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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 BP⋅CQ=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=2aAssuming 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 ⇒(h−a)(k−a)=a2⇒hk−ak−ah+a2=a2⇒ak+ha=hk⇒1h+1k=1a⇒1α+1β=2a (∵h=2α,k=2β) Hence, the locus of the midpoint is 1x+1y=2a

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