Question

x/b+c=y/c+a=z/a+b show that (b-c)x+(c-a)y+(a-b)z​

Answer 1

Answer:

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