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(C) The centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\righ

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$(p, q)$

Vedclass · Answer

(C) The centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.For the given vertices,the centroid is $\left(\frac{\alpha+\beta+\gamma}{3}, \frac{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}}{3}\right)$.From the equation $x^3 - 3px^2 + 3qx - 1 = 0$,by Vieta's formulas:$\alpha+\beta+\gamma = 3p$$\alpha\beta + \beta\gamma + \gamma\alpha = 3q$$\alpha\beta\gamma = 1$Substituting these values into the centroid formula:$x$-coordinate $= \frac{3p}{3} = p$$y$-coordinate $= \frac{\frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma}}{3} = \frac{3q}{3(1)} = q$Thus,the centroid is $(p, q)$.

If $\alpha, \beta, \gamma$ are the real roots of the equation $x^3 - 3px^2 + 3qx - 1 = 0$,then find the centroid of the triangle whose vertices are $(\alpha, \frac{1}{\alpha}), (\beta, \frac{1}{\beta})$ and $(\gamma, \frac{1}{\gamma})$.

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