Question

If the lines $x+ay+a=0,bx+y+b=0,cx+cy+1=0(a\ne b\ne c\ne 1)$ are concurrent, then the value of $\frac{a}{a-1}+\frac{b}{b-1}+\frac{c}{c-1}$, is

• A. $-1$
• B. $0$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

Given $x+ay+a=0$ ...(1)

$bx+y+b=0$ ...(2)

$cx+cy+1=0$ ...(3)

These lines are concurrent means they have only one intersection point

From 1 & 2,

$(ab-1)y+ab-b=0$

$y=\frac{b-ab}{ab-1}$ & $x=-a\left(\frac{b-1}{ab-1}\right)$

From 1 & 3,

$(ac-c)y+ac-1=0$

$y=\frac{1-ac}{ac-c}$ & $x=-a\left(\frac{1-c}{ac-c}\right)$

So,

$\frac{b-ab}{ab-1}=\frac{1-ac}{ac-c}$

$\Rightarrow abc-cb+abc-a^{2}bc=ab-a^{2}bc-1+ac$

$\Rightarrow 1+2abc=ab+ac+bc$ ...(4)

Now,

$\frac{a}{a-1}+\frac{b}{b-1}+\frac{c}{c-1}=\frac{ab-a+ab-b}{ab-a-b+1}+\frac{c}{(c-1)}$

$=\frac{2abc-ac-bc-2ab+a+b+abc-ac-bc+c}{abc-ac-bc+c-ab+a+b-1}$

$=\frac{3abc-2(ac+bc+ab)+a+b+c}{abc-(ac+bc+ab)+a+b+c-1}$

$=\frac{abc-(ac+bc+ab)+(a+b+c)-1}{abc-(ac+bc+ab)+(a+b+c)-1}=1$ [From (4)]

Hence, option C.