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 the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is:

• A. $2$
• B. $0$
• C. $3$
• D. $1$

Answer 1

The correct option is D $1$

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 the transformation $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$
$\Rightarrow \left|\begin{array}{ccc}a& 1-a& 1-a\\ 1& b-1& 0\\ 1& 0& c-1\end{array}\right|=0$
$\Rightarrow a(b-1)(c-1)-(c-1)(1-a)-(b-1)(1-a)=0$
Dividing by $(1-a)(1-b)(1-c)$, we have:
$\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$
$\Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$