The locus of the orthocenter of the triangle formed by the lines \((1 + p)x - py + p(1 + p) = 0\), \((1 + q)x - qy + q(1 + q) = 0\), and \(y = 0\), where \(p \neq q\), is:
Step-by-step Solution:
Intersection Points:
1. Intersection point of \( y = 0 \) with the first line is \( B(-p, 0) \).
2. Intersection point of \( y = 0 \) with the second line is \( A(-q, 0) \).
3. Intersection point of the two lines is \( C(pq, (p+1)(q+1)) \).
Altitudes:
1. Altitude from \( C \) to \( AB \) is \( x = pq \).
2. Altitude from \( B \) to \( AC \) is \( y = -\frac{q}{1+q}(x + p) \).
Solving the Two Lines:
Solving the two lines, we get:
- \( x = pq \)
- \( y = -pq \)
Locus of Orthocenter:
The locus of the orthocenter is \( x + y = 0 \), which is a straight line.
This confirms that the correct answer is D. A straight line.