Find the area of the triangle whose vertices are :
(i) (2, 3), (–1, 0), (2, – 4)
(ii) (–5, –1), (3, –5), (5, 2)

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Detailed 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)]

(i) 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

(ii) Let A(x₁, y₁) = (- 5, - 1), B(x₂, y₂) = (3, - 5) and C(x₃, y₃) = (5, 2)

Area of a triangle = 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 [(- 5)((- 5) - 2) + 3(2 - ( -1)) + 5(- 1 - ( - 5))]

= 1/2 (35 + 9 + 20)

= 32 square units

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