6. Let points $P$ and $Q$ move on the lines
$$
3 x-y+5=0 \text{ and } 3 x-y-13=0
$$

respectively, the midpoint of segment $P Q$ is $M\left(x_{0}, y_{0}\right)$, and $x_{0}+y_{0} \geqslant 4$. Then the range of $\frac{y_{0}}{x_{1}}$ is $\qquad$
6. $[1,3)$.

Solution 1 Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}+x_{2}=2 x_{0}, y_{1}+y_{2}=2 y_{0} \text {, }
$$

and $3 x_{1}-y_{1}+5=0,3 x_{2}-y_{2}-13=0$.
Adding the two equations, we get
$$
\begin{array}{l}
3\left(x_{1}+x_{2}\right)-\left(y_{1}+y_{2}\right)-8=0 \\
\Rightarrow 3 x_{0}-y_{0}-4=0 \\
\Rightarrow 3 x_{0}=y_{0}+4 \geqslant\left(4-x_{0}\right)+4=8-x_{0} \\
\Rightarrow x_{0} \geqslant 2 . \\
\text { Therefore, } \frac{y_{0}}{x_{0}}=\frac{3 x_{0}-4}{x_{0}}=3-\frac{4}{x_{0}} \text { is increasing on }[2,+\infty] \text {, }
\end{array}
$$

so $1 \leqslant \frac{y_{0}}{x_{0}}<3$.
Solution 2 From Solution 1, we know that point $M$ lies on the ray $M_{0} M: y=3 x-4(x \geqslant 2)$, where point $M_{0}(2,2)$.

Since $\frac{y_{0}}{x_{0}}$ represents the slope of the line connecting point $M$ on the ray $M_{0} M$ to the origin $O$, we have
$$
1=k_{O M_{0}} \leqslant \frac{y_{0}}{x_{0}}<3 .
$$