<p>If the three distinct lines \(x + 2ay + a = 0\), \(x + 3by + b = 0\) and \(x + 4ay + a = 0\) are concurrent, then the point \((a, b)\) lies on a</p>
Step-by-Step Solution
Key Concept: Three lines are concurrent if they pass through a common point. Substitute the intersection point of two lines into the third line equation to find the relationship between parameters a and b.
<p><strong>Step 1:</strong> Find the intersection point of lines 1 and 3.</p><p>Line 1: x + 2ay + a = 0</p><p>Line 3: x + 4ay + a = 0</p><p>Subtracting: -2ay = 0, so y = 0 (since a ≠ 0 for distinct lines)</p><p>Substituting back: x + a = 0, so x = -a</p><p>Intersection point: (-a, 0)</p><p><strong>Step 2:</strong> For concurrency, line 2 must pass through (-a, 0).</p><p>Line 2: x + 3by + b = 0</p><p>Substituting point (-a, 0): -a + 3b(0) + b = 0</p><p>-a + b = 0</p><p>b = a</p><p><strong>Step 3:</strong> The point (a, b) satisfies b = a, which is the line y = x.</p><p>∴ Answer: D (The point (a, b) lies on the line y = x)</p>
Correct Answer: D