Area of a Triangle

Area of a Triangle

The area of a triangle, whose vertices are (x₁, y₁), (x₂, y₂), and (x₃, y₃) is = ½ |x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)|.

Proof:

Let ABC be a triangle whose vertices are A (x₁, y₁), B (x₂, y₂), and C (x₃, y₃). Draw AL, BM and CN as perpendiculars form A, B and C, respectively, on the x – axis. Clearly, ABML, ALNC and BMNC are all trapeziums. We have

Area of ΔABC = Area of trapezium ABML + Area of trapezium ALNC – Area of trapezium BMNC

= ½ (BM + AL) (ML) + ½ (AL + CN) (LN) – ½ (BM + CN) (MN)

= ½ (y₂ + y₁) (x₁ – x₂) + ½ (y₁ + y₃) (x₃ – x₁) – ½ (y₂ + y₃) (x₃ – x₂)

= ½ |x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)|

\(=\frac{1}{2}\left\| \begin{matrix}   {{x}_{1}} & {{y}_{1}} & 1  \\   {{x}_{2}} & {{y}_{2}} & 1  \\   {{x}_{3}} & {{y}_{3}} & 1  \\\end{matrix} \right\|\).