The locus of the orthocentre of the triangle formed by the lines $(1+p)x - py + p(1+p) = 0$, $(1+q)x - qy + q(1+q) = 0$, and $y = 0$, where $p \neq q$, is

$\text{Given: } L_1: (1+p)x - py + p(1+p) = 0$ $L_2: (1+q)x - qy + q(1+q) = 0$ $L_3: y = 0, \quad p \neq q$ $\text{To find: locus of the orthocentre of the triangle}$ $1.\ \text{find } L_1 \cap L_3 \text{ and } L_2 \cap L_3$ $2.\ \text{find } C = L_1 \cap L_2$ $3.\ \text{find the orthocentre, then eliminate } p, q$ $L_1 \cap L_3:\ (1+p)(x+p) = 0 \Rightarrow x = -p$ $\Rightarrow A(-p, 0)$ $L_2 \cap L_3:\ x = -q \Rightarrow B(-q, 0)$ $AB \text{ lies on } y = 0 \text{ (the } x\text{-axis)}$ $L_1 \cap L_2:\ x = pq, \quad y = (1+p)(1+q)$ $\Rightarrow C\left(pq,\ (1+p)(1+q)\right)$ $\text{Altitude from } C \perp AB:\quad x = pq$ $H \text{ lies on } x = pq \Rightarrow X_H = pq$ $m_{BC} = \dfrac{(1+p)(1+q) - 0}{pq - (-q)} = \dfrac{(1+p)(1+q)}{q(1+p)} = \dfrac{1+q}{q}$ $m_{\perp} = -\dfrac{q}{1+q}$ $\text{Altitude from } A:\ y = -\dfrac{q}{1+q}\,(x+p)$ $y = -\dfrac{q}{1+q}\,(pq+p) = -\dfrac{q}{1+q}\cdot p(q+1) = -pq$ $\Rightarrow H(pq,\ -pq)$ $X = pq, \quad Y = -pq \Rightarrow Y = -X$ $\Rightarrow x + y = 0$ $p=1,\ q=2:\ H(2,-2) \Rightarrow 2 + (-2) = 0\ \checkmark$ $\boxed{\text{Answer: (4) a straight line: } x + y = 0}$