Let A (a, b), B (b, c) and C (c, a) be the vertices of Δ ABC whose medians are AD, BE and CF respectively.
So, D, E and F are the mid points of BC, AC and AB respectively.
∴ Co-ordinates of D are
Centroid G divides median AD in the ratio 2:1 i.e. AG : GD = 2 : 1
∴ co-ordinates of G are
But, it is given that centroid G are (0, 0) ____(2)
Equating (1) & (2)
Now, using identity, if a+b+c = 0, then a3 + b3 + c3 = 3abc.
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alessre
Hello, the coordinates of the centroid of the triangle are x=(x₁+x₂+x₃)/3 , y=(y₁+y₂+y₃)/3 so,according to the given question, (a+b+c)/3=0 a+b+c=0 we could use the rule/law that if a +b+c =0, then a³+b³+c³ = 3abc.
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Here is the answer to your question.
Let A (a, b), B (b, c) and C (c, a) be the vertices of Δ ABC whose medians are AD, BE and CF respectively.
So, D, E and F are the mid points of BC, AC and AB respectively.
∴ Co-ordinates of D are
Centroid G divides median AD in the ratio 2:1 i.e. AG : GD = 2 : 1
∴ co-ordinates of G are
But, it is given that centroid G are (0, 0) ____(2)
Equating (1) & (2)
Now, using identity, if a+b+c = 0, then a3 + b3 + c3 = 3abc.
the coordinates of the centroid of the triangle are
x=(x₁+x₂+x₃)/3 , y=(y₁+y₂+y₃)/3
so,according to the given question,
(a+b+c)/3=0
a+b+c=0
we could use the rule/law that if a +b+c =0,
then a³+b³+c³ = 3abc.
bye :-)