Area of a Triangle and Collinearity of Points Using Determinants

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

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

Condition for collinearity. If the three points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ all lie on a single straight line, the "triangle" they form has zero area — it has been flattened into a line. 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 often faster than comparing the slopes between pairs of points, especially once a student is already comfortable evaluating $3\times 3$ determinants using the properties from the earlier sections of this chapter (for instance, spotting that one row is a combination of the other two, without expanding the determinant fully at all). …