If the lines `a x+y+1=0,x+b y+1=0` and `x+y+c=0` are concurrent `(a!=b!=c!=1)` , prove that `1/(1-a)+1/(1-b)+1/(1-c)=1` .

Explanation

Given,the lines
`ax+y+1=0,x+by+1=0, and x+y+c=0` are concurrent, then

`|[ a ,1 , 1 ],[1,b,1],[1,1,c]|`

Performing the operations `R_2->R_2−R_1` and `R_3->R_3−R_1`

`|[a,1-a,1-a],[1,b-1,0],[1,0,c-1]|=0`

Expanding,

`⇒a(b−1)(c−1)−(1−a)(c−1)+(1−a)(−(b−1))=0`

Dividing the equation by `(1−a)(1−b)(1−c)`

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