Find the area of the triangle whose vertices are (a,b+c),(b,c+a) and (c,a+b).
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Answered by
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concept : if three points A , B and C form a triangle ABC.
then, area of triangle ABC =
here given, three points (a, b + c), (b, c + a) and (c, a + b)
then, area of triangle formed by these points = 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 + ba - bc + cb - ac ]
= 0
hence, area of triangle whose vertices are (a, b + c), (b, c + a) and (c, a + b) = 0
[important point : if area of triangle becomes zero, it means vertices of triangle is collinear. I mean, they lie in same line . ]
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33
Answer:
here is ur answer.........
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