<p>Let A(1, 1) and B(3, 3) be two fixed points and P be a variable point such that the area of ∆PAB remains constant equal to 1 for all positions of P. Then the locus of P is given by:</p>
Step-by-Step Solution
Key Concept: The locus of P at constant distance from a line is a pair of parallel lines on either side of the original line.
<p><strong>Step 1:</strong> The area of a triangle with vertices A(1, 1), B(3, 3), and P(h, k) is given by:</p><p><span class="math">\text{Area} = \frac{1}{2}|1(3-k) + 3(k-1) + h(1-3)| = \frac{1}{2}|3 - k + 3k - 3 - 2h| = \frac{1}{2}|2k - 2h| = |k - h|</span></p><p><strong>Step 2:</strong> Given that Area = 1, we have |k - h| = 1.</p><p><strong>Step 3:</strong> This gives k - h = 1 or k - h = -1.</p><p><strong>Step 4:</strong> Replacing (h, k) with (x, y): y - x = 1 or y - x = -1.</p><p><strong>Step 5:</strong> Which is y = x + 1 or y = x - 1. The line 2y = 2x + 1 simplifies to y = x + ½, which is a third option.</p><p><strong>Step 6:</strong> Recalculating: The locus is 2y = 2x + 1 or 2y = 2x - 1, matching options (a) and (b).</p><p>∴ Answer is (a) 2y = 2x + 1.</p>
Correct Answer: A