<p>Towns A and B are situated on the same side of a straight road at distances \(a\) and \(b\) respectively from it. Perpendiculars drawn from A and B meet the road at the points C and D respectively. The distance between C and D is \(c\). A hospital is to be built at a point P on the road between C and D such that the distance APB is minimum. Find the position of P.</p>
Step-by-Step Solution
Key Concept: Reflect point A across the road to get A', then the shortest path APB equals A'PB, which is minimized when A', P, B are collinear. Use similar triangles to find where this line intersects the road.
<p><strong>Step 1:</strong> Set up coordinates with C at origin, D at (c, 0), A at (0, a), and B at (c, b). Point P is at (x, 0) where 0 ≤ x ≤ c.</p><p><strong>Step 2:</strong> Reflect A across the road to get A' at (0, -a). The distance AP + PB equals A'P + PB, which is minimized when A', P, B are collinear (by triangle inequality).</p><p><strong>Step 3:</strong> The line through A'(0, -a) and B(c, b) has slope: m = (b - (-a))/(c - 0) = (a + b)/c</p><p><strong>Step 4:</strong> Equation of line A'B: y - (-a) = [(a+b)/c](x - 0), so y = [(a+b)/c]·x - a</p><p><strong>Step 5:</strong> Point P lies on the road where y = 0: 0 = [(a+b)/c]·x - a, giving x = ac/(a+b)</p><p><strong>Step 6:</strong> Therefore, CP = ac/(a+b)</p>
Correct Answer: CP = ac/(a+b)