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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 value of \({1\over 1-a}+{1\over 1-b}+{1\over 1-c}\) 

a) -1
b) 0
c) 1
d) 2
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The correct answer is option c) 1

if the given lines are concurrent  , then \( \begin{array}{|ccc|cc|} a& 1 & 1 \\ 1 & b & 1\\ 1 &1 & c\\ \end{array}=0 \)  

Applying { \(C_2\rightarrow C_2-C_1 and \space C_3\rightarrow C_3-C_1\)

\( \begin{array}{|ccc|cc|} a& 1 -a& 1 -a\\ 1 & b-1 & 0\\ 1 &0 & c-1\\ \end{array}=0 \)

= a(b-1)(c-1) -(c-1)(1-a)-(b-1)(1-a)=0

=\({a\over 1-a}+{1\over 1-b}+{1\over 1-c}\) 

(dividing by (1-a )(1-b)(1-c)) 

\(={1\over 1-a}+{1\over 1-b}+{1\over 1-c}=1\)

 

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