If $\tan \alpha, \tan \beta$, tan $\gamma$ are the roots of the equation $x^{3}-3 a x^{2}+3 b x-1=0$. Find the centroid of the triangle whose vertices are $(\tan \alpha, \cot \alpha),(\tan \beta, \cos \beta),(\tan \gamma, \cot \gamma)$.

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The centroid of the triangle is at point $(a, b)$.

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Step 1: Find the roots of the cubic equation

The roots of the cubic equation are given as $\tan{\alpha}, \tan{\beta}, \tan{\gamma}$. According to Vieta's formulas, the sum of the roots of a cubic equation $x^{3}-3 a x^{2}+3 b x-1=0$ is equal to $3a$. By the addition formula for tangent function, $\tan{\alpha} + \tan{\beta} + \tan{\gamma} = \tan{\alpha} + \tan{\beta} - \tan{(\pi - \gamma)} = \tan{(\alpha +\beta)} = 3a$.

Step 2: Calculate the $x$-coordinate of centroid

Substituting the values into the formula for the $x$-coordinate of the centroid, we get $\frac{\tan \alpha + \tan \beta + \tan \gamma}{3} = a$.

Step 3: Calculate the $y$-coordinate of centroid

Similarly, we can follow the same steps to find the $y$-coordinates of the triangle vertices. We know that $\cot(\alpha) = \frac{1}{\tan(\alpha)}$, so we can compute the $y$-coordinate of the centroid as $\frac{\frac{1}{\tan \alpha} + \frac{1}{\tan \beta} + \frac{1}{\tan \gamma}}{3}$. But knowing that $\frac{1}{\tan \alpha} + \frac{1}{\tan \beta} + \frac{1}{\tan \gamma} = 3b$, the $y$-coordinate of the centroid simplifies to $b$.

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Chapter 1: Chapter 1

If \(\tan \alpha, \tan \beta\), tan \(\gamma\) are the roots of the equation \(x^{3}-3 a x^{2}+3 b x-1=0\). Find the centroid of the triangle whose vertices are $(\tan \alpha, \cot \alpha),(\tan \beta, \cos \beta),(\tan \gamma, \cot \gamma)$.

Short Answer

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Step 1: Find the roots of the cubic equation

Step 2: Calculate the \(x\)-coordinate of centroid

Step 3: Calculate the \(y\)-coordinate of centroid

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More chapters from the book ‘Skills in Mathematics for All Engineering Entrance Examinations: Coordinate Geometry’

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Chapter 3

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Chapter 6

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