<s>[INST] Find locus of $\Delta ABC$ centroid with orthocentre at origin and side slopes 2, 3 and 5 Let $ABC$ be a triangle with slopes of the sides $AB$, $BC$, $CA$ are $2,3,5$ respectively. Given origin is the orthocentre of the triangle $ABC$. Then find the locus of the centroid of the triangle $ABC$.    
Since sides have slopes $2,3,5$ then altitudes must have slopes $\frac{-1}{2}, \frac{-1}{3}$ and $\frac{-1}{5}$ respectively. Then equations of the altitudes are respectively  \begin{align*} & 2y+x=0 \\ & 3y+x=0 \\ & 5y+x=0. \end{align*} If $(\alpha, \beta)$ be the coordinates of the centroid, how can I find the locus of the point $(\alpha, \beta)$ from here?
 [/INST] Hint:  Fix a triangle $ABC$ with the required property.  Show that any other triangle $A'B'C'$ with the required property is obtained from $\triangle ABC$ by a dilation about the origin.  This proves that the locus of the centroid is a line passing through the origin (and the centroid $G$ of the fixed triangle $ABC$).  
For example, let $D$ be the base of the altitude from $A$ of $\triangle ABC$.  If $D$ has coordinates $(x_D,y_D)$, then you know that $3y_D+x_D=0$.  So you may assume wlog that $x_D=3\cdot 168=504$ and $y_D=-168$ (the number $168$ is chosen so that $A$, $B$, $C$, $D$, and $G$ have integer coordinates).  Therefore the equation for $BC$ is $y-y_D=3(x-x_D)$ or $y=3x-10$.  Thus, the points $B$ and $C$ are the intersections of $y=3x-10$ with $5y+x=0$ and with $2y+x=0$, respectively.  Now you know where $B$ and $C$ are, it should be easy to find $A$, and then $G$.

  If $D$ is chosen as above, then $A=(495,-165)$, $B=(525,-105)$, and $C=(480,-240)$.  The centroid is then $$G=\frac{A+B+C}{3}=\left(500,-170\right).$$

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