1Rahul let me make it more easy....
Let,
x/b-c = y/c-a = z/a-b = k
So, x =(b-c)k, y=(c-a)k, z=(a-b)k
Now,
L.H.S = ax+by+cz
= a(b-c)k + b(c-a)k + c(a-b)k
= abk-ack + bck-abk + ack-bck
= 0
If
x = y = z
b - c c - a a - b
then, prove that, ax + by + cz = 0
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UP 0 DOWN 0 0 5
5 Answers
1Easy using Componendo and dividendo
see first xb-c = yc-a = za-b ............1
As
xb-c = yc-a = x+yb-a ..........2
using 1 and 2
we get x+yb-a = za-b
which gives x+y = -z ....................3
Again
xb-c = yc-a
gives xc -xa = yb-yc
≡ (x+y) c = xa +yb
But from (3) x+y= - z
therefore -zc = ax+by
≡ ax+by+ cz = 0
:)
1
36@Neeraj - surely it is... but the way i wanted it is the way how ninepointcircle has proved it in..
or else its a general method...
1@ninepointcircle---can u xplain how did u get d eqn 2....dat componendo nd dividendo one.