35. On the coordinate plane, there is a line $l, O \perp l$. Consider a vector $\overrightarrow{O P}=\vec{p}$, where $P \in l$; then $p=|\vec{p}|$ is the distance from the line to the origin. Denote by $\varphi$ the angle that $\vec{e}=\frac{\vec{p}}{p}$, the unit vector of $\vec{p}$, forms with the unit vector $\vec{i}$. Let $\vec{r}=\overrightarrow{O M}$ be the variable radius vector of an arbitrary point $M \in l$. Taking into account that the vector $\overrightarrow{P M}=\vec{r}-\vec{p}$ is perpendicular to the vector $\vec{p}$, derive the vector equation of the line $l$ in the form $\vec{r} \cdot \vec{e}=\vec{p} \cdot \vec{e}$ and then from this, obtain the normal equation of the line $l$ in coordinates: $x \cos \varphi + y \sin \varphi = p$.

Discuss the cases of different positions of the line relative to the coordinate axes.
35. Since vectors $\vec{p}$ and $\vec{r}-\vec{p}$ are perpendicular, we have: $(\vec{r}-\vec{p}) \vec{p}=0$. From this, given that $\vec{p}=\overrightarrow{p e}$, we obtain:

$$
\overrightarrow{r e}=p
$$

Here $\vec{r}=x \vec{i}+y \vec{j}, \vec{e}=\vec{i} \cos \varphi+\vec{j} \sin \varphi$. Multiplying, we get:

$$
x \cos \varphi+y \sin \varphi=p
$$

If in this equation $p=0$, it means that the line passes through the origin, forming an angle of $\varphi+90^{\circ}$ with the $O x$ axis. If $p \neq 0, \varphi=0$, the equation takes the form: $x=p$ - this is the equation of a line parallel to the $O y$ axis and passing at a distance $p$ from the origin. For $\varphi=90^{\circ}$, we have: $y=p$ - the equation of a line parallel to the $O x$ axis and passing at a distance $p$ from the origin. Finally, the equations $y=0$ and $x=0$ are the equations of the $O x$ and $O y$ axes.