area of a triangle in the real plane.

$A = \frac{1}{2} \, \left| \begin{matrix}x_1 & y_1 & 1 \\x_2 & y_2 & 1 \\x_3 & y_3 & 1 \\ \end{matrix} \right|$

Presumably the former equation was first derived from the latter, but the two are equivalent.

Now suppose you have a quadrilateral whose vertices are numbered in clockwise or counterclockwise order. Then the area is given by

$A = \frac{1}{2} \, \left| \begin{matrix}x_1 & y_1 & 1 & 0\\x_2 & y_2 & 1 & 1\\x_3 & y_3 & 1 & 0\\x_4 & y_4 & 1 & 1\\\end{matrix} \right|$

The proof is easy. If you expand the determinant by minors on the last column, you get the sum of two 3 × 3 determinants corresponding to the areas of the two triangles formed by splitting the quadrilateral along the diagonal connecting the first and third points.

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