Question

If $\frac{a}{x-2y+3z}=\frac{b}{y-2z+3x}=\frac{c}{z-2x+3y}$ and
$x+y+z\ne 0$, then each ratio is also equal to ______.

• A. $\frac{a+b+c}{2(x+y+z)}$
• B. $\frac{a+b+c}{(x+y+z)}$
• C. $\frac{2(a+b+c)}{x+y+z}$
• D. $\frac{a+b+c}{x+y+z}$

Answer 1

The correct option is A $\frac{a+b+c}{2(x+y+z)}$

Let $\frac{a}{x-2y+3z}=\frac{b}{y-2z+3x}=\frac{c}{z-2x+3y}=k$

$\therefore$ by theorem of equal ratio;

$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\frac{a+c+e}{b+d+f}$

$\Rightarrow k=\frac{a+b+c}{(x-2y+3z)+(y-2z+3x)+(z-2x+3y)}$

$=\frac{a+b+c}{(2x+2y+2z)}$

$=\frac{a+b+c}{2(x+y+z)}$

$\therefore \frac{a}{x-2y+3z}=\frac{b}{y-2z+3x}=\frac{c}{z-2x+3y}=\frac{a+b+c}{2(x+y+z)}$