The area of a triangle in the coordinate plane

The area of a triangle with known coordinates can be computed by a formula. Consider the triangle with vertices $A(1, 1)$, $B(5, 1)$, $C(2, 4)$.

When a base and height are easy to read off, you can use them directly. The side $AB$ lies on $y = 1$ with length $5 - 1 = 4$. The height from $C$ is its distance from this base, the difference between $C$'s $y$-coordinate $4$ and the base's $y = 1$, namely $4 - 1 = 3$. So the area is $\dfrac{1}{2} \times 4 \times 3 = 6$.

For a general triangle whose base is not parallel to an axis, use this formula. The area of the triangle $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$ is

Substituting gives $\dfrac{1}{2}|1(1 - 4) + 5(4 - 1) + 2(1 - 1)| = \dfrac{1}{2}|-3 + 15 + 0| = \dfrac{1}{2} \times 12 = 6$, matching the value above. The absolute value keeps the area positive, and the result is the same in whatever order the vertices are listed. The three lines on the graph are the sides $AB$, $BC$, $CA$, and the large dots are the vertices $A(1, 1)$, $B(5, 1)$, $C(2, 4)$.