Question

If $a\ne p,b\ne q,c\ne r$ and $\left|\begin{array}{ccc}p& b& c\\ a& q& c\\ a& b& r\end{array}\right|=0$, then the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is

• A. $0$
• B. $1$
• C. $-1$
• D. $-2$

Answer 1

The correct option is D $-2$

$\left|\begin{array}{ccc}p& b& c\\ a& q& c\\ a& b& r\end{array}\right|=0$

$R_1\to R_1-R_3$

$\left|\begin{array}{ccc}p-a& 0& c-r\\ a& q& c\\ a& b& r\end{array}\right|=0$

$R_2\to R_2-R_3$

$\left|\begin{array}{ccc}p-a& 0& c-r\\ 0& q-b& c-r\\ a& b& r\end{array}\right|$

$p-a\left(r\right(q-b)-b(c-r)+(c-r\left)\right(-a(q-b))$

$R_3\to R_1+R_3$

$\left|\begin{array}{ccc}p-a& 0& c-r\\ 0& q-b& c-r\\ p& b& c\end{array}\right|=0$

$\left|\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ \frac{p}{p-a}& \frac{b}{q-b}& \frac{c}{c-r}\end{array}\right|=0$

$(\frac{c}{c-r}-\frac{b}{q-b})+\left(\frac{-p}{p-a}\right)=0$

$\frac{p}{p-a}=\frac{c}{c-r}-\frac{b}{q-b}$ substituting gets value $=-2$