<p>Let \(a\), \(b\), \(c\) and \(d\) be non-zero numbers. If the point of intersection of the lines \(4ax + 2ay + c = 0\) and \(5bx + 2by + d = 0\) lies in the fourth quadrant and is equidistant from the two axes then</p>
Step-by-Step Solution
Key Concept: If a point is equidistant from both axes and lies in the fourth quadrant, it must have coordinates (k, -k) for some k > 0. Substitute this into both line equations to establish relationships between the coefficients.
<p><strong>Step 1:</strong> A point equidistant from both axes satisfies |x| = |y|. In the fourth quadrant (x > 0, y < 0), the point has form (k, -k) where k > 0.</p><p><strong>Step 2:</strong> Substitute (k, -k) into the first line equation: 4ak + 2a(-k) + c = 0 → 4ak - 2ak + c = 0 → 2ak + c = 0, so c = -2ak.</p><p><strong>Step 3:</strong> Substitute (k, -k) into the second line equation: 5bk + 2b(-k) + d = 0 → 5bk - 2bk + d = 0 → 3bk + d = 0, so d = -3bk.</p><p><strong>Step 4:</strong> From Step 2: c/a = -2k and from Step 3: d/b = -3k. Therefore: c/a ÷ d/b = -2k ÷ -3k = 2/3, which gives 2b·c = 3a·d (or equivalent form).</p><p>∴ Answer: A</p>
Correct Answer: A