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If the lines $ax + y + 1 = 0$, $x + by + 1 = 0$ and $x + y + c = 0$ (where $a, b$ and $c$ are distinct and different from $1$ ) are concurrent, then the value of $\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}} =$
$-1$
$1$
$0$
$2$
Solution
For concurrent lines, $\left| {\begin{array}{*{20}{c}}
a&1&1 \\
1&b&1 \\
1&1&c
\end{array}} \right|\, = \,0$
Applyling $c_2-c_1$ and $c_3-c_1$, we get
$\Delta \, = \,\left| {\begin{array}{*{20}{c}}
a&{1 – a}&{1 – a} \\
1&{b – 1}&0 \\
1&0&{c – 1}
\end{array}} \right|\, = \,0$
$a(b – 1)\,(c – 1)\, – \,(1 – a)\,(c – 1)$ $\, – (1 – a)\,(b – 1)\, = \,0$
Dividing $(1 – a)\,(1 – b)\,(1 – c)$, we get
$\frac{a}{{1 – a}}\, + \,\frac{1}{{1 – b}}\, + \,\frac{1}{{1 – c}} = 0$
$\frac{1}{{1 – a}}\, + \,\frac{1}{{1 – b}}\, + \,\frac{1}{{1 – c}} = \frac{{1 – a}}{{1 – a}}\, = 1$
