Area of a Triangle Using Determinants

Setting Up the Formula

Given a triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$, elementary coordinate geometry (splitting the triangle into trapezoids against the $x$-axis, or using the shoelace method) gives its area as

This same expression can be written far more compactly, and remembered far more easily, as a $3\times3$ determinant:

Expanding this determinant along the third column (Section 3) reproduces exactly the expression above, confirming the two forms are the same formula.

The Absolute Value Is Essential

Unlike a length or an area computed directly, the determinant itself can come out negative depending on the order in which the three vertices are listed (listing them clockwise versus anticlockwise flips the sign, by Property 2 of Section 2 -- swapping two rows of the determinant corresponds to swapping two vertices). Since an area can never be negative, the formula always takes the absolute value of the determinant before halving it:

Corollary: A Test for Collinearity

Three points are collinear (all lie on one straight line) exactly when the "triangle" they would form has zero area -- geometrically, the triangle collapses to a line segment. This gives an immediate determinant test for collinearity: the points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are collinear if and only if

This test is often faster than computing and comparing slopes, especially when the coordinates involve unknowns, since it reduces directly to the row-operation techniques of Section 2 (a column becoming proportional to another, for instance, is an instant proof of collinearity without evaluating a single numeric slope).

Using the Formula to Find an Unknown Coordinate …