If the lines `a x+y+1=0,x+b y+1=0a n dx+y+c=0(a ,b ,c` being distinct and different from `1)` are concurrent, then prove that `1/(1-a)+1/(1-b)+1/(1-c)=1.`

Options

A. 0
B. 1
C. 1/(a+b+c)
D. none of these

Answer

B

Explanation

If the given lines are concurrent, then
`|{:(a,1,1),(1,b,1),(1,1,c):}| = 0`
`"or "|{:(a,1-a,1-a),(1,b-1,0),(1,0,c-1):}| = 0`
`" "("Applying "C_(2)to C_(2)-C_(1) "and "C_(3)toC_(3)-C_(1))`
or a(b-1)(c-1)-(c-1)(1-a)-(b-1)(1-a)=0
`"or "(a)/(1-a) + (1)/(1-b)+(1)/(1-c) = 0`
`" "["Dividing by"(1-a)(1-b)(1-c)]`
Adding 1 on both sides, we get
`(1)/(1-a) + (1)/(1-b)+(1)/(1-c) =1 `