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The vertices of a triangle are $(pq, 1 / (pq)), (qr, 1 / (qr)) and (rq, 1 / (rq)),$ where p,q and r are the roots of the equation $y^{3} - 3 y^{2} + 6 y + 1 = 0$ . The coordinates of its centroid are

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(1,2)

(2,-1)

(1,-1)

(2,3)

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Correct option is B

p,q,r are the roots of equation y3−3y2+6y+1=0. So, p+q+r=3,pq+qr+rp=6, and pqr=-1. Now, the centroid of the triangle is (pq+qr+rp3,1pq+1qr+1rp3)i.e., (pq+qr+rp3,p+q+r3pqr)≡(63,3−3)or (2,−1)

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The vertices of a triangle are (pq,1/(pq)),(qr,1/(qr)) and (rq,1/(rq)),where p,q and r are the roots of the equation y3−3y2+6y+1=0. The coordinates of its centroid are

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The vertices of a triangle are $(pq, 1 / (pq)), (qr, 1 / (qr)) and (rq, 1 / (rq)),$ where p,q and r are the roots of the equation $y^{3} - 3 y^{2} + 6 y + 1 = 0$ . The coordinates of its centroid are