<p>A straight line through a fixed point \((2, 3)\) intersects the coordinate axes at distinct points \(P\) and \(Q\). If \(O\) is the origin and the rectangle \(OPRQ\) is completed, then the locus of \(R\) is</p>
Step-by-Step Solution
Key Concept: When a line through (2,3) meets axes at P(a,0) and Q(0,b), the fourth vertex R of rectangle OPRQ is at (a,b). Use the collinearity condition for P, (2,3), Q to find the relationship between a and b, then eliminate parameters.
<p><strong>Step 1:</strong> Let the line intersect x-axis at P(a, 0) and y-axis at Q(0, b), where a ≠ 0 and b ≠ 0 (distinct points).</p><p><strong>Step 2:</strong> The equation of line PQ in intercept form is: $\frac{x}{a} + \frac{y}{b} = 1$</p><p><strong>Step 3:</strong> Since the line passes through (2, 3), substitute this point: $\frac{2}{a} + \frac{3}{b} = 1$</p><p><strong>Step 4:</strong> For rectangle OPRQ with O at origin, P(a,0), and Q(0,b), the fourth vertex R is at (a, b).</p><p><strong>Step 5:</strong> Let R = (h, k), so a = h and b = k. Substitute into the condition from Step 3:</p><p>$$\frac{2}{h} + \frac{3}{k} = 1$$</p><p><strong>Step 6:</strong> Replace (h, k) with (x, y) to get the locus:</p><p>$$\frac{2}{x} + \frac{3}{y} = 1$$</p><p>Or equivalently: $2y + 3x = xy$ with x ≠ 0, y ≠ 0, and x ≠ 2, y ≠ 3</p><p>∴ Answer: B</p>
Correct Answer: B