Question

If the lines $ax+y+1=0,x+by+1=0$ & $x+y+c=0$ where a, b & c are distinct real numbers different from 1 are concurrent, then the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (D)

1

Lines are concurrent when area of triangle formed with these lines is zero
$\left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0\Rightarrow a(bc-1)-1(c-1)+1(1-b)=0\Rightarrow abc-a-c+1+1-b=0\Rightarrow a+b+c-2=abc\Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$