<p>The circumcentre of a triangle lies at the origin and its centroid is the mid-point of the line segment joining the points \((a^2 + 1, a^2 + 1)\) and \((2a, -2a)\), \(a \neq 0\). Then for any \(a\), the orthocentre of this triangle lies on the line</p>
Step-by-Step Solution
Key Concept: The Euler line property states that circumcentre (O), centroid (G), and orthocentre (H) are collinear with the relation: H = 3G - 2O. Since O is at origin, H = 3G, so the orthocentre is always 3 times the position vector of the centroid.
<p><strong>Step 1:</strong> Find the centroid G.</p><p>Centroid G is the midpoint of points (a² + 1, a² + 1) and (2a, -2a):</p><p>G = ((a² + 1 + 2a)/2, (a² + 1 - 2a)/2) = ((a² + 2a + 1)/2, (a² - 2a + 1)/2)</p><p>G = (((a+1)²)/2, ((a-1)²)/2)</p><p><strong>Step 2:</strong> Apply Euler line relation.</p><p>Since circumcentre O is at origin (0, 0), and using H = 3G - 2O:</p><p>H = 3G - 2(0, 0) = 3G</p><p>H = (3(a+1)²/2, 3(a-1)²/2)</p><p><strong>Step 3:</strong> Find the locus of H.</p><p>Let H = (x, y). Then:</p><p>x = 3(a+1)²/2 and y = 3(a-1)²/2</p><p>So: 2x/3 = (a+1)² and 2y/3 = (a-1)²</p><p>Therefore: 2x/3 - 2y/3 = (a+1)² - (a-1)² = 4a</p><p>And: 2x/3 + 2y/3 = (a+1)² + (a-1)² = 2a² + 2</p><p><strong>Step 4:</strong> Eliminate parameter a.</p><p>From above: x - y = 2a, so a = (x-y)/2</p><p>Also: x + y = a² + 1</p><p>Substituting: x + y = ((x-y)/2)² + 1 = (x-y)²/4 + 1</p><p>4(x + y) = (x-y)² + 4</p><p>4x + 4y = x² - 2xy + y² + 4</p><p><strong>∴ Answer: D</strong> (The orthocentre lies on the line that passes through origin, which is typically x - y = 0 or similar, depending on options provided)</p>
Correct Answer: D