2. When point $P(x, y)$ is any point on the line $l$, point $Q(4 x+2 y, x+3 y)$ is also a point on this line. Then the equation of the line $l$ is $\qquad$ .
2. $x+y=0$ or $x-2 y=0$.

Let points $P$ and $Q$ coincide (invariant points under transformation), we get
$$
\left\{\begin{array}{l}
x=4 x+2 y, \\
y=x+3 y .
\end{array}\right.
$$

Thus, the origin $(0,0)$ must lie on the line. According to this, by the collinearity of points $O, P, Q$, we get
$$
\frac{y}{x}=\frac{x+3 y}{4 x+2 y} \Rightarrow x^{2}-x y-2 y^{2}=0,
$$

which is $(x+y)(x-2 y)=0$.
Therefore, the equation of the line $l$ that satisfies the condition is
$$
x+y=0 \text { or } x-2 y=0 \text {. }
$$