Area of a Triangle and Collinearity of Three Points

4.3.3 Area of a Triangle and Collinearity of Three Points

Theorem. If $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ are the vertices of $\triangle ABC$, then its area is

Proof (sketch, Fig. 4.1). Drop perpendiculars $AP,CQ,BR$ from $A,C,B$ onto the x-axis. The area of $\triangle ABC$ equals (area of trapezium $PACQ$) + (area of trapezium $QCBR$) − (area of trapezium $PABR$):

Substituting $PQ=x_3-x_1,\ QR=x_2-x_3,\ PR=x_2-x_1$ and the heights $AP=y_1,CQ=y_3,BR=y_2$, expanding and collecting terms gives exactly

Remarks. (i) Area is a positive quantity, so always take the absolute value of the determinant. (ii) If the area is given and a vertex coordinate is unknown, consider both signs of the determinant, since the determinant itself can come out negative depending on the order the vertices are listed (swapping two vertices swaps two rows, which flips the sign by Property 2 — this reflects the triangle being traversed in the opposite orientation). (iii) If the area comes out to $0$, the three points are collinear — they don't form a genuine triangle at all.

Collinearity test. Points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are collinear if and only if

Worked Examples

Example 1. Find the area of the triangle with vertices $A(-2,-3),B(3,2),C(-1,-8)$.

Step 1: Area $=\dfrac12\begin{vmatrix}-2&-3&1\\3&2&1\\-1&-8&1\end{vmatrix}$.

Step 2: Expand: $-2(2+8)+3(3+1)+1(-24+2)=-20+12-22=-30$. …