Area of a triangles with vertices at (2, 3), (-1, 0) and (2, -4) is :

Solution

Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃).   The area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]   Let A(x₁, y₁) = (2, 3),  B(x₂, y₂) = (- 1 , 0), and C(x₃, y₃) = (2, - 4)   Area of a triangle is given by  1/2 [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]   By substituting the values of vertices, A, B, C in the formula.   Area of the given triangle = 1/2 [2(0 - (-4))  + (-1)((- 4) - 3) + 2(3 - 0)]   = 1/2 (8 + 7 + 6)   = 21/2 square units