If we assume that two vertices of the triangle lie on the $x$-axis (which can be obtained by rotation and translation) then the vertices are $(0,0)$, $(x_1,0)$ and $(x_2,y_2)$. Let us try to find the equations of the altitudes.
The first one is easy
$$x=x_2 \qquad (1).$$
What about second one? We want to find a line with normal vector $(x_2-x_1,y_2)$ which goes trough the point $(0,0)$. (We know the normal vector since the line has to be to perpendicular on the segment connecting the points $(x_1,0)$ and $(x_2,y_2)$.) Using standard form of the equation of line we get
$$(x_2-x_1)x+y_2y=0 \qquad (2).$$
The third of them: We are looking for a line with normal vector $(x_2,y_2)$ which goes through the point $(x_1,0)$. This gives the equation
$$x_2x+y_2y=x_2x_1 \qquad (3).$$
The only one thing is missing - to show that the system of linear equations (1), (2) and (3) has a solution.
I think that the solution would not be much more complicated even without the assumption that $y_1=0$ (i.e. without using the rotation). But I would say that moving one of the vertices to $(0,0)$ simplifies things a little. (EDIT: But I agree with Srivatsan's comment that working out the details of the actual computation could be messy.)