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 :

Step-by-step solution

Given:Area of triangle formed by 5 , 6 , 3 , 2   &   α , β = 12   square   units ∴ 1 2 α β 1 5 6 1 3 2 1 = 12 ⇒ α β 1 5 6 1 3 2 1 = 24 ⇒ α 6 - 2 - β 5 - 3 + 1 10 - 18 = ± 24 ⇒ 4 α - 2 β - 8 = ± 24 ⇒ 4 α - 2 β = 32 ,   4 α - 2 β + 16 = 0 ⇒ 2 α - β - 16 = 0 ,   2 α - β + 8 = 0 So point will be either α , 2 α + 8   ,   α , 2 &