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Questions  

The y-axis and the lines  (a5  2a3 )x + (a + 2)y + 3a = 0  and (a5  3a2 )x + 4y + a  2 = 0 are concurrent for 

a
Two values of a
b
Three values of a
c
Five values of a
d
no value of a

detailed solution

Correct option is A

Let the lines intersect at the point k then (a+2)k+3a=0 and 4k+a−2=0 ⇒3aa+2=a−24⇒a2−12a+4=0which gives us two values of a

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