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Question
Find the area of triangles whose vertices are A(−1, 2), B(2, 4), C(0, 0).
Solution
Here, A(x1, y1) ≡ A(–1, 2), B(x2, y2) ≡ B(2, 4), C(x3, y3) ≡ C(0, 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|(-1, 2, 1),(2, 4, 1),(0, 0, 1)|`
= `1/2[-1(4 - 0) - 2(2 - 0) + 1(0 - 0)]`
= `1/2(-4 - 4)`
= `1/2(- 8)`
= –4
Since the area cannot be negative.
∴ A(ΔABC) = 4 sq. units
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