Area of a triangle

The triangle with corners (x₁, y₁), (x₂, y₂), (x₃, y₃) has area ½[x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)], which is ½ times the determinant with rows (x₁, y₁, 1), (x₂, y₂, 1), (x₃, y₃, 1).

An area cannot be negative, so the absolute value of that expression is taken. The sign only records the order in which the corners were listed.

When the area is given and a coordinate is unknown, set the determinant expression equal to both +area and −area; each sign can give a valid answer.

Worked example: corners (3, 8), (−4, 2), (5, 1) give ½[3(2 − 1) − 8(−4 − 5) + 1(−4 − 10)] = ½(3 + 72 − 14) = 61/2 square units.

Three points are collinear exactly when the triangle they form has zero area, that is when the determinant is 0.

The line through two points: a point (x, y) is on the line through (1, 3) and (0, 0) when the triangle it makes with them has zero area, which gives ½(y − 3x) = 0, the line y = 3x.

Worked example: with the corners listed as A(1, 3), B(0, 0), D(k, 0), the determinant works out to 3k, so an area of 3 needs 3k/2 = ±3, giving k = 2 or k = −2.