Let the triangle PQR be the image of the triangle with vertices $ (1, 3), (3, 1) $ and $ (2, 4) $ in the line $ x + 2y = 2 $. If the centroid of $ \Delta PQR $ is the point $ (\alpha, \beta) $, then $ 15(\alpha - \beta) $ is equal to :

Question

If eccentricity of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, which passes through the point $(3, 4)$ is $\frac{\sqrt{5}}{3}$, then length of latus rectum of ellipse, is :

Answer 1

Let $ G $ be the centroid of $ \Delta PQR $, with vertices at $ (1, 3), (3, 1), $ and $ (2, 4) $.The centroid $ G $ is calculated as the average of the vertex coordinates:\[G = \left( \frac{1+3+2}{3}, \frac{3+1+4}{3} \right) = \left( \frac{6}{3}, \frac{8}{3} \right) = \left( 2, \frac{8}{3} \right).\]We then determine the image of $ G $ under the transformation $ x + 2y = 2 $. The corresponding transformation matrix yields:\[\alpha - 2 = \frac{-2}{5}, \quad \beta - \frac{8}{3} = \frac{-32}{15} + 2,\]from which we derive $ \alpha = -\frac{2}{5} $ and $ \beta = -\frac{24}{15} $.Consequently, $ 15(\alpha - \beta) = 15(-\frac{2}{5} - \frac{24}{15}) = 22 $.Thus, $ 15(\alpha - \beta) = \boxed{22} $.

Let the triangle PQR be the image of the triangle with vertices \( (1, 3), (3, 1) \) and \( (2, 4) \) in the line \( x + 2y = 2 \). If the centroid of \( \Delta PQR \) is the point \( (\alpha, \beta) \), then \( 15(\alpha - \beta) \) is equal to :

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The Correct Option is D

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