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?
 A: 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
$$y=-\frac{17}{50}x$$
A: 
Please note that the figure given above is not drawn to scale.
The altitudes $HA_1$, $HB_1$, and $HC_1$ of the triangle $A_1B_1C_1$ are the three green lines. $H$ and $G_1$ are its orthocenter and centroid respectively. Note that the three equations shown alongside the altitudes are from your text. In addition to them we need the equations of the three sides of the triangles. $u$, $v$, and $w$ can be considered as parameters and they are equal to the intercepts, which each individual side makes with the $y$-axis.  
It is possible to express the coordinates of each of the vertices of the triangle $A_1B_1C_1$ as the intersection of a pair of lines in two different ways. For instance, vertex $C_1$ can be seen as the intersect of $HC_1$ and $A_1 C_1$ or $HA_1$ and $B_1 C_1$. Therefore,
$$C_1=\left(-\frac{2w}{11},\frac{w}{11}\right)=\left(-\frac{2v}{7},\frac{v}{7}\right).$$
Because both set of coordinates represents the same point $C_1$, we have,
$$v=\frac{7w}{11}.\tag{1}$$
In the case of the vertices $A_1$ and $B_1$ we need to express their respective coordinates only in one way as shown below.
$$A_1=\left(-\frac{3w}{16},\frac{w}{16}\right),\space\space\space\space B_1=\left(-\frac{5v}{16},\frac{v}{16}\right)$$ 
We substitute the parameter $v$ in the coordinates of $B_1$ with the parameter $w$ using the relationship (1) to get,
$$B_1=\left(-\frac{35w}{176},\frac{7w}{176}\right).$$
Since we were able to express the coordinates of all three vertices as functions of just one parameter, i.e. $w$, we can now find a simple expression for the centroid $G_1$. For this we use the information given in Maxim's comment.
$$G_1=\left(-\frac{50w}{264},\frac{17w}{264}\right)$$ 
Therefore, the parametric form of the locus is
$$
    \begin{matrix}
    x & = & -\frac{50w}{264} \\
    y & = &  \space\space\space\space \frac{17w}{264}. \\
    \end{matrix}
$$
To find an equation, which relates $y$ to $x$, we have to do away with $w$. 
$$y+\frac{17}{50}x=0$$ 
This is the equation of a straight line in the slope-intercept form. It is evident from this that the locus is a line passing through the origin.
Here is a question for you to find the answer if you are interested in. Since the locus passes through the origin, the origin is itself the median of a certain triangle. What is this triangle?
A: 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 $$G=\frac{A+B+C}{3}=\left(500,-170\right).$$

