Example 7 Given the function $f(x)=\frac{x-5}{2 x+m}$ whose graph is symmetric with respect to the line $y=x$, find the real number $m$.

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Example 7 Given the function $f(x)=\frac{x-5}{2 x+m}$ whose graph is symmetric with respect to the line $y=x$, find the real number $m$.
Solution: Clearly, $(5,0)$ is a point on the graph of the function $y=\frac{x-5}{2 x+m}$. Therefore, the point $(5,0)$'s reflection across the line $y=x$, which is $(0,5)$, is also on the graph of this function. Thus, $5=-\frac{5}{m}$, which means $m=-1$.

Let $\left(x_{0}, y_{0}\right)$ be any point on the graph of the function $y=\frac{x-5}{2 x-1}$, then $y_{0}=\frac{x_{0}-5}{2 x_{0}-1}$. Its reflection across the line $y=x$ is $\left(y_{0}, x_{0}\right)$, so
$$
\frac{y_{0}-5}{2 y_{0}-1}=\frac{\frac{x_{0}-5}{2 x_{0}-1}-5}{2 \times \frac{x_{0}-5}{2 x_{0}-1}-1}=x_{0} \text {, }
$$

Therefore, the point $\left(y_{0}, x_{0}\right)$ is also on the graph of the function $y=\frac{x-5}{2 x-1}$, meaning the graph of the function $y=\frac{x-5}{2 x-1}$ is symmetric about the line $y=x$.