Let the triangle PQR be the image of the tria | vedclass

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(D) Let $G$ be the centroid of the triangle formed by vertices $(1,3), (3,1)$ and $(2,4)$.$G = \left(\frac{1+3+2}{3}, \frac{3+1+4}{3}\right) = \left(2

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$22$

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(D) Let $G$ be the centroid of the triangle formed by vertices $(1,3), (3,1)$ and $(2,4)$.$G = \left(\frac{1+3+2}{3}, \frac{3+1+4}{3}\right) = \left(2, \frac{8}{3}\right)$.Let $(\alpha, \beta)$ be the image of $G$ with respect to the line $x+2y-2=0$.The formula for the image $(x', y')$ of a point $(x_0, y_0)$ in the line $ax+by+c=0$ is $\frac{x'-x_0}{a} = \frac{y'-y_0}{b} = -2\frac{ax_0+by_0+c}{a^2+b^2}$.Substituting the values: $\frac{\alpha-2}{1} = \frac{\beta-8/3}{2} = -2\frac{2+2(8/3)-2}{1^2+2^2}$.$\frac{\alpha-2}{1} = \frac{\beta-8/3}{2} = -2\frac{16/3}{5} = -\frac{32}{15}$.Thus,$\alpha = 2 - \frac{32}{15} = -\frac{2}{15}$ and $\beta = \frac{8}{3} - \frac{64}{15} = \frac{40-64}{15} = -\frac{24}{15}$.Finally,$15(\alpha-\beta) = 15\left(-\frac{2}{15} - (-\frac{24}{15})\right) = 15\left(\frac{22}{15}\right) = 22$.

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 $\triangle PQR$ is the point $(\alpha, \beta)$,then $15(\alpha-\beta)$ is equal to:

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