Find the locus of the orthocenter of the triangle formed by the lines $(1 + p)x–py+p(1 + p)=0,(1+q)x–qy+q(1+q)=0,y=0$ where $p\ne q$ Find the locus of the orthocenter of the triangle formed by the lines
$(1 + p)x – py + p(1 + p) = 0$, $(1 + q)x – qy + q(1 + q) = 0, y = 0$ where $p\ne q$
My attempt is as follows:-
 
Let orthocenter be $(h,k)$
Equation of perpendicular from $B$ to $AC$ will be $x=pq$
As orthocenter will lie on it, so $h=pq\tag{1}$
Equation of perpendicular from $C$ to $AB$ will be $(1+p)y+px+pq=0$
As orthocenter will lie on it, so $(1+p)k+ph+pq=0\tag{2}$
Equation of perpendicular from $A$ to $BC$ will be $(1+q)y+qx+pq=0$
As orthocenter will lie on it, so $(1+q)k+qh+pq=0\tag{3}$
Subtracting equation $(3)$ from $(2)$
$$(p-q)k+(p-q)h=0$$
As $p\ne q$,so $k+h=0$
So locus of orthocenter is $h+k=0$
But is this correct? Actual answer is just "straight line", so that's why I am not sure whether this is the correct locus or not.
 A: The ortho-center is at $(pq,-pq)$ which lies on $x+y=0$ for any value of $p$ and $q$.
So your answer is correct.
A: I propose here to settle this question into a more general framework.
The point is that the three equations you have given belong to the same parametric family with general equation 
$$\text{line} \ L_p : \ \ \ \ (1+p)x+(-p)y+p(1+p)=0 \tag{1}$$
(for the third one, take $p=-1$), .
Take a look at the following picture making obvious the fact that lines $L_p$ are exactly the tangents to a certain parabola (see remark below for a proof).

(one sees in particular that straight line with equation $y=0$ is one of these tangents).
Now, we can use a classical (well, rather classical) property of a parabola :
The orthocenter of the triangle determined by three tangents to a parabola lies on the so-called "directrix line" of the parabola with equation $x+y=0$ drawn in red on the picture (see for example property 7 of paragraph 4 in http://users.math.uoc.gr/~pamfilos/eGallery/problems/ParaProblems.htm#tri or lemma of page 200 in https://epub.ub.uni-muenchen.de/4550/1/Fritsch_Rudolf_4550.pdf  (reference : Kiepert's parabola) or http://jwilson.coe.uga.edu/EMT669/Student.Folders/Giddings.Jemma/Parabolas/Parabolas.html). 
Remarks : 
1) All the facts given here can be established, starting from the fact that this parabola has implicit equation $(x-y)^2-2(x+y)+1=0$.
2) There is a somewhat similar "dual" property for the rectangular hyperbola : Prove: A triangle inscribed in a rectangular hyperbola has its orthocenter on that hyperbola ("dual" because it deals with chords instead of tangents). 
