Question

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

• A. $0$
• B. $1$
• C. $\frac{1}{a+b+c}$
• D. $a+b+c$

Answer 1

The correct option is B $1$

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

$\left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0$
Performing the operations $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$\Rightarrow \left|\begin{array}{ccc}a& 1-a& 1-a\\ 1& b-1& 0\\ 1& 0& c-1\end{array}\right|=0\Rightarrow a(b-1)(c-1)-(1-a)(c-1)+(1-a)(-(b-1\left)\right)=0$
Dividing the equation by $(1-a)(1-b)(1-c)$
$\Rightarrow \frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0\Rightarrow -\frac{1-a-1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0\Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$