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Step-by-Step Solution
Step 1: Identify the coordinates of intersections and the rectangle’s vertices
Let the straight line passing through the fixed point $(2,\,3)$ intersect the $x$-axis at
$P(a,\,0)$ and the $y$-axis at $Q(0,\,b)$. The origin is $O(0,\,0)$. When a rectangle $OPRQ$ is
completed using these points, the fourth vertex $R$ has coordinates $(a,\,b)$ because:
$O$ to $P$ is along the $x$-axis.
$O$ to $Q$ is along the $y$-axis.
Hence, $R$ is the point diagonally opposite $O$, giving $R(a,\,b)$.
Step 2: Use the equation of the line through $(2,\,3)$
The line intersecting the axes at $P(a,\,0)$ and $Q(0,\,b)$ can be written as
$$
\frac{x}{a} + \frac{y}{b} = 1.
$$
Since $(2,\,3)$ lies on this line, it must satisfy the equation:
$$
\frac{2}{a} + \frac{3}{b} = 1.
$$
Step 3: Express the relation in terms of $R(a,\,b)$
Note that the coordinates of $R$ are $(x,\,y) = (a,\,b)$. Substituting $x = a$ and $y = b$
into the condition $\frac{2}{a} + \frac{3}{b} = 1$, we get:
$$
\frac{2}{x} + \frac{3}{y} = 1.
$$
Step 4: Simplify to find the locus
Multiply both sides by $xy$ to clear the denominators:
$$
x y \left(\frac{2}{x} + \frac{3}{y}\right) = x y \cdot 1 \\
2y + 3x = xy.
$$
Rearrange to find a direct relation between $x$ and $y$:
$$
xy - 3x - 2y = 0
\quad \Longrightarrow \quad
xy = 3x + 2y
\quad \Longrightarrow \quad
3x + 2y = xy.
$$
This is the required locus of $R$.
Final Answer
The locus of $R$ is given by $3x + 2y = xy.$
