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Question

If the lines x+y+1=0, 4x+3y+4=0 and x+αy+β=0, where α2+β2=2, are concurrent then

A
α=1,β=1
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B
α=1,β=±1
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C
α=1,β=±1
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D
α=±1,β=1
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Solution

The correct option is D α=1,β=1

Consider the given lines are

x+y+1=0......(1)

4x+3y+4=0.......(2)

x+αy+β=0.......(3)

andwhereα2+β2=2

Then solve by matrix form,(1), (2) and (3) to, we get

1 1 1

4 3 4

1 α β

1(3β4α)1(4β4)+1(4α3)=0

3β4α4β+4+4α3=0

β+1=0

β=1

Put the value of β in given equation.

α2+β2=2

α2+12=2

α2=1

α=±1

Hence, this is the amswer.

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