Locus of Orthocentre of Triangle Formed by Variable Lines

To find the locus of the orthocentre of the triangle formed by the given lines, let the three lines representing the sides of the triangle be:
$L_1: (1+p)x - py + p(1+p) = 0$
$L_2: (1+q)x - qy + q(1+q) = 0$
$L_3: y = 0$

Step 1: Find the vertices of the triangle

1. Vertex $A$ (intersection of $L_1$ and $L_3$):
   Setting $y = 0$ in the equation for $L_1$:
   $(1+p)x + p(1+p) = 0 \implies (1+p)(x + p) = 0$
   Assuming $p \neq -1$, we get $x = -p$.
   Thus, the coordinates of vertex $A$ are:
   $A = (-p, 0)$

2. Vertex $B$ (intersection of $L_2$ and $L_3$):
   Setting $y = 0$ in the equation for $L_2$:
   $(1+q)x + q(1+q) = 0 \implies (1+q)(x + q) = 0$
   Assuming $q \neq -1$, we get $x = -q$.
   Thus, the coordinates of vertex $B$ are:
   $B = (-q, 0)$

3. Vertex $C$ (intersection of $L_1$ and $L_2$):
   Subtracting the equation of $L_2$ from $L_1$:
   $[(1+p) - (1+q)]x - (p - q)y + [p(1+p) - q(1+q)] = 0$
   $(p - q)x - (p - q)y + (p - q)(p + q + 1) = 0$
   Since $p \neq q$, we can divide throughout by $(p - q)$:
   $x - y + (p + q + 1) = 0 \implies y = x + p + q + 1$

   Substituting this expression for $y$ into $L_1$:
   $(1+p)x - p(x + p + q + 1) + p(1+p) = 0$
   $x + px - px - p^2 - pq - p + p + p^2 = 0$
   $x - pq = 0 \implies x = pq$

   Hence, the $x$-coordinate of vertex $C$ is:
   $x_C = pq$

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Step 2: Find the equations of the altitudes

1. Altitude from vertex $C$ to side $AB$ (which lies along the line $y = 0$):
   Since the line $y = 0$ is the $x$-axis (horizontal line), the altitude drawn from vertex $C$ perpendicular to it is a vertical line passing through $C$:
   $x = x_C \implies x = pq$

2. Altitude from vertex $A(-p, 0)$ to side $L_2$:
   The slope of the line $L_2$ is:
   $m_2 = \frac{1+q}{q}$
   The slope of the altitude perpendicular to $L_2$ is:
   $m_{\perp} = -\frac{1}{m_2} = -\frac{q}{1+q}$
   Therefore, the equation of the altitude passing through $A(-p, 0)$ is:
   $y - 0 = -\frac{q}{1+q}(x - (-p))$
   $y = -\frac{q}{1+q}(x + p)$

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Step 3: Determine the coordinates of the orthocentre $H(h, k)$

The orthocentre $H(h, k)$ is the point of intersection of these altitudes.

From the vertical altitude:
$h = pq$

Substituting $x = h = pq$ into the second altitude:
$k = -\frac{q}{1+q}(pq + p) = -\frac{q}{1+q} \cdot p(q + 1) = -pq$

Thus, the coordinates of the orthocentre are:
$h = pq, \quad k = -pq$

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Step 4: Find the locus of the orthocentre

Adding the coordinates of the orthocentre:
$h + k = pq + (-pq) = 0$

Replacing $(h, k)$ with $(x, y)$, the equation of the locus is:
$x + y = 0$

This represents a straight line.

Therefore, the locus of the orthocentre is a straight line.

Correct Answer: D