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I need to graph a rectangle on the Cartesian coordinate system. Is there an equation for a rectangle? I can't find it anywhere.

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  • $\begingroup$ 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 ? $\endgroup$ – Sasha Oct 1 '11 at 21:05
  • $\begingroup$ I'm looking for a cartesian equation of a rectangle. For example the equation of a circle is $x^2 + y^2=a^2$ $\endgroup$ – Cobold Oct 1 '11 at 21:11
  • $\begingroup$ An implicit Cartesian equation would be the one Peter gave. Methinks that ain't much. Maybe you want a parametric equation? $\endgroup$ – J. M. is a poor mathematician Oct 1 '11 at 22:28

11 Answers 11

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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. ;)

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  • $\begingroup$ 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?). $\endgroup$ – J. M. is a poor mathematician Oct 1 '11 at 23:32
  • $\begingroup$ 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. $\endgroup$ – Zev Eisenberg Jan 29 '14 at 1:53
  • $\begingroup$ Forgot link: math.stackexchange.com/questions/655369/… $\endgroup$ – Zev Eisenberg Jan 29 '14 at 2:00
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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

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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.

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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.

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  • $\begingroup$ +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 $\endgroup$ – Xoque55 Dec 10 '15 at 22:26
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Maybe you're looking for something like this: for $x\in(-1,2)$ plot $y=|x|$ and $y=3-|x-1|$.

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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.

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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.

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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.

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

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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 $4\alpha>b^2$.

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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.

$$ \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix} $$

$$ x_1 = x_0\cos \theta - y_0\sin\theta\\ y_1 = x_0\sin \theta + y_0\cos \theta\\ $$

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Lets say the rectangle is centred at $c(h,k)$ and half the length is $a$ and half the width is $b$.

The equation for this rectangle would be:

$$(\frac{x-h}{a})^\infty+(\frac{y-k}{b})^\infty=1$$

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  • $\begingroup$ $x^\infty$ is not a thing. $\endgroup$ – Rahul Jul 30 '18 at 14:09
  • $\begingroup$ @Rahul 1: Can you direct me to a proof that my equation is not a thing ? 2: Graphing software seems to think it is a thing, which is what the OP asked for. 3: A large even number can be used instead of $\infty$ for a close approximation. $\endgroup$ – Kantura Jul 31 '18 at 11:31

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