Question

Given $a=\frac{x}{(y-z)},b=\frac{y}{(z-x)},c=\frac{z}{(x-y)}$, where $x,y$ and $z$ are not all zero, then the value of $ab+bc+ca$ is

• A. $0$
• B. $1$
• C. $-1$
• D. none of these

Answer 1

The correct option is C $-1$

Given $a=\frac{x}{(y-z)},b=\frac{y}{(z-x)},c=\frac{z}{(x-y)}$
These equation can be written as
$x-ay+az=0$
$bx+y-bz=0$
$-cx+cy+z=0$
Since, $x,y,z$ are not all 0, so for the consistency of system
$D=0$
$\left|\begin{array}{ccc}1& -a& a\\ b& 1& -b\\ -c& c& 1\end{array}\right|=0$
$\Rightarrow ab+bc+ac+1=0$
$\Rightarrow ab+bc+ac=-1$