The vertices of a triangle are `(p q ,1/(p q)),(p q)),(q r ,1/(q r)),` and `(r q ,1/(r p)),` where `p ,q` and `r` are the roots of the equation `y^3=3y^2+6y+1=0` . The coordinates of its centroid are `(1,2)` (b) `2,-1)` `(1,-1)` (d) `2,3)`

Options

A. `(1,2)`
B. `(2,-1)`
C. `(1,-1)`
D. `(2,3)`

Answer

B

Explanation

p,q,r, are the roots of equation `y^3-3y^2+6y+1=0`. So, `p+q+r=3,pq+qr+rp=6`, and `pqr=-1`. Now, the centroid of the triangle is
`((pq_+qr+rp)/(3),((1)/(pq)+(1)/(qr)+(1)/(rp))/(3))`
i.e., `((pq_+qr+rp)/(3),(p_+q+r)/(3pqr))-=((6)/(3),(3)/(6))or (2,-1)`