A straight line passing through P(3,1) meets the coordinates axes at A and B. It is given that distance of this straight line from the origin O is maximum. Area of triangle △OAB is equal to :
503sq. units
253sq. units
203sq. units
1003sq. units
A
503sq. units
B
253sq. units
C
203sq. units
D
1003sq. units
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Solution
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Line AB will be farthest from the origin if OP is right angled to loine drawn mOP=13⇒mAB=−3 Thus, the equation of AB is (y−1)=−3(x−3) ⇒A=(103,0),B=(0,10) ⇒ΔOAB=12(OA)(OB)=12×103×10=1006=503
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