8. Let $k, l, m$ be real numbers, $m \neq 0$. In the Cartesian coordinate system, the graph of the function $y=f(x)=k+\frac{m}{x-l}$ is the curve $C_{1}$, and the graph of another function $y=g(x)$ is the curve $C_{2}$, which satisfies that $C_{2}$ is symmetric to $C_{1}$ with respect to the line $y=x$. If the points $(1,4),(2,3),(2,4)$ are all on the curve $C_{1}$ or $C_{2}$, then the value of $f(k+l+m)$ is $\qquad$
Notice that the function $f(x)$ is one-to-one, then there are two cases for the points on curve $C_{1}$: $(1,4),(2,3),(4,2)$ or $(2,4),(3,2),(4,1)$.
Thus $\left\{\begin{array}{l}4=k+\frac{m}{1-l}, \\ 3=k+\frac{m}{2-l}, \\ 2=k+\frac{m}{4-l}\end{array} \Rightarrow\left\{\begin{array}{l}k=0, \\ l=-2, \\ m=12\end{array} \quad\right.\right.$ or $\left\{\begin{array}{l}4=k+\frac{m}{2-l}, \\ 2=k+\frac{m}{3-l}, \\ 1=k+\frac{m}{4-l}\end{array} \Rightarrow\left\{\begin{array}{l}k=-2, \\ l=0, \\ m=12 .\end{array}\right.\right.$
Then $f(x)=\frac{12}{x+2}$ or $f(x)=-2+\frac{12}{x}$. Therefore, $f(10)=1$ or $-\frac{4}{5}$.