Question

Let $p,q$ and $r$ be the roots of the equation $y^3-3y^2+6y+1=0$. If the vertices of a triangle are $(pq,\frac{1}{pq}),$ $(qr,\frac{1}{qr})$ and $(rp,\frac{1}{rp}),$ then the coordinates of its centroid are

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

Answer 1

The correct option is B $(2,-1)$

Given : $p,q,r$ are the roots of equation $y^3-3y^2+6y+1=0.$
So, we get
$p+q+r=3pq+qr+rp=6pqr=-1$

Vertices of the triangle are $(pq,\frac{1}{pq}),(qr,\frac{1}{qr})$ and $(rp,\frac{1}{rp})$

Now, the centroid of the triangle
$=(\frac{pq+qr+rp}{3},\frac{\frac{1}{pq}+\frac{1}{qr}+\frac{1}{rp}}{3})=(\frac{6}{3},\frac{p+q+r}{3pqr})=(2,\frac{3}{-3})=(2,-1)$