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 $\Delta ABC$, where $\alpha ,\beta$ are the roots of the equation $x^2-6p_{1}x+2=0,\beta ,\gamma$ are the roots of the equation $x^2-6p_{2}x+3=0$ and $\gamma ,\alpha$ are the roots of the equation $x^2-6p_{3}x+6=0,p_1,p_2,p_3$ being positive. Then, the coordinates of the centroid of $\Delta ABC$ 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
$x^2-6p_{1}x+2=0$
$\therefore \alpha +\beta =6p_1,\alpha \beta =2\dots$(i)
$\beta ,\gamma$ are the roots of the equation
$x^2-6p_{2}x+3=0$.
$\therefore \beta +\gamma =6p_2,\beta \gamma =3\dots$(ii)
$\gamma ,\alpha$ are the roots of the equation
$x^2-6p_{3}x+6=0$
$\therefore \gamma +\alpha =6p_3,\gamma \alpha =6$
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 $\alpha \beta \gamma =6$ ;
$\alpha =\alpha =6\alpha \beta \gamma =6$

$\Rightarrow \beta =1$
$\therefore \alpha +\beta =6p_1$
$\Rightarrow 3=6p_1$
$\Rightarrow p_1=\frac{1}{2}$
$\beta +\gamma =6p_2$
$\Rightarrow 4=6p_2$
$\Rightarrow p_2=\frac{2}{3}$
and $\gamma +\alpha =6p_3$
$\Rightarrow 5=6p_3$
$\Rightarrow p_3=\frac{5}{6}$
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 $\left(2,\frac{11}{18}\right)$