Question
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
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