2. Let the real constant $k$ be such that the equation
$$
2 x^{2}+2 y^{2}-5 x y+x+y+k=0
$$

represents two intersecting lines in the plane coordinate system $x O y$, with the intersection point $P$. If points $A$ and $B$ lie on these two lines respectively, and $|\overrightarrow{P A}|=|\overrightarrow{P B}|=1$, then $\overrightarrow{P A} \cdot \overrightarrow{P B}=$ $\qquad$
2. $\pm \frac{4}{5}$.

Notice,
$$
\begin{array}{l}
2 x^{2}+2 y^{2}-5 x y+x+y+k \\
=(2 x-y)(x-2 y)+(2 x-y)-(x-2 y)+k \\
=(2 x-y-1)(x-2 y+1)+k+1=0 .
\end{array}
$$

Therefore, $k=-1$, the two lines are $2 x-y-1=0$ and $x-2 y+1=0$, their intersection point is $P(1,1)$.

The slopes of the two lines are $k_{1}=2, k_{2}=\frac{1}{2}$, the angle $\theta$ between them satisfies
$$
\tan \theta=\left|\frac{2-\frac{1}{2}}{2}\right|=\frac{3}{4} \Rightarrow \cos \theta= \pm \frac{4}{5} .
$$

Thus, $\overrightarrow{P A} \cdot \overrightarrow{P B}= \pm \frac{4}{5}$.