Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: How to obtain an equation of a polygon
Replies: 16   Last Post: Aug 22, 2014 12:10 AM

 Messages: [ Previous | Next ]
 Gary Tupper Posts: 544 Registered: 12/3/04
Re: How to obtain an equation of a polygon
Posted: Aug 19, 2014 11:12 PM

Greetings, Dave:

My interest in graphing software derives from my having created a
document to display how my son's graphing program could be utilized in
the study of elementary algebra. I am of the opinion that the software
can provide independent verification of the user's derivation of some
particular shape or locus. Some of the notions you alluded to are
included in the document. Perhaps of particular interest: the equation
of a regular polygon can be expressed concisely using the modulo operator.

http://www.peda.com/gary/GrafEqApplications.pdf

Gary Tupper

On 19/08/2014 1:13 PM, Dave L. Renfro wrote:
> Gary Tupper wrote (in part):
>
> http://mathforum.org/kb/message.jspa?messageID=9558358
>

>> Well, if you have used the software to graph say (y-x+3)(x-y^2)=0
>> or (y-|x|)(y+3)=0 & some similar such {exprA * exprB * exprC}=0 etc.,
>> then it would not be unreasonable to ask the student on a test to
>> provide the equation of the '+' coordinate axes.

> Note: I'm starting a new thread because what I'm posting might be
> of sufficient interest to others (lurking now or those stumbling
> on this at some future time) that I'd rather not burry it in the
> thread "Software We've Liked" that Gary Tupper's post is in.
>
> This reminds me of something neat I came across a few years ago.
> This was after I left teaching, so I never tried it in a classroom,
> but it seems to me that it could be used to make a worksheet for a
> possibly engaging one-day project for precalculus students to work
> on in groups. The method below shows how you can obtain an explicit
> equation whose graph is a specified polygon in the coordinate plane.
>
> I'll begin by making a couple of observations on how one can
> build new graphs from existing graphs.
>
> Let F and G be the graphs (considered as subsets of the coordinate
> plane) of f(x,y) = 0 and g(x,y) = 0. Then, using * for multiplication,
> we have the following two useful ways of obtaining other sets
> as graphs of an equation:
>
> "F union G" is the graph of f(x,y) * g(x,y) = 0
>
> "F intersect G" is the graph of |f(x,y)| + |g(x,y)| = 0
>
> For intersection you can also use the graph of
> [f(x,y)]^2 + [g(x,y)]^2 = 0 if you want to avoid using
> the absolute value function (e.g., if you wanted the
> end result to be the zero-set of a 2-variable polynomial).
>
> The basic idea is to determine how to obtain an equation
> whose graph is any specified line segment, and then use
> the first building principle above to obtain an equation
> whose graph is any specified set that can be expressed as
> a finite union of line segments.
>
> I'll show how to obtain an equation for the line segment
> with endpoints (1,0) and (0,1) by a method that can be easily
> adapted to obtain an equation for a line segment with
> any specified endpoints (a,b) and (c,d).
>
> First, consider the line passing through (1,0) and (0,1).
>
> Write this in the form of f(x,y) = 0. One possibility
> is x + y - 1 = 0.
>
> Consider what additional conditions need to be imposed so
> that we get the line segment with endpoints (1,0) and (0,1),
> rather than the whole line. In the case of x + y - 1 = 0,
> we want x >= 0 and x <= 1.
>
> Now let's express these inequality conditions as equations.
> One way to do this is |x| - x = 0 (for x >= 0)
> and |x - 1| + (x - 1) = 0 (for x <= 1).
>
> This comes directly from the definition of absolute value.
> Recall |u| = u iff u >= 0, and |u| = -u iff u <= 0.
>
> The line segment with endpoints (1,0) and (0,1) arises as
> the intersection of these three sets -- the graph of
> x + y - 1 = 0, the graph of |x| - x = 0 (gives a half-plane),
> the graph of |x - 1| + (x - 1) = 0 (gives another half-plane).
> Therefore, this intersection (i.e. the desired line segment)
> can be obtained as the graph of
>
> |x + y - 1| + ||x| - x| + ||x - 1| + (x - 1)| = 0
>
> A similar method can be used to obtain the line segment
> with endpoints (a,b) and (c,d) as the graph of an equation
> of the form f(x,y) = 0.
>
> For polygons (or anything that can be expressed as a union
> of finitely many line segments in the plane), you just obtain
> equations f_1(x,y) = 0, f_2(x,y) = 0, ..., f_n(x,y) = 0 for
> each of the line segments, and then the desired planar set
> will be the graph of
>
> f_1(x,y) * f_2(x,y) * ... * f_n(x,y) = 0.
>
> Dave L. Renfro

Date Subject Author
8/19/14 Dave L. Renfro
8/19/14 Robert Hansen
8/19/14 Robert Hansen
8/19/14 Gary Tupper
8/20/14 GS Chandy
8/20/14 Robert Hansen
8/20/14 Gary Tupper
8/20/14 Robert Hansen
8/20/14 GS Chandy
8/20/14 Robert Hansen
8/20/14 Dave L. Renfro
8/20/14 GS Chandy
8/20/14 GS Chandy
8/21/14 Dave L. Renfro
8/21/14 Robert Hansen
8/22/14 Bishop, Wayne
8/22/14 Bishop, Wayne