4. When point $P(x, y)$ is any point on line $l$, point $Q(4 x+2 y$, $x+3 y)$ is also a point on this line, then the equation of line $l$ is $\qquad$
4. $x+y=0$ or $x-2 y=0$ Detailed explanation: Let the equation of the line be $y=k x+b$, then points $P, Q$ satisfy the equation as follows:
$$
\left\{\begin{array} { l } 
{ y = k x + b } \\
{ x + 3 y = k ( 4 x + 2 y ) + b }
\end{array} \Rightarrow \left\{\begin{array}{l}
y=k x+b \\
(3-2 k) y=(4 k-1) x+b
\end{array}\right.\right.
$$

If $b \neq 0$, then $\frac{1}{3-2 k}=\frac{k}{4 k-1}=1$, no solution; if $b=0$, then $\frac{1}{3-2 k}=\frac{k}{4 k-1}$, solving yields $k=-1, \frac{1}{2}$. Therefore, the equation of line $l$ is $x+y=0$ or $x-2 y=0$.