# 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.

• Do you mean the like of $0 \le x \le 1 \land 0 \le y \le 1$ ? Or do you need to parametrize the boundary of the rectangle ? Oct 1, 2011 at 21:05
• I'm looking for a cartesian equation of a rectangle. For example the equation of a circle is $x^2 + y^2=a^2$ Oct 1, 2011 at 21:11
• An implicit Cartesian equation would be the one Peter gave. Methinks that ain't much. Maybe you want a parametric equation? Oct 1, 2011 at 22:28

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. ;)

• I did say "for plotting purposes", @Peter. There are things like GrafEq and Mathematica for plotting implicit Cartesian equations, but in general they require way much more effort on the part of the computer to plot than parametric equations (have you seen the algorithms behind implicit equation plotters?). Oct 1, 2011 at 23:32
• I posted a new question based on this one. I want the equation of a square where each point is at the same angle as the input angle $t$, which is not true of the parametric equation above. Jan 29, 2014 at 1:53
• Forgot link: math.stackexchange.com/questions/655369/… Jan 29, 2014 at 2:00

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

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.

• +1. This is a great generalization of "rectangular-ish" shapes since $\lim_{n \to \infty}$ of $x^n+y^n=p^n$ actually gives the Max Norm: en.wikipedia.org/wiki/Uniform_norm Dec 10, 2015 at 22:26

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.

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.

Maybe you're looking for something like this: for $x\in(-1,2)$ plot $y=|x|$ and $y=3-|x-1|$.

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.

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.

$$|\frac {\sqrt{2}}{2}x+\frac {\sqrt{2}}{2}y|+|\frac {\sqrt{2}}{2}x-\frac {\sqrt{2}}{2}y| = c$$

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 $4\alpha>b^2$.