Question

Let $A\left(\alpha, \frac{1}{\alpha}\right), B\left(\beta, \frac{1}{\beta}\right), C\left(\gamma, \frac{1}{\gamma}\right)$ be the vertices of a $\Delta A B C$, where $\alpha, \beta$ are the roots of the equation $x^{2}-6 p_{1} x+2=0, \beta, \gamma$ are the roots of the equation $x^{2}-6 p_{2} x+3=0$ and $\gamma, \alpha$ are the roots of the equation $x^{2}-6 p_{3} x+6=0, p_{1}, p_{2}, p_{3}$ being positive. Then, the coordinates of the centroid of $\Delta A B C$ is

• A. $\left(1, \frac{11}{18}\right)$
• B. $\left(0, \frac{11}{8}\right)$
• C. $\left(2, \frac{11}{8}\right)$
• D. None of these

Answer 1

Answer: (C)

It is given that $\alpha, \beta$ are the roots of the equation
$x^{2}-6 p_{1} x+2=0$
$\therefore \alpha+\beta=6 p _{1}, \alpha \,\beta=2 \dots$(i)
$\beta, \gamma$ are the roots of the equation
$x^{2}-6 p_{2} x+3=0$.
$\therefore \beta+\gamma=6 p _{2}, \beta \gamma=3 \dots$(ii)
$\gamma, \alpha$ are the roots of the equation
$x^{2}-6 p_{3} x+6=0$
$\therefore \gamma+\alpha=6 p _{3}, \gamma \alpha=6$
From Eqs. (i), (ii) and (iii), we get
$\Rightarrow \alpha\, \beta\, \gamma=6$
$[\therefore \alpha, \beta, \gamma > 0]$
Now, $\alpha \beta=2$ and $\alpha \beta \gamma=6 $
$\Rightarrow \gamma=3$
$\beta \gamma=3$ and $\alpha \beta \gamma=6$ ;
$ \alpha=\alpha=6 \alpha \beta \gamma=6 $

$\Rightarrow \beta=1$
$\therefore \alpha+\beta=6 p _{1} $
$\Rightarrow 3=6 p _{1}$
$ \Rightarrow p _{1}=\frac{1}{2}$
$\beta+\gamma=6 p _{2} $
$\Rightarrow 4=6 p _{2} $
$\Rightarrow p _{2}=\frac{2}{3}$
and $\gamma+\alpha=6 p _{3}$
$ \Rightarrow 5=6 p _{3}$
$ \Rightarrow p _{3}=\frac{5}{6}$
The coordinates of the centroid of triangle are
$\left(\frac{\alpha+\beta+\gamma}{3}, \frac{1}{3}\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right)\right) $ or
$\left(\frac{6}{3}, \frac{1}{3}\left(\frac{1}{2}+1+\frac{1}{3}\right)\right)$ or $\left(2, \frac{11}{18}\right)$