Question

If $a\ne p,b\ne q,c\ne r$ and $\left|\begin{array}{ccc}p& b& c\\ p+a& q+b& 2c\\ a& b& r\end{array}\right|=0$, then $\frac{p}{p-q}+\frac{q}{q-b}+\frac{r}{r-c}=$

• A. 3
• B. 2
• C. 1

Answer 1

The correct option is B 2

$\left|\begin{array}{ccc}p& b& c\\ p+a& q+b& 2c\\ a& b& r\end{array}\right|=0$ Applying $R_2\to R_2-R_1$
$=\left|\begin{array}{ccc}p& b& c\\ a& q& c\\ a& b& r\end{array}\right|=0$
Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$=\left|\begin{array}{ccc}p& b& c\\ a-p& q-b& 0\\ a-p& 0& r-c\end{array}\right|=0$
On expansion we get,
$p(q-b)(r-c)-b(a-p)(r-c)-c(q-b)(a-p)=0\Rightarrow (p-a)(q-b)(r-c)[\frac{p}{(p-a)}+\frac{b}{(q-b)}+\frac{c}{(r-c)}]=0\Rightarrow (p-a)(q-b)(r-c)[\frac{p}{(p-a)}+\frac{q}{(q-b)}-1+\frac{r}{(r-c)}-1]=0\because p\ne a,q\ne b,r\ne c\therefore \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=2.$