If the lines x+ay+a=0, bx+y+b=0, cx+cy+1=0 (a,b,c being distinct and ≠ 1) are concurrent,then the value of aa−1+bb−1+cc−1 is

Given lines x+ay+a=0 ⇒ a(xa+y+1)=0⇒ xa+y+1=0−−(1)bx+y+b=0⇒ x+yb+1=0−−(2)cx+cy+1=0⇒ x+y+1c=0−−(3)(1)−(3)⇒ xa−x+1−1c=0⇒ x(1−aa)=(1−cc)⇒ x=(1−c1−a)(ac)(2)−(3)⇒ yb−y+a−1c=0⇒ y(1−bb)=(1−cc)y=(1−c1−b)(bc)from(3) (1−c1−a)(ac)+(1−c1−b)(bc)+1c=0⇒ (c−1a−1)(ac)+(c−1b−1)(bc)+1c=0⇒ c−1c(aa−1+bb−1+1c−1)=0⇒ aa−1+bb−1+cc−1+1=1⇒ aa−1+bb−1=cc−1=1

If the lines x+ay+a=0, bx+y+b=0, cx+cy+1=0 (a,b,c being distinct and ≠ 1) are concurrent,then the value of aa−1+bb−1+cc−1 is

Given lines x+ay+a=0 ⇒ a(xa+y+1)=0⇒ xa+y+1=0−−(1)bx+y+b=0⇒ x+yb+1=0−−(2)cx+cy+1=0⇒ x+y+1c=0−−(3)(1)−(3)⇒ xa−x+1−1c=0⇒ x(1−aa)=(1−cc)⇒ x=(1−c1−a)(ac)(2)−(3)⇒ yb−y+a−1c=0⇒ y(1−bb)=(1−cc)y=(1−c1−b)(bc)from(3) (1−c1−a)(ac)+(1−c1−b)(bc)+1c=0⇒ (c−1a−1)(ac)+(c−1b−1)(bc)+1c=0⇒ c−1c(aa−1+bb−1+1c−1)=0⇒ aa−1+bb−1+cc−1+1=1⇒ aa−1+bb−1=cc−1=1

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$I f t h e l i n e s x + a y + a = 0, b x + y + b = 0, c x + c y + 1 = 0 (a, b, c b e i n g d i s t i n c t a n d \neq 1) a r e c o n c u r r e n t, t h e n t h e v a l u e o f \frac{a}{a - 1} + \frac{b}{b - 1} + \frac{c}{c - 1} i s$

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Detailed Solution

$\begin{array}{l} G i v e n l i n e s x + a y + a = 0 \Rightarrow a (\frac{x}{a} + y + 1) = 0 \\ \Rightarrow \frac{x}{a} + y + 1 = 0 - - (1) \\ b x + y + b = 0 \Rightarrow x + \frac{y}{b} + 1 = 0 - - (2) \\ c x + c y + 1 = 0 \Rightarrow x + y + \frac{1}{c} = 0 - - (3) \\ (1) - (3) \Rightarrow \frac{x}{a} - x + 1 - \frac{1}{c} = 0 \\ \Rightarrow x (\frac{1 - a}{a}) = (\frac{1 - c}{c}) \\ \Rightarrow x = (\frac{1 - c}{1 - a}) (\frac{a}{c}) \end{array}$

$\begin{array}{l} (2) - (3) \Rightarrow \frac{y}{b} - y + a - \frac{1}{c} = 0 \\ \Rightarrow y (\frac{1 - b}{b}) = (\frac{1 - c}{c}) \\ y = (\frac{1 - c}{1 - b}) (\frac{b}{c}) \\ f r o m (3) (\frac{1 - c}{1 - a}) (\frac{a}{c}) + (\frac{1 - c}{1 - b}) (\frac{b}{c}) + \frac{1}{c} = 0 \\ \Rightarrow (\frac{c - 1}{a - 1}) (\frac{a}{c}) + (\frac{c - 1}{b - 1}) (\frac{b}{c}) + \frac{1}{c} = 0 \\ \Rightarrow \frac{c - 1}{c} (\frac{a}{a - 1} + \frac{b}{b - 1} + \frac{1}{c - 1}) = 0 \\ \Rightarrow \frac{a}{a - 1} + \frac{b}{b - 1} + \frac{c}{c - 1} + 1 = 1 \\ \Rightarrow \frac{a}{a - 1} + \frac{b}{b - 1} = \frac{c}{c - 1} = 1 \end{array}$