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

Question

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

Aicharya · Accepted Answer

The centroid $G^{\prime \prime}(\alpha, \beta)$ of $\triangle P Q R$ be image of centroid of given triangle $P^{\prime} Q^{\prime} R^{\prime}$.

Centroid of $\Delta P^{\prime} Q^{\prime} R^{\prime}=\left(\frac{1+3+2}{3}, \frac{3+1+4}{3}\right)=$ $G^{\prime}\left(2, \frac{8}{3}\right)$.
Image of $G\left(2, \frac{8}{3}\right)$, w.r.t. line $x+2 y=2$ is $(\alpha, \beta)$
Then $\frac{\alpha-2}{1}=\frac{\beta-\frac{8}{3}}{2}=\frac{-2\left(2+\frac{16}{3}-2\right)}{1+4}$
$\therefore \quad \frac{\alpha-2}{1}=\frac{\beta-\frac{8}{3}}{2}=-\frac{32}{15}$
$\therefore \quad \alpha=-\frac{2}{15}$ and $\beta=-\frac{8}{5}$
Then $15(\alpha-\beta)=15\left(-\frac{2}{15}+\frac{24}{15}\right)=22$ ^

Let the triangle PQR be the image of the triangle with vertices $(1,3),(3,1)$ and $(2,4)$ in the line $x+2…

Solution