The straight line through a fixed point (2,3) intersects the coordinate axes at distinct point P and Q. If O is the origin and the rectangle OPRQ is completed then the locus of R is

Options

A. `3x+2y=6xy`
B. `3x+2y=6`
C. `2x+3y=xy`
D. `3x+2y=xy`

Explanation

To solve the problem step by step, we need to find the locus of point R, which is formed by the rectangle OPRQ, where O is the origin, P is on the x-axis, Q is on the y-axis, and the line passes through the fixed point (2, 3).

### Step-by-Step Solution:

1. **Understanding the Points**:
- Let P be the point where the line intersects the x-axis, so P = (h, 0).
- Let Q be the point where the line intersects the y-axis, so Q = (0, k).
- The fixed point through which the line passes is (2, 3).

2. **Equation of the Line**:
- The line can be expressed in intercept form as:
\[ \frac{x}{h} + \frac{y}{k} = 1 \]

3. **Substituting the Fixed Point**:
- Since the line passes through the point (2, 3), we substitute x = 2 and y = 3 into the line equation:
\[ \frac{2}{h} + \frac{3}{k} = 1 \]

4. **Clearing the Denominators**:
- Multiply through by hk to eliminate the denominators:
\[ 2k + 3h = hk \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ hk - 2k - 3h = 0 \]
- This can be rewritten as:
\[ hk - 3h - 2k = 0 \]

6. **Expressing in Terms of x and y**:
- Since we want the locus of point R, we replace h with x and k with y:
\[ xy - 3x - 2y = 0 \]

7. **Final Form**:
- Rearranging gives us:
\[ 3x + 2y = xy \]

### Conclusion:
The locus of point R is given by the equation:
\[ 3x + 2y = xy \]