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Question

In a right angled isosceles ABC, where A is at origin and side lengths AB and AC are equal to a units. If point B and C are produced to P and Q respectively such that BPCQ=AB2, then the locus of the midpoint of PQ is

A
1x+1y=4a
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B
1x+1y=12a
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C
1x+1y=1a
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D
1x+1y=2a
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Solution

The correct option is D 1x+1y=2a
Assuming A as the origin, B=(a,0) and C=(0,a) as shown in the below figure,
Let the coordinates of points P=(h,0) and Q=(0,k)
Midpoint of PQ is
M=(α,β)=(h2,k2)

Given, BP.CQ=AB2
(ha)(ka)=a2hkakah+a2=a2ak+ha=hk1h+1k=1a1α+1β=2a (h=2α,k=2β)

Hence, the locus of the midpoint is
1x+1y=2a

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