Sum
Find the area of the triangle whose vertices are: (3, 2), (–1, 5), (–2, –3)
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Solution
Here, A(x1, y1) ≡ A(3, 2), B(x2, y2) ≡ B(–1, 5), C(x3, y3) ≡ C(–2, –3)
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|(3, 2, 1),(-1, 5, 1),(-2, -3, 1)|`
= `1/2[3(5 + 3) - 2(-1 + 2) + 1(3 + 10)]`
= `1/2(24 - 2 + 13)`
∴ A(ΔABC) = `35/2` sq.unitts.
Concept: Application of Determinants - Area of a Triangle
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