Why Straight Lines? Uncovering the Theory Behind f(x,y)=0

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Discussion Overview

The discussion revolves around the function f(x,y) = x^2 - 4xy + 3y^2 and its representation as a graph showing two straight lines. Participants explore the reasons behind the emergence of straight lines from the equation f(x,y) = 0, including the implications of substituting y=mx and the nature of the function's factors.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that the equation can be factored into linear components, suggesting that the zero locus consists of the union of two lines.
  • Another participant mentions the homogeneity of the function, indicating that it is a degree 2 curve with a double point, which typically results in two lines.
  • A follow-up question arises regarding the nature of the locus f(x,y)=c for positive constants, which is initially described as an ellipse, prompting curiosity about its transition to two lines when c=0.
  • One participant corrects their earlier confusion about the shape of the locus, clarifying that it is a hyperbola rather than an ellipse due to the coefficients in the equation.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and interpretation of the mathematical properties involved. There is no consensus on the visualization of the transition from an ellipse to two lines, and some confusion persists regarding the nature of the curves described.

Contextual Notes

Participants reference algebraic geometry concepts and the implications of degree and homogeneity in polynomial equations. There are unresolved questions about the geometric interpretation of the function's behavior across different values of c.

dodo
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Given the real function [tex]f(x,y) = x^2 - 4xy + 3y^2[/tex], the equation f(x,y) = 0 shows in a graph as 2 straight lines, y=x and y=x/3. For pairs (x,y) between the lines, f(x,y) < 0; for (x,y) outside the lines, f(x,y) > 0.

It is easy to prove the above, by substituting y=mx in the equation f(x,y)=0 and finding the values of m (giving either 1 or 1/3). All the behavior above follows by playing with m.

The question is, why straight lines? What's the theory for choosing y=mx as the appropriate substitution?
 
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that equation is reducible, i.e. x^2 - 4xy +3y^2 = (x-y)(x-3y), and both factors are linear, i.e.describe lines. so the zero locus of the product is the union of the zero loci of those 2 lines.

one way to "know" it is going to consist of lines is to note that it is homogeneous of degree 2, i.e. all terms have total degree 2, so the origin is a double point, and the only degree 2 curve with a double point is the union of 2 lines.

thus in fact any curve with equation aX^2 + bXY + c Y^2 factors into 2 linear equations (over the complex numbers).

this is the very simplest beginning example of the algebraic geometry of plane curves. take a look at walker,
algebraic plane curves, for an elementary and concrete, yet deep and expert discussion.
 
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Hey, thanks. I have to read (or google) some things to fully understand that, but it's a good start.

I take you mean "the product of two linear factors, (ax+by)", when you say "the union of 2 lines", here,
and the only degree 2 curve with a double point is the union of 2 lines.
(so they are actually two lines having a zero intercept, i.e. both passing through the origin).

One little follow-up: the locus of f(x,y)=c, for c a positive constant, is an ellipse, no matter how small c is. How come this degrades to two lines when =0? I'm trying to visualize horizontal slices on a 3D shape, but this beats me.
 
Dodo said:
One little follow-up: the locus of f(x,y)=c, for c a positive constant, is an ellipse, no matter how small c is. How come this degrades to two lines when =0? I'm trying to visualize horizontal slices on a 3D shape, but this beats me.
Oh, sorry, forget this last paragraph. Since B^2 > 4AC, this is an hyperbola, not an ellipse. I was confused due to a graph I made a few days ago, but in that case the cross-term coefficient B was -3, not -4.
 
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