Question

If $x\ne a,y\ne b,z\ne c$ and $\left|\begin{array}{ccc}a& y& z\\ x& b& z\\ x& y& c\end{array}\right|=0$ then $\left(\frac{x+a}{x-a}+\frac{y+b}{y-b}+\frac{z+c}{z-c}\right)$ is equal to

• A. $-1$
• B. $1$
• C. $2$
• D. $0$

Answer 1

The correct option is A $-1$

$\left|\begin{array}{ccc}a& y& z\\ x& b& z\\ x& y& c\end{array}\right|=0$

$R_2\to R_2-R_1.R_3\to R_3-R_1$

$\Rightarrow \left|\begin{array}{ccc}a& y& z\\ x-a& -(y-b)& 0\\ (x-a)& 0& -(z-c)\end{array}\right|=0$

$\Rightarrow a(y-b)(z-c)+y(x-a)(z-c)+z(0+(x-a\left)\right(y-b\left)\right)=0$
$\Rightarrow \frac{a}{x-a}+\frac{y}{y-b}+\frac{z}{z-c}=0$

$\Rightarrow \frac{a}{x-a}+\frac{y-b+b}{y-b}+\frac{z-c+c}{z-c}=0$
$\Rightarrow \frac{a}{x-a}+1+\frac{b}{y-b}+1+\frac{c}{z-c}=0$
$\Rightarrow \frac{a}{x-a}+\frac{b}{y-b}+\frac{c}{z-c}=-2$ .... $\left(i\right)$

Now $\frac{x+a}{x-a}+\frac{y+b}{y-b}+\frac{z+c}{z-c}$
$=\left(\frac{x+a}{x-a}-1\right)+\left(\frac{y+b}{y-b}-1\right)+\left(\frac{z+c}{z-c}-1\right)+3$
$=2\left(\frac{a}{x-a}+\frac{b}{y-b}+\frac{c}{z-c}\right)+3$
$=2(-2)+3=-1$ ........... From $\left(i\right)$