Question

If the lines ax +y +1 =0, x +by +1 =0 and x +y +c = 0(a, b, c being distinct and different from 1) are concurrent, then $\left(\frac{1}{1-a}\right)+\left(\frac{1}{1-b}\right)+\left(\frac{1}{1-c}\right)$=

Answer 1

If the given lines are concurrent, then
$\left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0$
Applying $C_2\to C_2-C_1\text{and}{C}_3\to C_3-C_1$
$\Rightarrow \left|\begin{array}{ccc}a& 1-a& 1-a\\ 1& b-1& 0\\ 1& 0& c-1\end{array}\right|=0$
Expanding along $C_1$
$\Rightarrow a(b-1)(c-1)-(c-1)(1-a)-b(b-1)(1-a)=0$
$\Rightarrow \frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$
[Dividing by (1-a)(1-b)(1-c)]
Adding 1 on both sides, we get
$\Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$
Hence, the correct option is (B).