Intercept Theorem

Date: 06/02/2003 at 23:10:10
From: Travis
Subject: My own Theorem

A few teachers helped me prove my own theorem. My teachers have told 
me they have never seen a theorem like it before.

My theorem states that by dividing (B) by the opposite of (M) or (-M) 
you get the x-intercept.

Date: 06/03/2003 at 08:56:13
From: Doctor Peterson
Subject: Re: My own Theorem

Hi, Travis.

This is not new, but is certainly a useful fact. You can prove it 
easily:

  y = mx + b

  0 = mx + b  for x-intercept

  -b = mx     subtracting b from both sides

  -b/m = x    dividing both sides by m

So it doesn't save a lot of work, but if you are working on a lot of 
these problems, it can add up. I myself probably derive this fact the 
way I just showed every time I solve one of these; that is, rather 
than memorizing it, I am aware in the back of my mind that there is 
such a trick, go through the bit of algebra in my head quickly to 
remind myself what is divided by what, and then apply it to each 
problem. That's a good way to keep your mind uncluttered but be able 
to save time on repetitive work.

I'm a little surprised your teachers haven't seen it; but that may be 
because it is not generally taught as a fact by itself. Since you and 
I solve these problems on our own (you as a student, I in helping 
students without having an answer book!), we are more likely than a 
teacher to do them enough to want and discover a shortcut like this.

Keep looking for easier ways to do things, because that's how a lot 
of useful math is discovered. If after every problem that's at all 
hard, you look back and ask yourself, "How could I have done this 
more quickly?" - or, "How can I do the next one with less work?" - 
you will be using your mind at its best.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/