A straight line passing through a fixed point (2, 3) intersects the coordinate axes at points \( P \) and \( Q \). If \( O \) is the origin and \( R \) is a variable point such that \( OPRQ \) is a rectangle, then the locus of \( R \) is
Show Hint
- \( 3x + 2y = xy \)
- \( 2x + 3y = xy \)
- \( 3x + 2y = 6 \)
- \( 3x + 2y = 6xy \)
The Correct Option is A
Solution and Explanation
Step 1: General form of line intersecting axes
Let the required line intersect the x-axis at \( P = (a, 0) \) and the y-axis at \( Q = (0, b) \). The line passes through the point \( (2, 3) \), so we use the intercept form of the line: \[ \frac{x}{a} + \frac{y}{b} = 1 \] Step 2: Substitute the known point
The point \( (2, 3) \) lies on this line, so it must satisfy the equation: \[ \frac{2}{a} + \frac{3}{b} = 1 \tag{1} \] Step 3: Define point \( R \)
The rectangle \( OPRQ \) has vertices at: - \( O = (0, 0) \) - \( P = (a, 0) \) - \( Q = (0, b) \) - \( R = (a, b) \) So, point \( R \) is the opposite corner of the rectangle, and has coordinates \( (x, y) = (a, b) \).
Step 4: Express in terms of \( x \) and \( y \)
Substitute \( a = x \), \( b = y \) into equation (1): \[ \frac{2}{x} + \frac{3}{y} = 1 \] Step 5: Eliminate denominators
Multiply both sides by \( xy \) to eliminate the denominators: \[ 2y + 3x = xy \Rightarrow \boxed{3x + 2y = xy} \] So, the locus of \( R \) is: \[ \boxed{3x + 2y = xy} \]
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