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Question

If the lines x+ay+a=0, bx+y+b=0, cx+cy+1=0(abc1) are concurrent, then the value of aa1+bb1+cc1, is

A
1
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B
0
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C
1
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D
3
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Solution

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,

(ab1)y+abb=0

y=babab1 & x=a(b1ab1)

From 1 & 3,

(acc)y+ac1=0

y=1acacc & x=a(1cacc)

So,

babab1=1acacc

abccb+abca2bc=aba2bc1+ac

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

Now,

aa1+bb1+cc1=aba+abbabab+1+c(c1)

=2abcacbc2ab+a+b+abcacbc+cabcacbc+cab+a+b1

=3abc2(ac+bc+ab)+a+b+cabc(ac+bc+ab)+a+b+c1

=abc(ac+bc+ab)+(a+b+c)1abc(ac+bc+ab)+(a+b+c)1=1 [From (4)]


Hence, option C.


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