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 If the straight lines x+y2=0,2xy+1=0, and ax+byc=0 are concurrent, then the family of lines 

2ax+3by+c=0(a,b,c are nonzero ) is concurrent at 

a
(2,3)
b
(1/2,1/3)
c
(-1/6 ,- 5/9)
d
(2/3,-7/5)

detailed solution

Correct option is C

Given that three lines are concurrent hence,11−22−11ab−c=0Expand the determinant, we get or a+5b−3c=0 or −a3−53b+c=0 ---(1)2ax+3by+c=0----(2)  comparing (1) and (2)   hence the straight lines are concurrent at 2x=−13 and  3y=−53 So, x=−16,y=−59

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