Area of a triangle

Determinants give a quick formula for the area of a triangle whose corners are known.

For corners (x₁, y₁), (x₂, y₂), (x₃, y₃):

Area = ½ |x₁, y₁, 1; x₂, y₂, 1; x₃, y₃, 1|

Write each point as a row, add a column of 1s, expand, and halve. If the answer comes out negative, take its positive value: area is never negative.

Example: (0, 0), (4, 0), (0, 3). Expanding along row 1 (two zeros) leaves only the 1 × |4, 0; 0, 3| = 12, so the area is ½ × 12 = 6. That matches ½ × base 4 × height 3.

When the area is given and a corner has an unknown, set the determinant equal to +2 × area and –2 × area, because the sign was dropped. That usually gives two answers.

Collinear points. If three points lie on one straight line, the 'triangle' is flat and has zero area. So three points are collinear exactly when the determinant is 0. Example: (1, 1), (2, 3), (3, 5): 1(3 – 5) – 1(2 – 3) + 1(10 – 9) = –2 + 1 + 1 = 0. Collinear.

Equation of a line through two points. Take a general point (x, y) on the line as the third corner. It is collinear with the other two, so the determinant is 0. For (0, 1) and (2, 5): |x, y, 1; 0, 1, 1; 2, 5, 1| = –4x + 2y – 2 = 0, which gives y = 2x + 1.