Question

If x : a = y : b = z : c then
$\frac{ax-by}{(a+b)(x-y)}+\frac{by-cz}{(b+c)(y-z)}+\frac{cz-ax}{(c+a)(z-x)}$ is equal to

• A. 1
• B. 2
• C. 3
• D. 0

Answer 1

The correct option is C 3

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\left(say\right)\Rightarrow x=ak,ynl.z=ck$

$\therefore \frac{ax-by}{(a+b)(x-y)}+\frac{by-cz}{\left((b+c)\right)(y-c)}+\frac{cz-ax}{(c+a)(z-x)}$

$=\frac{a\times ak-b\times bk}{\left((a+b)\right)(ak-bk)}+\frac{b\times bk-c\times ck}{(b+c)(bk-ck)}+\frac{c\times ck-c\times ak}{(c+a)(ck-ak)}$

$=\frac{k(a^2-b^2)}{k(a^2-b^2)}+\frac{k(b^2-c^2)}{k(b^2-c^2)}+\frac{k(c^2-a^2)}{k(c^2-a^2)}=1+1+1=3$