<p>Find the minimum value of \(|z-2-i|+|z+1+3i|+|z-2+i|\)</p>
Step-by-Step Solution
Key Concept: The minimum sum of distances from a point z to three fixed points occurs when z lies on the line segment connecting the two extreme points (Fermat point concept). Use geometric interpretation: the minimum occurs when the three points are collinear and z is positioned optimally on this configuration.
<p><strong>Step 1:</strong> Identify the three fixed points: A(2,1), B(-1,-3), C(2,-1)</p><p><strong>Step 2:</strong> Check if points are collinear by verifying if C lies on line AB.</p><p>Slope from A to B: $\frac{-3-1}{-1-2} = \frac{-4}{-3} = \frac{4}{3}$</p><p>Slope from A to C: $\frac{-1-1}{2-2} = \frac{-2}{0}$ (undefined - vertical line)</p><p><strong>Step 3:</strong> Since A and C have the same x-coordinate (x=2), and B(-1,-3) is not on this vertical line, the three points form a triangle.</p><p><strong>Step 4:</strong> For a triangle, apply the principle: $|z-A| + |z-B| + |z-C| \geq |A-C| + |B-C|$ when z is positioned such that it lies on segment AC.</p><p><strong>Step 5:</strong> Calculate distances between points:</p><p>$|A-C| = |2+i-(2-i)| = |2i| = 2$</p><p>$|B-C| = |-1-3i-(2-i)| = |-3-2i| = \sqrt{9+4} = \sqrt{13}$</p><p><strong>Step 6:</strong> When z lies on segment AC (the vertical line x=2), the sum becomes $|z-A| + |z-C| + |z-B| = 2 + \sqrt{13}$</p><p>∴ Answer: $2 + \sqrt{13}$</p>
Correct Answer: A