Equation of a rectangle I need to graph a rectangle on the Cartesian coordinate system. Is there an equation for a rectangle? I can't find it anywhere. 
 A: Try plotting $x^n + y^n = p^n$ where $p$ is the side length and $n$ is an even number. The larger $n$ is, the sharper the sides are.
A: I found recently a new parametric form for a rectangle, that I did not know earlier: 
$$
\begin{align}
x(u) &= \frac{1}{2}\cdot w\cdot \mathrm{sgn}(\cos(u)),\\
y(u) &= \frac{1}{2}\cdot h\cdot \mathrm{sgn}(\sin(u)),\quad  (0 \leq u \leq 2\pi)
\end{align}
$$
where $w$ is the width of the rectangle and $h$ is its height.  
I have used this in modelling parametric ruled surfaces, where it seems to be rather handy.
A: If the equations of the diagonals of the rectangle are Ax + By + C = 0 and Dx + Ey + F = 0 then an equation for the rectangle is:
M|Ax + By + C| + N|Dx + Ey + F| = 1
M and N can be found by substituting the coordinates of two adjacent vertices of the rectangle.
In fact, this equation can be used to describe any parallelogram.  Roughly speaking, M (together with A and B) and N (together with D and E) give the size of the diagonals of the parallelogram. 
A: Based on Raskolnikov's answer here, one can build an implicit Cartesian equation for a $2p \times 2q$ rectangle:
$$\left(\frac{x}{p}\right)^2+\left(\frac{y}{q}\right)^2=\sec\left(\arctan\left(\frac{x}{p},\frac{y}{q}\right)-\frac{\pi}{2}\left\lfloor\frac2{\pi}\arctan\left(\frac{x}{p},\frac{y}{q}\right)+\frac12\right\rfloor\right)^2$$
Another one is based on modifying the implicit equation of a Lamé curve:
$$\left|\frac{x}{p}+\frac{y}{q}\right|+\left|\frac{x}{p}-\frac{y}{q}\right|=2$$

For purposes of plotting with a computer, the implicit equation isn't terribly convenient to handle, so I'll throw in a set of parametric Cartesian equations for free, based on the parametric equations of the Lamé curve:
$$\begin{align*}x&=p\left(|\cos\,t|\cos\,t+|\sin\,t|\sin\,t\right)\\y&=q\left(|\cos\,t|\cos\,t-|\sin\,t|\sin\,t\right)\end{align*}$$
Here's another one, based on a special case of the parametric equations given in this answer:
$$\begin{align*}x&=p\left(\cos\left(\frac{\pi}{2}\lfloor u\rfloor\right)-(2u-2\lfloor u\rfloor-1)\sin\left(\frac{\pi}{2}\lfloor u\rfloor\right)\right)\\y&=q\left(\sin\left(\frac{\pi}{2}\lfloor u\rfloor\right)+(2u-2\lfloor u\rfloor-1)\cos \left(\frac{\pi}{2}\lfloor u\rfloor\right)\right)\end{align*}$$
...and another one:
$$\begin{align*}x&=p\max\left(-1,\min\left(\frac4{\pi}\arcsin\left(\sin\left(\frac{\pi u}{2}+\frac{\pi}{4}\right)\right),1\right)\right)\\y&=q\max\left(-1,\min\left(-\frac4{\pi}\arcsin\left(\cos\left(\frac{\pi  u}{2}+\frac{\pi}{4}\right)\right),1\right)\right)\end{align*}$$
...and I suppose I should stop here. ;)
A: Maybe you're looking for something like this: for $x\in(-1,2)$ plot $y=|x|$ and $y=3-|x-1|$.
A: In general, the implicit formula for a rectangle a la $x^2 + y^2 = a^2$ for circles is not going to be well defined.  This should be at least somewhat clear, as the boundary of a rectangle is not analytic (smooth) like the boundary of a circle is.  I suppose we could generate a piece wise function to graph the edges, something like:  $$f(x,y) = \begin {cases} (x,b) , (x,0) &  0 \leq x \leq b \\

(0,y) , (a,y) &  0 \leq y \leq a \end {cases}$$
For a rectangle with its bottom left corner at (0,0) and sides a,b.  Such a function is messy, still non-analytic and doesn't help you that much.    Ultimately, I think searching for a good implicit function of a rectangle is going to be nonproductive. What problem are you trying to apply this to?  Any comment as to your next steps / applications for the equation you're searching for will prove helpful.
A: This is very easy. Instead of using all of that complex math, you can instead just use the rotation matrix to rotate a simple absolute value function.
$$
     \begin{pmatrix}
     \cos \theta & -\sin \theta \\
     \sin \theta & \cos \theta \\
     \end{pmatrix} 
     \begin{pmatrix}
     x \\
     y \\
     \end{pmatrix}
$$
$$
      x_1 = x_0\cos \theta - y_0\sin\theta\\
      y_1 = x_0\sin \theta + y_0\cos \theta\\
$$
Afterwards substitute the angle to be 45°. The original equation is |x|+|y|=c
This is because the absolute value is at a 45 degree angle.
$$
      |\frac {\sqrt{2}}{2}x+\frac {\sqrt{2}}{2}y|+|\frac {\sqrt{2}}{2}x-\frac {\sqrt{2}}{2}y| = c
$$
A: This is an equation for a rectangle which has corners at $(a,b)$ and $(c,d)$
$$(x-a)(x-c)(y-b)(y-d)=0$$
but it extends a little beyond the corners, so instead 
$$\sqrt{(a-x)(x-c)}\sqrt{(b-y)(y-d)}=0$$
which would throw an error for square roots of negative numbers 
A: There is another interesting form using the Heaviside step function: $\theta(x)$. 
If the sides are $a$ and $b$ and the rectangle is centered at $(x_0,y_0)$ then:
$$(y-y_0)^2+\alpha\theta\left[(x-x_0)^2-\frac{a^2}{4}\right]=\frac{b^2}{4}$$
where $4\alpha>b^2$.
