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Question

If the straight line x(a+2b)+y(a+3b)=a+b passes through a fixed point for different values of a and b, then the fixed point is

A
(2,1)
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B
(2,1)
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C
(1,2)
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D
(1,2)
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Solution

The correct option is A (2,1)
Given equation is
x(a+2b)+y(a+3b)=a+ba(x+y1)+b(2x+3y1)=0(1)

λ1L1+λ2L2=0 represents family of straight lines which will always pass through the point of intersection of lines L1=0,L2=0
i.e., x+y1=0 and 2x+3y1=0,
Solving above equations, we get
y=1,x=2

Hence, the fixed point is (2,1).

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