Equation of a line using parameters $\theta$ and $\rho$ or $r$ I am curious as to how the below two equations of a line are derived:
$$x\sin(t)-y\cos(t)+p=0\tag{1}\label{1}$$
where $t$ is the angle the line makes from $+x$-axis and $p$ is the shortest distance from the line to the origin (length of vector from origin to line orthogonally).
$t$ is given by $\arctan(y/x)$ and $p$ can be found by knowing a point $(x,y)$ on the line after finding $t$.
Another equation of the line is:
$$x\cos(t)+y\sin(t)=r\tag{2}\label{2}$$
It is basically the same idea. $r$ is the shortest distance from the line to the origin, which is again the length of a vector pointing orthogonally to the line.
The Question:
The two equations are different yet similar. I am not sure how they are derived. I have attempted to derive the equation \eqref{2} below, but I am stuck with regards to deriving equation \eqref{1}
I tried to derive the second equation as follows (I need verification):
The dot product of two vectors is $0$ if and only if they are perpendicular. Given a line containing two points pointed to by two vectors $\langle x,y\rangle$ and $\langle x_0,y_0\rangle$. 
The vector from $\langle x_0,y_0\rangle$ to $\langle x,y\rangle$ is the vector $\langle x-x_0,y-y_0\rangle$.
We can use the unit circle to define a unit vector, $n$, (contained in the unit circle) normal to our line,
$$\langle nx,ny\rangle=\langle\cos t,\sin t\rangle$$
If a point $(x,y)$ lies on our line, it must satisfy the equation $\langle\cos t,\sin t\rangle*\langle x-x_0,y-y_0\rangle=0$, where $*$ is the dot product.
Multiplying out, you get:
$$\langle\cos t,\sin t\rangle*\langle x-x_0,y-y_0\rangle=x\cos t-x_0\cos t+y\sin t-y_0\sin t=x\cos t+y\sin t-(x_0\cos t+y_0\sin t)$$
Let $r=x_0\cos t+y_0\sin t$. Then equation can be rewritten as:
$$x\sin t+y\cos t=r$$
where $r$ is the shortest distance from origin to the line, i.e, the length of the vector pointing from origin to the line orthogonally. Is this correct?
 A: Let the foot of normal from the origin have coordinates $(p\cos t,p\sin t)$. The parameter $t$ is the direction angle of the normal, while $p$ is the distance of the origin to the line.
Then you express that the segment from the foot of normal to any point on the line is orthogonal to the normal,
$$(x-p\cos t)\cos t+(y-p\sin t)\sin t=0$$ or
$$\color{green}{x\cos t +y\sin t=p}.$$
If you instead parameterize with the direction angle of the line itself, $t'=t+\pi/2$ and
$$-x\sin t'+y\cos t'=p,$$ also written
$$\color{green}{x\sin t'-y\cos t'+p=0}.$$
A: HINT:
It is recommended to use first letters of alphabet set as constants and last letters as variables.
One need understand only one derivation and what the constants mean in the polar/normal form of a straight line.
 
The simple sketch shows two components of $ p = px+ py $ as  green and brown projected segments' sum in terms of $x,y,\alpha$.
In Equn (2), $t=\alpha $ is the angle made with normal on  straight line with shortest pedal/normal length $p$.
$$ x \cos \alpha + y \sin \alpha = p $$
is rotated through $\pi/2$ four times to get 4 lines forming a square around origin... even though $t=\alpha $ is fixed for a single line.
$$x  \sin \alpha  - y \cos \alpha = p, $$
$$-x \cos \alpha  - y \sin \alpha  = p, $$
and
$$- x  \sin \alpha  + y \cos \alpha  = p $$
This is also same as applying standard rotation matrix through $\pi/2,$ four times over.
