Key Concepts and Formulas

• Equation of a Line (Point-Slope Form): The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
• Intercepts: The x-intercept is the point where the line intersects the x-axis (y=0), and the y-intercept is the point where the line intersects the y-axis (x=0).
• Rectangle Properties: If a rectangle has vertices at (0,0), (a,0), and (0,b), then the fourth vertex is (a,b).

Step-by-Step Solution

1. Define the Problem and Variables: We are given a fixed point A(2,3). A line passes through A and intersects the x and y axes at points P and Q, respectively. O is the origin. OPRQ forms a rectangle. Our goal is to find the locus of point R. Let R have coordinates (h, k).

2. Equation of the Line: Since the line passes through (2,3), we can express its equation in point-slope form using slope m: $y - 3 = m(x - 2)$ Here, m is the parameter.

3. Find Coordinates of P and Q:

• Finding P (x-intercept): The point P lies on the x-axis, so its y-coordinate is 0. Substituting y=0 into the line equation: $0 - 3 = m(x_P - 2)$ $-3 = m(x_P - 2)$ Solving for $x_P$: $x_P - 2 = -\frac{3}{m}$ $x_P = 2 - \frac{3}{m}$ Therefore, the coordinates of P are $\left(2 - \frac{3}{m}, 0\right)$. Note that we must have $m \neq 0$ because if $m=0$ then the line would be horizontal and would not intersect the x-axis at a finite point.

• Finding Q (y-intercept): The point Q lies on the y-axis, so its x-coordinate is 0. Substituting x=0 into the line equation: $y_Q - 3 = m(0 - 2)$ $y_Q - 3 = -2m$ $y_Q = 3 - 2m$ Therefore, the coordinates of Q are $(0, 3 - 2m)$.

4. Relate P, Q, and R: Since OPRQ is a rectangle and O is the origin, the coordinates of R are determined by the x-coordinate of P and the y-coordinate of Q. Hence: $h = x_P = 2 - \frac{3}{m}$ $k = y_Q = 3 - 2m$

5. Express h and k in terms of the Parameter 'm': We have the equations: $h = 2 - \frac{3}{m} \quad \ldots (1)$ $k = 3 - 2m \quad \ldots (2)$

6. Eliminate the Parameter 'm': Solve equation (1) for m: $h = 2 - \frac{3}{m}$ $\frac{3}{m} = 2 - h$ $m = \frac{3}{2 - h} \quad \ldots (3)$ Solve equation (2) for m: $k = 3 - 2m$ $2m = 3 - k$ $m = \frac{3 - k}{2} \quad \ldots (4)$ Equate the two expressions for m: $\frac{3}{2 - h} = \frac{3 - k}{2}$

7. Simplify the Equation: Cross-multiply: $3(2) = (3 - k)(2 - h)$ $6 = 6 - 3h - 2k + hk$ $0 = -3h - 2k + hk$ $3h + 2k = hk$ Replace (h, k) with (x, y) to find the locus: $3x + 2y = xy$

Common Mistakes & Tips

• Parameter Elimination: Ensure you solve for the parameter m correctly before equating the expressions.
• Rectangle Properties: Understanding how the coordinates of R relate to P and Q based on the rectangle's properties is crucial.
• Algebraic Errors: Double-check your algebraic manipulations to avoid sign errors or incorrect simplifications.

Summary By expressing the coordinates of point R in terms of the parameter m, and then eliminating m, we found the locus of R to be $3x + 2y = xy$.

The final answer is \boxed{3x + 2y = xy}, which corresponds to option (A).