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 equal to

• A. 1
• B. 2
• C. 3
• D. c

Answer 1

Answer: (B)

2

$\left|\begin{array}{ccc}p& b& c\\ a& q& c\\ a& b& r\end{array}\right|=0$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$\left|\begin{array}{ccc}p-a& b-q& 0\\ 0& q-b& c-r\\ a& b& r\end{array}\right|=0$
$(p-a)\left[r\right(q-b)-b(c-r\left)\right]-(b-q)[-a(c-r\left)\right]=0$
$r(p-a)(q-b)+b(p-a)(r-c)+a(q-b)(r-c)=0$
$\Rightarrow \frac{r}{r-c}+\frac{b}{q-b}+\frac{a}{p-a}=0$
$\Rightarrow \frac{r}{r-c}=-\frac{b}{q-b}-\frac{a}{p-a}$
Now, consider $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$
$=\frac{p}{p-a}+\frac{q}{q-b}-\frac{b}{q-b}-\frac{a}{p-a}$
$=1+1=2$