Hi

Here is a proposal (hope you won't find it too long ^^)

Let ABC the triangle with A and C lying on the line y=0

(AB): (1+p)x-py+p(1+p)=0 therefore one direction vector is (p;1+p)

(BC): (1+q)x-qy+q(1+q)=0 therefore one direction vector is (q;1+q)

It is easy to show that A(-p;0) and C(-q;0)

Let H(x;y) be the orthocentre

gives p(x+q)+(1+p)y=0 [1]

gives q(x+p)+(1+q)y=0 [2]

[1]-[2] gives x+y=0, which means that the locus of H is a straight line