<p>\(ABC\) is a variable triangle with the fixed vertex \(C(1, 2)\) and \(A\), \(B\) having coordinates \((\cos t, \sin t)\), \((\sin t, -\cos t)\) respectively, where \(t\) is a parameter. Find the locus of the centroid of \(\triangle ABC\).</p>
Step-by-Step Solution
Key Concept: The centroid of a triangle with vertices at (x₁,y₁), (x₂,y₂), (x₃,y₃) is ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3). Express the centroid coordinates in terms of parameter t, then eliminate t to find the locus by recognizing that cos²t + sin²t = 1.
<p><strong>Step 1: Find centroid coordinates.</strong> With A(cos t, sin t), B(sin t, -cos t), C(1, 2), the centroid G is:</p><p>G = ((cos t + sin t + 1)/3, (sin t - cos t + 2)/3)</p><p><strong>Step 2: Set h and k as centroid coordinates.</strong></p><p>3h = cos t + sin t + 1 ⟹ cos t + sin t = 3h - 1</p><p>3k = sin t - cos t + 2 ⟹ sin t - cos t = 3k - 2</p><p><strong>Step 3: Use the identity constraint.</strong> Square both equations and add:</p><p>(cos t + sin t)² + (sin t - cos t)² = (3h - 1)² + (3k - 2)²</p><p><strong>Step 4: Expand the left side.</strong></p><p>cos²t + 2sin t cos t + sin²t + sin²t - 2sin t cos t + cos²t = 2(cos²t + sin²t) = 2</p><p><strong>Step 5: Equate both sides.</strong></p><p>(3h - 1)² + (3k - 2)² = 2</p><p>∴ <strong>Locus: (3x - 1)² + (3y - 2)² = 2</strong></p><p>This represents a circle with center (1/3, 2/3) and radius √2/3.</p>
Correct Answer: (3x-1)² + (3y-2)² = 2