Question

x/b+c-a = y/c+a-b = z/a+b-c then (b-c)x + (c-a)y + (a-b)z=
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Answer 1

Answer:

Step-by-step explanation:

xb+c−a=yc+a−b=za+b−c=k

x=k(b+c-a)

y=k(c+a-b)

z=k(a+b-c)

(b-c)x+(c-a)y+(a-b)z

=k{(b+c-a)(b-c)+(c+a-b)(c-a)+(a+b-c)(a-b)}

=k{b^2-c^2-a(b-c)+c^2-a^2-b(c-a)+a^2-b^2-c(a+b)}

=k{a(c-b)+b(a-c)+c(b-a)}

=k{0}=0.