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A straight line L with negative slope passes through the point (8,2) and cuts the positive coordinate axes at points P and Q. The absolute minimum value of OP + OQ as L varies, where O is the origin, is

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answer is 18.

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Detailed Solution

Let the equation of line be (y -2) - m(x - 8) where m <0⇒ P≡(8−2m,0)andQ≡(0,2−8m) Now,OP+OQ=|8−2m|+|2−8m| =10+2−m+(−8m) ≥10+22−m×(−8m)≥18

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