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 $\left(\frac{1}{1-a}\right)+\left(\frac{1}{1-b}\right)+\left(\frac{1}{1-c}\right)=$

• A. $0$
• B. $1$
• C. $\frac{1}{(a+b+c)}$
• D. None of these

Answer 1

Answer: (B)

$1$

The given lines are concurrent. Therefore

$\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=0$

$\left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0$
Applying $C_2\Rightarrow C_2-C_1$ and $C_3\Rightarrow C_3-C_1$
we get
$\left|\begin{array}{ccc}a& 1-a& 1-a\\ 1& b-1& 0\\ 1& 0& c-1\end{array}\right|$
Expanding, we get
$a(b-1)(c-1)-(c-1)(1-a)-(b-1)(1-a)=0$
Dividing by $(1-a)(1-b)(1-c)$ we get
$\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$
Adding $1$ on both the sides we get
$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$