Equation of a line using parameters $\theta$ and $\rho$ or $r$

Question

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?

Answer 1

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

$$\color{green}{x\sin t'-y\cos t'+p=0}.$$

Answer 2

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.