Question

If $\frac{x}{xa+yb+zc}=\frac{y}{ya+zb+xc}=\frac{z}{za+xb+yc}$ and $x+y+z\ne 0$, then each ratio is:

• A. $\frac{1}{a-b-c}$
• B. $\frac{1}{a+b-c}$
• C. $\frac{1}{a-b+c}$
• D. $\frac{1}{a+b+c}$

Answer 1

The correct option is D $\frac{1}{a+b+c}$

Let $\frac{x}{xa+yb+zc}=\frac{y}{ya+zb+xc}=\frac{z}{za+xb+yc}=k$

We can write this as

$\Rightarrow k=\frac{x+y+z}{xa+yb+zc+ya+zb+xc+za+xb+yc}$

$\Rightarrow k=\frac{x+y+z}{xa+ya+za+ya+yb+yc+za+zb+zc}$

$\Rightarrow k=\frac{x+y+z}{a(x+y+z)+b(x+y+z)+c(x+y+z)}$

$\Rightarrow k=\frac{x+y+z}{(a+b+c)(x+y+z)}$

$\Rightarrow k=\frac{1}{a+b+c}$

Therefore each ratio is $\Rightarrow \frac{1}{a+b+c}$