Equation of a rectangle

Question

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

Answer 1

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:

$$\left|\frac{x}{p}+\frac{y}{q}\right|+\left|\frac{x}{p}-\frac{y}{q}\right|=2$$

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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. ;)

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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

Answer 7

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.

Answer 8

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.

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

Answer 9

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 $4\alpha>b^2$.