I am learning about the straight line and its slope was defined as "rise over run".
Why not "run over rise"? Was "rise over run" chosen arbitrarily or it was derived based on some physical relationship?
Thanks in advance
2026-03-25 17:21:33.1774459293
On
On
On
On
On
Definition of a slope of a straight line
423 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
6
There are 6 best solutions below
0
On
With the conventional definition, "large slope" corresponds to a very steep uphill incline, and "zero slope" corresponds to a horizontal line. Both of those line up naturally with how we talk about a physical incline.
0
On
Rise is the difference in y and run is the difference in x. If you are given 2 point point1= $(x_1,y_1)$ and point2= $(x_2,y_2)$ the difference can be written as $(x_1-x_2\,,\,y_1-y_2)$ if the two points are on the same line than this difference contains the ratio of rise to run for the line. (run,rise) and the slope of the line can therefor be written as $y_1-y_2 \over x_1-x_2$ or as it is often said $rise \over run$.
Because we do math on a chalk board where we typically have x on the horizontal and y on the vertical doing slope this way is sensible as it makes the slope of a horizontal line 0 and a vertical line $\infty$, which is intuitive.
0
On
We need to go back to the linear function $y=mx+b$, $m$ is the slope and the y-intercept is $(0,b)$. What does that all mean? Well, let's replace $x$ by $x+1$ and plug that in the equation: $y=m(x+1)+b=mx+b+m.$ (Watch that order: I wrote $b+m$ instead of $m+b$) With respect to the given function, if you go one unit to the right from $(0,b)$, you will now have to go $m$units up to get back on the line. (Draw it out on a paper!) So here we have established the most important property of the slope: It is the vertical displacement needed to get back on the line, if you move one unit away horizontally from a point on the line. The imaginary triangle, which you often see on laminated cheat sheets for Algebra 1 is depicted with unit $1$ on the horizontal side, $m$ on the vertical side and the line passing though the hypothenuse. Now how does that relate to $\frac{y2-y1}{x2-x1}?$ This is essentially "scaling down" a right triangle between two given points in the xy plane to the triangle that has sides $1$ and $m$, think of similar right triangles. This is where the "rise over run" comes from. Not a definition, but a well crafted consequence of what the slope truly means.
0
On
There are are just two choices: rise over run or run over rise. Choosing is a matter of convention and then one sticks to convention.
Anyway let me attempt to give some reasoning:
Imagine a civil engineer's problem to construct a road bridge to cross a railway line. She/he takes into consideration the height of trains and arrives at some number like 9 metres above the ground level would be the appropriate maximum height for the bridge.
Having arrived at that number how much height to clear the engineer has to decide what run is needed to reach that height that will be comfortable to drivers while climbing. Approximating the slope as rational number $m/n$ would mean $m$ metres of rise needs $n$ metres of run. So the bridge will start $n$ metres from the first point where height $m$ is required.
0
On
It depends on what you mean by “rise” and “run”. If you take “run” to mean “horizontal distance”, then it's arbitrary, because (this part of) mathematics at this level doesn't care about the difference between horizontal and vertical.
If by “run” you mean “independent variable” (say $x$) and by “rise” mean “dependent variable” ($y$, $f(x)$, and such), then there is a reason to choose this definition of slope over another:
With this definition, the slope of $f$ where $f(x) = 0$ is $0$, which is a real number.
With the “run over rise” definition, the slope of $f$ is undefined, or at least not a real number if you choose to use infinities.
Since we like to avoid dealing with $\infty$, which has fewer useful properties than regular real numbers, the usual definition is more useful.
(Also, in “run over rise” there is no function which has a slope of $0$. So the usual definition has fewer oddities all around.)
Slope is a word that is used to describe many things, but I think one thing is most pertinent here..."effort" as it relates to moving up an incline. It's intuitive. That's probably why they chose "rise over run". Of course, they could have chosen "drop over run", but that just doesn't work as well.