11. Define the difference of two sets $A$ and $B$ as $A-B=\{x \mid x \in A$, but $x \notin B\}$, and the symmetric difference $A \nabla B=(A-B) \cup(B-A)$. If $A=\{(x, y) \mid y=x+1\}, B=$ $\left\{(x, y) \left\lvert\, \frac{x-5}{x-4}=1\right.\right\}$, then $A \nabla B=$ $\qquad$
11. $\{(4,5)\} . B=\{(x, y) \mid y=x+1$, and $x \neq 4\}$.