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Mathematics · Ch 4 — Determinants

Area of a Triangle

4.3

Area of a Triangle

The Determinant Formula for Area

You already know the area of a triangle with vertices (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), and (x3,y3)(x_3, y_3) is given by:

Area=12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]\text{Area} = \frac{1}{2} \bigl[ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \bigr]

This expression can be written compactly as a determinant. The area Δ\Delta is:

Δ=12∣x1y11x2y21x3y31∣\Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}

The absolute value of this determinant gives the area, because area is always positive.

Watch out

If the problem gives you the area and asks for a missing coordinate, you must consider both the positive and negative values of the determinant. For example, if area = 3, then 12∣…∣=±3\frac{1}{2} \begin{vmatrix} \dots \end{vmatrix} = \pm 3.

Why Collinear Points Give Zero Area

If three points lie on the same straight line, they are collinear. A triangle formed by collinear points has no area — it is degenerate. Therefore, the determinant in the area formula must be zero. …