 
                 
                InterviewSolution
 Saved Bookmarks
    				| 1. | 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)`A. `(1,2)`B. `(2,-1)`C. `(1,-1)`D. `(2,3)` | 
| Answer» Correct Answer - B 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)` | |