If x/(b-c) = y/(c-a) = z/(a-b), Prove that: ax+by+cz = 0
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Given, x/(b-c) = y/(c-a) = z/(a-b) k(say)
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Step-by-step explanation:
x=k(b-c) , y=k(c-a) , z=k(a-b)
ax+by+cz
=ak(b-c)+bk(c-a)+ck(a-b)
= abk-ack+bck-abk+ack-bck (cut the common factors)
=0
LHS=RHS
hence proved
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