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Question

The value of a for which the lines x=1, y=2 and a2x+2y20=0 are concurrent is:


A

1

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B

4

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C

-1

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D

-2

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Solution

The correct option is B

4


Since x=1, y=2 and a2x+2y20=0 are concurrent
x=1 and y=2 lie on the line a2x+2y20=0
x=1 and y=2 is a solution of given equation.
On substituting x=1 and y=2 in equation a2x+2y20=0, we get
a2×1+2×220=0a216=0
a2=16a=4 or 4


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