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 y3−3y2+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)=(2,3−3)=(2,−1)