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What makes slope rise over run? What makes the standard equation for a line use a slope of rise over run as opposed to run over rise?

What would the standard equation of a line look like if m was equal to run/rise $\frac{\Delta x}{\Delta y}$

user1551
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user16795
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    We normally write $y$ as a function of $x$. For example, $x$ might be time and $y$ might be position in yards. Then $\Delta y/\Delta x$ would be in yards per second, but $\Delta x/\Delta y$ would be in seconds per yard. – Empy2 Sep 30 '14 at 15:07
  • Think about the physics: if you look at a steep hill, it has a steeper slope using the usual formula. If you used run/rise, a very steep hill would have a smaller slope. – rogerl Sep 30 '14 at 15:11
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    That is all on the logical level, what makes sense logically. I am trying to figure out where this comes from fundamentally. – user16795 Sep 30 '14 at 15:16
  • @user16795 The definition $\frac{\Delta y}{\Delta x}$ instead of $\frac{\Delta x}{\Delta y}$ is because we have a very good use of that definition, because of the connection between the definition and the "logical level". Nothing stops you from mathematically defining an "reciprocal derivative" or something with your alternative definition. – Alice Ryhl Sep 30 '14 at 15:50
  • If you called x y, and y x, nothing in math would change. Only the notation. And for that matter, why is the positive direction around the unit circle counterclockwise? Why not clockwise? That also would not cause any fundamental change in mathematics. – user4894 Sep 30 '14 at 22:33
  • @Darksonn I understand that, what would the equation for a line be if slope was inversed? Would it be more "complex"? – user16795 Oct 01 '14 at 18:13
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    @user16795 Depends on what you mean with line, if you mean something which, when drawn represents a line, it would still be $y=ax+b$ since we didn't define it relative to the derivative. The slope by this new formula, would be $\frac1a$ since $$\lim_{h\to0}\frac h{f(x+h)-f(x)}=\frac1a$$ for any $f(x)=ax+b$. If you however mean that an equation is a line if the derivative (or in this case our alternative derivative) is a constant value, the equation would simply be $y=\frac xa+b$ where $a$ is the slope. Actually that's equivalent to $y=ax+b$ however $a$ just isn't the slope anymore – Alice Ryhl Oct 01 '14 at 18:20

4 Answers4

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If you write the equation of a line as $$y = mx+b$$ then it comes from this naturally. That's because if you move along the line a little bit from the point $(x,y)$ to the point $(x+\Delta x, y+\Delta y)$, then the new point satisfies the equation too. So you have that $$y+\Delta y = m(x+\Delta x) + b = mx + m\,\Delta x + b$$ Subtracting these gives $$\Delta y = m\,\Delta x$$ which means that $$\frac{\textrm{rise}}{\textrm{run}} =\dfrac{\Delta y}{\Delta x} = m$$ You can call it what you want, but the coefficient of $x$ in the original equation is the rise over the run. If you want to call run/rise the slope, then its reciprocal would be the coefficient of $x$.

Michael Hardy
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MPW
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Graph of a linear function

I think it has something to do with how we write from left to right and therefore interpret that graph and the definition of slope "from left to right". Look at graph 1 normally and you will observe a slope of 3, RISE over RUN. Our reference axis is x.

Same graph but from a rotated view

Look at graph 2 now and if you read the axis from right to left instead, with the positive and negative inverted, you can instead observe a slope of 1/3, RUN over RISE. Our reference axis here is y.

Hope that helped.

Shlok Utchanah
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Just like a linear function $y=ax+b$ shows the relation between the two variables (which is dependent is determined by the problem), the slope of the line (linear function) shows the relation between changes of the two variables: $\frac{\Delta y}{\Delta x}$ or $\frac{\Delta x}{\Delta y}$.

The key is which variable is dependent.

For $y=ax+b$, if $y$ is dependent on $x$, then the slope is: $$\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}=\frac{(ax_2+b)-(ax_1+b)}{x_2-x_1}=a;$$ if $x$ is dependent on $y$, then the slope is: $$\frac{\Delta x}{\Delta y}=\cdots=\frac{1}{a}.$$

farruhota
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We conventionally draw graphs like this: $$ \begin{array}{c|ccccccc} 10 & & & & & & & \bullet \\ 9 \\ 8 & & & & & & \bullet \\ 7 \\ 6 \\ 5 & & & & \bullet & \bullet \\ 4 \\ 3 & & & \bullet \\ 2 \\ 1 & \bullet \\ 0 & & \bullet \\ \hline & \text{sunday} & \text{monday} & \text{tuesday} & \text{wednesday} & \text{thursday} & \text{friday} & \text{saturday} \end{array} $$ The slope should measue how steep the graph is, which in this example is how fast this changing quantity is changing as time passes. As you see, this quantity went from $1$ up to $10$ in $6$ days. That means its average rate of change is $9/6=1.5\text{ per day}$. That is the slope of the line through the first and the last of the plotted points. That is how steep it is, and thus it is how fast the quantity changes.

Michael Hardy
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