If $a
eq p, b
eq q, c
eq r$ and $\left|\begin{array}{lll}p & b & c \ a & q & c \ a & b & r\e

Question

If $a 
eq p, b 
eq q, c 
eq r$ and $\left|\begin{array}{lll}p & b & c \ a & q & c \ a & b & r\end{array}ight|=0$ Then find the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$

Professor PC Thomas Classes · Answer

$\left|\begin{array}{lll}p & b & c \ a & q & c \ a & b & r\end{array}ight|=0$ The operation $R_{1} ightarrow R_{1}-R_{3}$ and $R_{2} ightarrow R_{2}-R_{3}$ gives $$ \left|\begin{array}{ccc} p-a & 0 & c-r \ 0 & q-b & c-r \ a & b & r \end{array}ight|=0 $$ On solving this determinant, we get $$ \begin{aligned} & (q-b)[(p-a) r-a(c-r)] \ & -[(p-a)(c-r)-0] 	imes b=0 \ & (q-b)(p-a) r-(q-b)(c-r) a \ & -(p-a)(c-r) b=0 \end{aligned} $$ Now $b=(q-(q-b))$ and $a=(p-(p-a))$ $$ \begin{aligned} &=(q-b)(p-a) r+(q-b)(r-c)(p-p+a) \ &+(p-a)(r-c)(q-q+b)=0 \ &(r-c)+(p-a)(r-c)(q) \ &-(q-b)(p-a)(r-c)=0 \end{aligned} $$ Now dividing throughout by\ $(p-a)(q-b)(r-c)$, we get $$ \Rightarrow \frac{r}{r-c}+\frac{p}{p-a}+\frac{q}{q-b}=2 $$