if x/(b+c-a)= y/(a+c-b)=z(a+b-c) then (b-c)x + (c-a)y + (a-b)z=
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Step-by-step explanation:
Since all the terms are equal it must be equal to some constant value. So say
x/(b+c) = y/(c+a) = z/(a+b) =p
equating separately we get
x=p(b+c)
y=p(c+a)
z=p(a+b)
Now consider the equation
(b-c)x+(c-a)y+(a-b)z=k
k=p(b-c)(b+c)+p(c-a)(c+a)+p(a-b)(a+b)
k=p(b²-c²+c²-a²+a²-b²)
k=p(0)
k=0
So as considered
(b-c)x +(c-a)y+(a-b)z=0
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