Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

This is something I've been wondering about. Namely, I've always accepted "on intuition" that the equation

$$ax + by = c$$

is, when graphed, a line. You can plot the points $(x, y)$ satisfying the equation and see that yeah, they do indeed form a line.

But when I came across this, I realized was that I had accepted this idea without proof:

There is enough material in the text to convince the students empirically that a line in the plane is represented by a linear equation, and that the graph of a linear equation is a line. However, these two fundamental theorems on linear functions are not justified mathematically. (pg. 5)


Important theorems on linear functions are not proved. Relevant to the above two standards are two fundamental theorems: A line in the plane is represented by a linear equation and the graph of a linear equation is a line. Neither of these theorems is proved. (pg. 16)

Which made me wonder (I haven't seen the texts) -- just how would you not only "justify" this mathematically, but in a way a high-schooler would understand? And not only that, but in a manner which is actually enlightening? In addition, this criticism is leveled against all four high school geometry/algebra books.

For example, consider if we were using, say, Hilbert's axioms as our axiom set for Euclidean geometry. Then we could show that the equation is a line by something like this: find three points A, B, and C satisfying it so that $A * B * C$ (the "betweenness" relation), then show that for any point $D$ which is not $A$ or $B$, then one of $D * A * B$, $A * D * B$, or $A * B * D$ must hold. Going the other way (the converse), to show the line is given by the equation, you'd show that for any points A, B, C with $A * B * C$, then every point $D$ with $D * A * B$, $A * D * B$, or $A * B * D$ satisfies some equation of the form $ax + by = c$. However, it seems this kind of proof is fairly tedious (you have to check three cases in both implications), and relies on quadratic functions and radicals since you have to use the distance formula as that's how you'd define the "betweenness" relation for three points.

I suppose the details would vary with regard to the axiom set we use (I don't know if Hilbert's would necessarily be the best for "high school geometry") -- but it seems no matter which one we use, we need some way to determine that three points A, B, and C "lie on the same line" (which is what the "betweenness" relation does, although it does more, since it also orders the points), and a way to express this with regards to coordinatized points as well as points in the axiomatic geometry which the coordinate plane models. It seems that any formula I've seen for that fact using Cartesian coordinates requires a quadratic polynomial expression, for one. Any proof along these lines seems like it would be tedious, or require additional motivation, and so might not be enlightening at this lower level of the person's mathematical development. The "message" seems to easily get lost as one gets bogged down in mechanics.

How would you solve this problem? What's a good way to justify this at such a level?

share|cite|improve this question

marked as duplicate by rschwieb, Thomas Andrews, user127.0.0.1, Henry Swanson, Blue Feb 8 '14 at 8:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Yes, in theory you can use hilbert's axioms to show that lines drawn with straightedges are exactly the solution sets of linear equations. This is bound up with the proof that the real line coordinatizes the complete Euclidean plane. This approach is rather unwieldy to teach to beginners. – rschwieb Feb 8 '14 at 4:04
@rschwieb: Thanks for that -- just wasn't able to find anything like that on the searches (bad keywords?). However, the top-rated answer on there involves fields and other concepts which are well beyond high-school level. – mike4ty4 Feb 8 '14 at 4:11
I think this is a quality and important question. I contacted Prof. Harel (author of the report you quote), and he sent me the proofs that he had in mind. In particular, he starts with the coordinate definition of a line being all the points equidistant from two starting points (perpendicular bisector of segment PQ). – Daniel R. Collins Oct 26 '15 at 14:38

I think of similar triangles. This is the foundation upon which slope is built. Can you proceed along these lines?

share|cite|improve this answer
Pun intended? ;) – rschwieb Feb 8 '14 at 3:55
Now @rscweib, you'd never expect that from me! Fie and shame! Stuff and nonsense! – ncmathsadist Feb 8 '14 at 3:59
Yes, I think this would probably work much better. Then you don't need to mess around with adding up distances, just checking ratios. – mike4ty4 Feb 8 '14 at 4:00
I'm not entirely sure this answers the question, but I also could have mistaken the intent of the question. As far as I know, the proof that lines drawn with straight edges coinciding with solution sets of linear equations is not as trivial as using similar triangles... – rschwieb Feb 8 '14 at 4:02
Similar triangles drives the principle of slope. Draw a diagram and see. – ncmathsadist Feb 8 '14 at 4:11

Could you use the pythagorean distances? I'm thinking aloud, so please forgive any lapses in logic, but if you select three points, p1, p2, p3 on a straight line, you should get d(p1, p2)+d(p2,p3) = d(p1, p3). This should not be true for any three points not on a straight line. Pythagorean distance is nice for this because it's possible they're familiar with it, it's easy to justify if not, and it's easy to calculate. Again - this is off the top of my head, so corrections are welcome.

EDIT: The reason I'm a little skeptical about this suggestion is that I don't see immediately how I'd prove that this fact about the distances holds for the linear situation, and just for the linear situation. It'll come to me, I suppose, but if I had that I'd be happier.

share|cite|improve this answer

I'd start with some definitions:

  • A point is a location without any extent.
  • Two distinct points determine a unique line. This line is straight and extends forever in both directions.
  • The line segment (the part of the line starting at one point and ending at the other) has the shortest distance between those two points.

I'd then assign coordinates to those two points. It might help to use numbers, like $(2,1)$ and $(3,5)$. I'd then say that there's an equation that describes all points on this line, but that we don't know it yet.

I'd then investigate what it means for a point to be "on the line." Then I'd develop the idea that $\Delta y = m \Delta x$ where $m$ is a constant. That it's constant is the key. Maybe an illustration of this would be to put yourself in the plane, and march straight from one point to the other. The shortest way (gets back to the definition) is never to change direction, and always have the other point straight in front of you.

Then I'd calculate this $m$ for the line I have: $m = (5-1)/(3-2) = 4.$

I might conclude that the line's equation is $y = 4x$ but I'd be wrong. (I plug in a $y$ and I don't get an $x$ on the line, or vice versa.) At this point I'd show a few more lines that have a slope (introducing the term) of $m=4$, and then conclude that I need some other point to determine the equation of the line uniquely. I might note that the place where the line crosses the $y$ axis changes for my different lines, and then develop the idea of the $y$ intercept.

Finally, after using this specific example, I'd go out and generalize to points $(x_1, y_1)$ and $(x_2, y_2)$ which would lead to an equation in the general form $ax + by = c$.

Is this proof? Ehhhh ... don't know, but it may be "proof enough" to dispel the assumptions.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.