Let A be the set of all points (α,β) such that the area of triangle formed by the points (5,6),(3,2) and (α,β) is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is
A
8√5
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B
16√5
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C
4√5
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D
12√5
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Solution
The correct option is A8√5 12∣∣
∣∣αβ1561321∣∣
∣∣=12 ⇒|4α−2β−8|=24 ⇒|2α−β−4|=12
Locus =2x−y−4=12,2x−y−4=−12 ⇒2x−y−16=0 or 2x−y+8=0
Required length = minimum perpendicular distance from origin =min{16√5,8√5}=8√5