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Question

Let p,q and r be the roots of the equation y3−3y2+6y+1=0. If the vertices of a triangle are (pq,1pq), (qr,1qr) and (rp,1rp), then the coordinates of its centroid are

A
(1,2)
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B
(2,1)
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C
(1,1)
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D
(2,3)
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Solution

The correct option is B (2,1)
Given : p,q,r are the roots of equation y33y2+6y+1=0.
So, we get
p+q+r=3pq+qr+rp=6pqr=1

Vertices of the triangle are (pq,1pq),(qr,1qr) and (rp,1rp)

Now, the centroid of the triangle
=⎜ ⎜ ⎜pq+qr+rp3,1pq+1qr+1rp3⎟ ⎟ ⎟=(63,p+q+r3pqr)=(2,33)=(2,1)

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