(9) In the Cartesian coordinate system, the "rectangular distance" between points $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$ is defined as $d(P, Q)=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right|$. If the "rectangular distance" from $C(x, y)$ to points $A(1,3)$ and $B(6,9)$ is equal, where real numbers $x$ and $y$ satisfy $0 \leqslant x \leqslant 10,0 \leqslant y \leqslant 10$, then the sum of the lengths of all the loci of points $C$ that satisfy the condition is $\qquad$.
(9) From the condition, we have
$$
|x-1|+|y-3|=|x-6|+|y-9| .
$$

When $y \geqslant 9$, (1) becomes $|x-1|+6=|x-6|$, which has no solution;
When $y \leqslant 3$, (1) becomes $|x-1|=6+|x-6|$, which has no solution;
When $3 \leqslant y \leqslant 9$, (1) becomes
$$
2 y-12=|x-6|-|x-1| .
$$

If $x \leqslant 1$, then $y=8.5$, the length of the line segment is 1; if $1 \leqslant x \leqslant 6$, then $x+y=9.5$, the length of the line segment is $5 \sqrt{2}$; if $x \geqslant 6$, then $y=3.5$, the length of the line segment is 4.
In summary, the sum of the lengths of the line segments that constitute the trajectory of point $C$ is
$$
1+5 \sqrt{2}+4=5(\sqrt{2}+1) .
$$