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Question

A straight line L with negative slope, passes through the point (12,3) and cuts the positive coordinate axes at points P and Q. As L varies, the absolute minimum value of OP+OQ is (O is the origin)

A
10
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B
18
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C
16
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D
27
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Solution

The correct option is D 27
Let m<0 be the slope of the line L
Equation of line L is :
y3=m(x12)
mx12m3y12m3=1
OP=123m, OQ=312m
OP+OQ=15+(12m3m)
15+2(12m)(3m) (AMGM)
=15+2×6
=27

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