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Question
The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is ______.
Options
(a + b + c)2
0
a + b + c
abc
Solution
The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is 0.
Explanation:
Let the vertices of a triangle are, A ≡ (x1, y1) ≡ (a, b + c)
B ≡ (x2, y2) ≡ (b, c + a) and C = (x3, y3) ≡ (c, a + b)
∵ Area of ΔABC = Δ = `1/2[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]`
∴ Δ = `1/2[a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a)]`
= `1/2[a(c - b) + b(a - c) + c(b - a)]`
= `1/2(ac - ab + ab - bc + bc - ac)`
= `1/2(0)`
= 0
Hence, the required area of triangle is 0.
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