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Question
The area of the triangle formed by (a, b + c), (b, c + a) and (c, a + b)
Options
a + b + c
abc
(a + b + c)2
0
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Solution
We have three non-collinear points A ( a,b + c) ; B ( b, c + a) ; C( c,a + b).
In general if `A (x_1 ,y_1 ) ; B (x_2 , y_2 ) ; C (x_3 , y_3) ` are non-collinear points then are of the triangle formed is given by-
`"ar" (Δ ABC ) = 1/2|x_1 (y_2 - y_3) + x_2 (y_3 - y_1) +x_3 (y_1 - y_2 )|`
So,
`"ar" (Δ ABC) = 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)]`
= 0
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