Linear Function with Coefficients in Arithmetic Progression

I found an engaging example at Chris Harrow's blog. Chris is the Mathematics Chair at the Hawken School. Chris illustrates the problem in the free online Desmos calculator. I undertook to illustrate it in GeoGebra. Here's the problem:

All linear functions \(ax+by+c=0\), where the coefficients \(a,b,c\) form an arithmetic sequence, share a certain property. What is it?

The applet below is assumed to be suggestive. There are two sliders: one defines the value of \(a\), there other of \(d\) such that \(b=a+d\) and \(c=a+2d\). (For idiosyncratic reasons, GeoGebra display the linear equation as \(ax+by=-c.\)

9 January 2013, Created with GeoGebra

If you play with the applet it is pretty hard to miss the point (pun intended): all straight lines that have coefficients in arithmetic progression pass through the point \((1,-2)\). Verification is straightforward:

\(a\cdot 1+(a+d)(-2)+(a+2d)=0.\)

Chris list several methods by which the students can arrive at this conclusion.

Related material

  • What Is Line?
  • Functions, what are they?
  • Cartesian Coordinate System
  • Addition and Subtraction of Functions
  • Function, Derivative and Integral
  • Graph of a Polynomial of arbitrary degree
  • Graph of a Polynomial Defined by Its Roots
  • Inflection Points of Fourth Degree Polynomials
  • Lagrange Interpolation (an Interactive Gizmo)
  • Equations of a Straight Line
  • Taylor Series Approximation to Cosine
  • Taylor Series Approximation to Cosine
  • Sine And Cosine Are Continuous Functions
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