# Why is the difference in rise equal to the difference in run*slope?

• Mar 28th 2013, 02:40 PM
MiguelTime
Why is the difference in rise equal to the difference in run*slope?
Point slope form: y - y1 = m(x - x1)

I realize this is a manipulation of the slope equation: m = rise/run, but I don't understand what is happening or why it works. The difference in your y values is equal to the difference of your x values multiplied by the slope. I'm having trouble seeing this. I don't exactly understand what is going on here. How and why?

I came to ask if someone could please clear this confusion up for me? I would really appreciate the help in understanding this! :)

• Mar 28th 2013, 02:45 PM
HallsofIvy
Re: Why is the difference in rise equal to the difference in run*slope?
You seem to know that "slope" is defined as m= rise/run. If you multiply both sides of that equation by "run" you get "rise= m*run". That is, because [tex]m= \frac{y_1- y_0}{x_1- x_0}[tex], multiplying both sides by $x_1- x_0$ you have $y_1- y_0= m(x_1- x_0)$
• Mar 28th 2013, 03:41 PM
Plato
Re: Why is the difference in rise equal to the difference in run*slope?
Quote:

Originally Posted by MiguelTime
Point slope form: y - y1 = m(x - x1)
I realize this is a manipulation of the slope equation: m = rise/run, but I don't understand what is happening or why it works.

I have never been a fan of that definition: the slope equation: m = rise/run.

Now I understand why members of the mathematics education community invented it. But I don't find is useful.

Given two points, the slope of the line they determine is the change is the second coordinates divided by the change is the first coordinates (provided the second is not zero).

Examples: If $P: (4,3)~\&~Q: (7,1)$ then the change is the second coordinates is $-2$ while the change is the first coordinates is $3$ so that slope is $\frac{-2}{3}$.

Now all other pairs of points on the line $\overleftrightarrow {PQ}$ share that same ratio.