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 $ riangle 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 $x+2y=2$. If the centroid of $	riangle PQR$ is the point $(\alpha,\beta)$, then $15(\alpha-\beta)$ is equal to :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The centroid of the original triangle is $(2,\frac83)$. Its image in $x+2y-2=0$ is $(2,\frac83)-\frac{2(16/3)}{5}(1,2)=\left(-\frac2{15},-\frac{24}{15}\right)$, so $15(\alpha-\beta)=-2+24=22$. Answer (4).

Q. 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 :

Explanation

Reviewed by Dr. Abubacker

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