$2$ arbitrary points in the Cartesian plane describe a line with:
$$\vec r(t) = (1-t)P_1 + tP_2 = \langle x,y \rangle$$
What is the exact same line described in polar coordinates? My intuition tells me there is some kind of asymptote associated with this.
An expression for a line in polar coordinates can given as
$$ r(\theta) = \rho_0 \sec (\theta - \theta_0) \tag{1} $$
where $\rho_0$ is the distance from the origin to this line and $\theta_0$ is the angle of the perpendicular line.
If the original line has the form $\boldsymbol{r}(t) = \boldsymbol{r}_0 + \boldsymbol{v}t$ where $\boldsymbol{r}_0 = (x_0, y_0)$ is an arbitrary point on the line and $\boldsymbol{v} = (a,b)$ is the direction vector, then the vector $\boldsymbol{\rho} = (\rho_0\cos \theta_0, \rho_0\sin\theta_0)$, which describes the point of tangency between the line and the circle of radius $\rho_0$, is given by
$$ \boldsymbol{\rho} = \boldsymbol{r}_0 - \left|\frac{\boldsymbol{r}_0\cdot \boldsymbol{v}}{\boldsymbol{v}\cdot\boldsymbol{v}}\right|\boldsymbol{v} \tag{2} $$
The second term in $(2)$ is just the projection from the vector $\boldsymbol{r}_0$ onto the direction vector $\boldsymbol{v}$
The expression might be complicated, but you can't avoid trig functions in polar coordinates.