Find the orthocentre of a triangle with vertices $A(0,0)$, $B(3,0)$ and $C(0,4)$.
... Slope of AB$=\dfrac {y_2 - y_1}{x_2 - x_1}$ $$=\dfrac {0-0}{3-0} =0$$ If $CZ$ is the altitude perpendicular to $AB$, slope of $CZ$ be $m$. Then, $$0.m=-1$$ $$m=\dfrac {-1}{0}$$.
For all right triangles, which this is, the orthocenter is at the vertex of the right angle, so in this case $(0,0)$.
You can see this without worrying about the altitude from the hypotenuse. The altitudes on the other sides are the same as the legs of the right triangle, which, obviously, intersect at a vertex.