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Question

If line x+αy+1=0 is perpendicular to the line 2xβy+1=0 and parallel to the line x(β3)y1=0, then the value of |αβ| is

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Solution

Let the lines be
l1:x+αy+1=0l2:2xβy+1=0l3:x(β3)y1=0
Slope of the lines are
m1=1α, m2=2β, m3=1β3
From the given conditions,
m1×m2=12αβ=1αβ=2(1)m1=m31α=1β3α+β=3(2)
Now,
|αβ|=(α+β)24αβ|αβ|=98=1

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