Let $a, b$ and $c$ be distinct and none of them is equal to 1 . If the lines $x+a y+a=0$, $b x+y+b=0$ and $c…
Let $a, b$ and $c$ be distinct and none of them is equal to 1 . If the lines $x+a y+a=0$, $b x+y+b=0$ and $c x+c y+1=0$ are concurrent, then the value of $\frac{a}{a-1}+\frac{b}{b-1}+\frac{c}{c-1}$ is
1
-1
2
0
Solution
Given equations are
$\begin{aligned}
& x+a y+a=0 \\
& b x+y+b=0 \text { and } c x+c y+1=0
\end{aligned}$
The above three lines are concurrent, so
$\begin{aligned}
& \left|\begin{array}{lll}
1 & a & a \\
b & 1 & b \\
c & c & 1
\end{array}\right|=0 \\
\Rightarrow & 1(1-b c)-a(b-b c)+a(b c-c)=0 \\
\Rightarrow & 1-b c-a b+a b c+a b c-a c=0 \\
\Rightarrow & 1+2 a b c=a b+b c+a c \\
\text { Now, } \frac{a}{a-1}+\frac{b}{b-1}+\frac{c}{c-1} & (a-1)(b-1)(c-1) \\
= & \frac{a(b-1)(c-1)+b(a-1)(c-1)+c(a-1)(b-1)}{a c-a b-a c+b+a b c} \\
= & \frac{a b c-a b-a c+a+a b c-a b-b c+b c+c}{a b c+a+b+c-a b-b c-a c-1} \\
= & \frac{3 a b c-2(a b+b c+a c)+a+b+c}{a b c+a+b+c-(a b+b c+a c)-1} \\
= & \frac{3 a b c-2(1+2 a b c)+a+b+c}{a b c+a+b+c-(1+2 a b c)-1} \\
= & \frac{3 a b c-2-4 a b c+a+b+c}{a b c+a+b+c-1-2 a b c-1} \\
= & \frac{-a b c+a+b+c-2}{-a b c+a+b+c-2}
\end{aligned}$