If $a,b,c$ are the roots of the equation
$x^3-3px^2+3qx-1=0$,
then the centroid of the triangle with vertices
$\left(a,\frac1a\right),\left(b,\frac1b\right),\left(c,\frac1c\right)$
is the point
🎥 Video solution / Text Solution of this question is given below:
Centroid $=\left(\dfrac{a+b+c}{3},\dfrac{\frac1a+\frac1b+\frac1c}{3}\right)$
From the equation,
$a+b+c=3p$
Also
$\dfrac1a+\dfrac1b+\dfrac1c=\dfrac{ab+bc+ca}{abc}=\dfrac{3q}{1}=3q$
Hence centroid $=(p,q)$
Answer: $\boxed{(p,q)}$