I have been told that any straight line in $\mathbb{R}^2$ can be represented by two parameters $(p,\theta)$ and the equation:
$$x\cos(\theta)+y\sin(\theta)-p=0$$
But am struggling to see intuitively why this equation gives a straight line. Can anyone explain to me in terms of geometry what the parameters $p$ and $\theta$ represent in terms of the line (i.e. like when we represent a line as $y=mx+c$, $m$ is the gradient and $c$ is the intercept, what is $\theta$ and $p$ in this setting?)?
On
You may rewrite the equation in the familiar form $y=mx+c$,
$$y=-\cot\theta\> x+p\csc\theta$$
to identify
$$m = -\cot\theta,\>\>\>\>\>p= c\sin\theta$$
Knowing that $m$ is the gradient, measured by $m=\tan\alpha$, with $\alpha$ being the angle between the line and the $x$-axis, you can rewrite $m=\tan(\frac\pi2 + \theta)$ and deduce that $\theta$ is the angle between the given line and the $y$-axis. (See the graph below.)
Similarly, knowing that $c$ is the $y$-intercept, then $p=c\cos\theta$ represents the distance of the line from the origin.
This is sometimes called a "normal" equation. Here, $(\cos\theta,\sin\theta)$ is a normal vector to the line, and therefore $\theta$ is the angle between the normal and the $OX$-axis. $p$ is the distance from the line to the origin.