
How to obtain an equation of a polygon
Posted:
Aug 19, 2014 4:13 PM


Gary Tupper wrote (in part):
http://mathforum.org/kb/message.jspa?messageID=9558358
> Well, if you have used the software to graph say (yx+3)(xy^2)=0 > or (yx)(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 oneday 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 zeroset of a 2variable 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 halfplane), the graph of x  1 + (x  1) = 0 (gives another halfplane). 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

