Question

The vertices of a triangle are $\left( pq,1/\left( pq \right) \right),\left( qr,1/\left( qr \right) \right)$ and $\left( rq,1/\left( rq \right) \right)$, where $p,q$ and $r$ are the roots of the equation ${{y}^{3}}-3{{y}^{2}}+6y+1=0$. The coordinates of its centroid are
$A.\text{ }\left( 1,2 \right)$
$B.\text{ }\left( 2,-1 \right)$
$C.\text{ }\left( 1,-1 \right)$
$D.\text{ }\left( 2,3 \right)$

Answer 1

Hint: In this question we have been given with the three vertices of a triangle which constitute $p,q,r$ which are the roots of the equation ${{y}^{3}}-3{{y}^{2}}+6y+1=0$. Given this information, we have to find the coordinates of the centroid of the triangle. We will use the formula $C\equiv \left( \dfrac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}}{3},\dfrac{{{b}_{1}}+{{b}_{2}}+{{b}_{3}}}{3} \right)$
Where $\left( {{a}_{1}},{{b}_{1}} \right),\left( {{a}_{2}},{{b}_{2}} \right)$ and $\left( {{a}_{3}},{{b}_{3}} \right)$ are the vertices of the triangle. We will also use the properties of roots of a polynomial and find the coordinates of the centroid.

Complete step by step answer:
We have the vertices of the triangle given to us as:
$\Rightarrow \left( pq,1/\left( pq \right) \right),\left( qr,1/\left( qr \right) \right)$ and $\left( rq,1/\left( rq \right) \right)$
On putting the values of the vertices in the centroid formula, we get:
$C\equiv \left( \dfrac{pq+qr+rp}{3},\dfrac{\dfrac{1}{pq}+\dfrac{1}{qr}+\dfrac{1}{rp}}{3} \right)$
On taking the lowest common multiple in the second coordinate, we get:
$C\equiv \left( \dfrac{pq+qr+rp}{3},\dfrac{\dfrac{p+q+r}{pqr}}{3} \right)$
On rearranging the terms, we get:
$\Rightarrow C\equiv \left( \dfrac{pq+qr+rp}{3},\dfrac{p+q+r}{3\left( pqr \right)} \right)\to \left( 1 \right)$
Now we know that $p,q,r$ are the roots of the equation ${{y}^{3}}-3{{y}^{2}}+6y+1=0$. Therefore, we know the relationship of the roots with the coefficient of the polynomial as:
$\Rightarrow p+q+r=3$
$\Rightarrow pq+qr+rp=6$
$\Rightarrow pqr=-1$
Now on substituting the values in equation $\left( 1 \right)$, we get:
$\Rightarrow C\equiv \left( \dfrac{6}{3},\dfrac{3}{3\left( -1 \right)} \right)$
On simplifying the terms, we get:
$\Rightarrow C\equiv \left( 2,-1 \right)$, which is the required coordinate.

So, the correct answer is “Option B”.

Note: the relationship between the roots of a cubic equation $p{{x}^{3}}+q{{x}^{2}}+rx+s=0$, where $p\ne 0$, and the roots are $\alpha ,\beta $ and $\gamma $, then we have the relation as $\alpha +\beta +\gamma =-\dfrac{q}{p}$, $\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{r}{p}$ and $\alpha \beta \gamma =-\dfrac{s}{p}$. The centroid of a triangle is the central point of the triangle.