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If the lines: ax+y+1=0, x+by+1=0, x+y+c=0 are concurrent, the value of a/(a-1) + b/(b-1) + c/(c-1) is

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2 years ago

```							Dear student ax+y+1=0....(1)orabx+by+b=0.....(3)x+by+1=0....(4)orax+aby+a=0....(2)Subtracting equation 1 and 2, (1−ab)y+1−a=0y = a−11−abSubtracting equation 3 and 4, (ab−1)x+b−1=0x = 1−bab−1Now if these three are concurrent so values of x and y will satisfy third equation, 1−bab−1+a−11−ab+c=0b−1+a−11−ab+c=0a+b−2+c−abc=0....(5)Now, 11−a+11−b+11−c=1−b−c+bc+1+ac−a−c+1+ab−a−b(1−a)(1−b)(1−c)=3−2(a+b+c)+ab+bc+ca(1−a)(1+bc−b−c)=3−2(2+abc)+ab+bc+ca(1+bc−(b+c+a)−abc+ab+ac) [Using equation 5]=−1−2abc+ab+bc+ca(1+bc−(2+abc)−abc+ab+ac)=−1−2abc+ab+bc+ca−1−2abc+ab+bc+ca=1  RegardsArun
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2 years ago
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