Area of a Triangle and Collinearity Using Determinants

Determinants also give a compact formula for the area of a triangle with known vertices, and — as an immediate consequence — a test for whether three points lie on one straight line.

Area of a triangle. For a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$, $\text{Area} = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}.$ Because an area must be positive, but the determinant may come out negative depending on the order the vertices are listed, the formula is applied by taking the absolute value: $\text{Area} = \frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|.$

Condition for collinearity. If the three points all lie on one straight line, the "triangle" they form is flattened and has zero area. So three points are collinear exactly when $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0.$ This determinant test is usually quicker than comparing the slopes of the pairs of points, especially once a student is comfortable using a row operation to create zeros before expanding. …