2. In the Cartesian coordinate system, a line segment $A B$ of length 1 moves on the $x$-axis (point $A$ is to the left of $B$), point $P(0,1)$ is connected to $A$ by a line, and point $Q(1,2)$ is connected to $B$ by a line. Then the equation of the trajectory of the intersection point $R$ of lines $P A$ and $Q B$ is $\qquad$
2. $y(x-2)=-2$.

Let $A(a, 0), B(a+1,0)$, the intersection point of line $P A$ and $Q B$ is $R(x, y)$.
When $a=0$, the two lines do not intersect, so $a \neq 0$.
At this time, $l_{P A}: \frac{x}{a}+y=1$,
$$
l_{Q B}: y=-\frac{2}{a}(x-a-1) \text {. }
$$

From equations (1) and (2), we get $x=a+2, y=-\frac{2}{a}$.
Eliminating $a$ yields the trajectory equation of point $R$ as
$$
y(x-2)=-2 \text {. }
$$