Example 7. A moving straight line forms a triangle with the coordinate axes whose area is a constant. Find the equation of the locus of the midpoint of the line segment intercepted by the coordinate axes.

 untranslated text:
例7. 一动直线与坐标轴所成的三角形的面积为定值, 求这条直线被坐标轴所截线段的中点轨迹方程。
Let the moving line intersect the $x$-axis and $y$-axis at points $A\left(x_{0}, 0\right)$ and $B\left(0, y_{0}\right)$, respectively, and let the midpoint of $A B$ be $P(x, y)$.
Since
$$
x=\frac{x_{0}}{2}, y=\frac{y_{0}}{2}.
$$

It follows that $\mathrm{x}_{0}=2 \mathrm{x}, \mathrm{y}_{0}=2 \mathrm{y}$.
Given that the area of $\triangle \circlearrowleft A B$ is $\left.=\frac{1}{2} j x_{0} y_{0} \right\rvert\,=k$ (a constant), we have $\frac{1}{2}|2 \mathrm{x} \cdot 2 \mathrm{y}|=\mathrm{k}$.

Simplifying, we get
$$
x y=\frac{k}{2} \text { or } x y=-\frac{k}{2}.
$$

This is the equation of the required locus. It is easy to verify that the simplification process is an equivalent transformation. Therefore, the equations $x y=\frac{k}{2}$ or $\mathrm{xy}=-\frac{\mathrm{k}}{2}$ satisfy the condition $\mathrm{I}$, and they are indeed the equations of the required locus.