<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] Respectively, the vertexes A, B and C are on the three altitude lines you obtained $3y+x=0$, $5y+x=0$ and $2y+x=0$. So, let their coordinates be $A(a,-\frac a3)$, $B(b,-\frac b5)$ and $C(c,-\frac c2)$. Then, use them to match the three side slopes 
$$\frac{-\frac b5 + \frac a3 }{b-a}=2,\>\>\>\>\>\>\>
\frac{-\frac b5 + \frac c2 }{b-c}=3,\>\>\>\>\>\>\>
\frac{-\frac a3 + \frac c2 }{a-c}=5$$
which lead to the ratio $a:b:c = 33:35:32$ and the corresponding vertexes in terms of a single parameter $t$
$$A(33t,-\frac {33t}3),\>\>\>\>\>B(35t,-\frac {35t}5),\>\>\>\>\>C(32t,-\frac {32t}2)$$ 
Then, the coordinates of the centroid are
$$x = \frac{A_x+B_x+C_x}3=\frac{100t}3,\>\>\>\>\>y = \frac{A_y+B_y+C_y}3=-\frac{34t}3$$
Eliminate $t$ to obtain its locus
$$y=-\frac{17}{50}x$$
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