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Question
Find the area of the triangle whose vertices are: (0, 5), (0, – 5), (5, 0)
Solution
Here, A(x1, y1) ≡ A(0, 5), B(x2, y2) ≡ B(0, – 5), C(x3, y3) ≡ C(5, 0)
Area of a triangle = `1/2|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`
∴ A(ΔABC) = `1/2|(0, 5, 1),(0, -5, 1),(5, 0, 1)|`
= `1/2[0(-5 - 0) - 5(0 - 5) + 1(0 + 25)]`
= `1/2(0 + 25 + 25)`
= `50/2`
∴ A(ΔABC) = 25 sq. units
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