Why is the slope of a line defined the way it is.

Question

This may seem like a silly question but why is it $ \dfrac{y_2 - y_1}{ x_2 - x_1} $ instead of $ \dfrac{x_2 - x_1}{ y_2 - y_1} $ . It is a rate of change so why is it defined they way it is?

Answer 1

Probably the formula is the way it is for historical reasons.

Generally speaking, our intuition of "steepness" matches the definition of slope: the steeper a line is, the higher is its slope in absolute value. This works with slope equal to $\text{rise} / \text{run}$, but would not work if the slope were equal to $\text{run} / \text{rise}$ as you suggest.

For a modern example, when you drive down a steep mountain highway, you will notice signs saying something like "Danger: 5% grade ahead. Trucks shift to low gear". This sign is informing the truck driver that the slope, as a ratio of rise over run, is equal to $-.05$ (which, to a truck driver, is dangerously steep). A higher percentage represents a higher absolute value of slope represents a steeper highway represents more danger for the truck driver.

Answer 2

It doesn't really matter whether we consider "change in $y$ over change in $x$", or the reciprocal - can mix up our definitions as long as we mix up the way we think about things, in the exact same way. In the reciprocal version, slopes closer to $0$ would be more nearly verical, rather than slopes nearer $\pm\infty$, as they are now.

That said, with $y$ typically being the dependent, or response variable, we have a good precedent for making it the top of the fraction. We measure automobile efficiency in miles per gallon, where miles are really the response variable. We calculate normalized costs as price per unit; it's more natural to think of the unit, the denominator, as stable; it doesn't depend on anything. Psychologically, that's just the way humans (at least in my neck of the woods) like things, in all the examples I can think of.

Answer 3

When we define slope, we first have to figure out what we mean by slope. Usually we want the slope to be a number which represents the "steepnes" of the line, and a more steep line should have a larger slope.

The way we defined the slope exactly is: The amount that $y$ changes by, when we change $x$ by $1$. This is just a definition we could have said "The amount that $y$ changes by, when we change $x$ by $2$." instead, however we like simplicity and $1$ is the simplest number to use here.

The above definition gives the formula you know: $$ \frac{y_2-y_1}{x_2-x_1} $$

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As for the similar formula $$ \frac{x_2-x_1}{y_2-y_1} $$ This actually doesn't serve us well as slope, because the steeper the line would be, the smaller the slope. That's the opposite of what we wanted.

Answer 4

The intuitive concept of ''slope'' can become a rigorous mathematical concept only if we define some straight line to be ''horizontal'' and some other to be ''vertical''. If you draw a line on a paper, and you does not fix a coordinate system, this line has no well defined slope. Only wen you define two orthogonal lines as axis of an orthogonal reference system, your intuitive concept can be well defined. Usually we represents the horizontal axis as $x$ axis and the vertical as $y$ axis ( this is matter of convention), so the slope, can be defined as the ratio between the vertical displacement and the corresponding horizontal displacement, i.e. $\dfrac{y_2-y_1}{x_2-x_1}$.

As a consequence: in a non orthogonal coordinate system, the concept of slope become less interesting.

Answer 5

To summarize what has already been said on this, it's a common definition based on convention. Common definitions and mathematical conventions are helpful because they allow us to express many ideas, sometimes quite complex ideas, without having to explain everything from the beginning. So if I say I want to graph the function $y = 2x$, people with sufficient mathematical training will generally understand that I meant $y$ to be a function of $x$, and if I show them the graph I drew they will recognize it as a graph of a function that maps $x$ to $2x$ and not a function that maps $x$ to $x/2$.

So we tend to follow the common definitions, except sometimes when we don't.

For example, in the United States someone might say their car gets $30$ miles per gallon. In other countries the same car would be said to consume $7.84$ liters per $100$ kilometers. So if distance on the odometer is $x$ and fuel in the tank is $y$, people in one country use $\dfrac{y_2-y_1}{x_2-x_1}$ to measure something that people in another country use $\dfrac{x_2-x_1}{y_2-y_1}$ to measure. So here we have two groups of people using opposite conventions for a related-rates question. (People in the second country would probably prefer to say fuel in the tank is $x$ and distance on the odometer is $y$, however. Some conventions are stronger than others!)

Answer 6

Commonly $y=f(x)$. This is traditionally the case, and so you have $y$ and $x$ axes. Since $y$ is a function of $x$, the rate of change of $y$ over $x$ is $dy/dx$.

Hence $(y2-y1)/(x2-x1)$. You could conceivably print a 2 axis graph based on a multi dimensional function with the axes representing 2 of the parameters, but you'd also be focused on trying to find the relationship of the parameter of the $y$ axis vs the $x$. The labels don't really matter, but that's the convention.

Answer 7

Geometric Answer

Intuitively, we, to some degree, would agree that steepness would depend on how slanted a line is, where an angle between the line determines the slantedness (or more generally, the line's orientation). The larger the angle, the more elevated or steep the line is, and vice versa.

Furthermore, if we now transition to a Cartesian coordinate system, it is ultimately the distance between the values of two dependent variables, or the change in the dependent variable, that determines the angle between the (slanted) line, if we wish to increase the angle. The larger the difference, the larger the angle, which means the more slanted the line in the counterclockwise direction. For this reason, elevation cannot be measured as the inverse of the change in the dependent variable, for if the dependent variable's change is large, the elevation is small, which is nonsense.

From the above paragraph, I believe we may perfect our intuitive understanding of slope through the following graph and derived equation for elevation that I constructed

$\hspace{3cm} $

where $\alpha$ satisfies $0 \leq \alpha \leq \pi$ and is measured in radians. For the following discourse, it is helpful to interpret $y$ as two-dimensional height.

To be clear, although $\Delta x$, $\Delta y$ $> 0$ in the image, equation $(1)$ below holds for positive or negative values of $\Delta x$ and $\Delta y$.

It is clear, therefore, that the elevation of the line with respect to tangent is

$$tan(\alpha) = \frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1} \tag{1}$$

Appealing to the graph, observe that if $\Delta y$ is large and $\Delta x < \Delta y$, then $\alpha$ grows larger. Geometrically, we mentally see that the elevation of the line rises in a counterclockwise direction. Hence, the steepness of the line increases, which is what $(1)$ renders.

However, if we calculated the angle using cotangent, then, obviously, $cot(\alpha) = \frac{\Delta x}{\Delta y}$. But our conditions on $\Delta x$ and $\Delta y$ necessitates that $cot(\alpha)$ calculates a small elevation, since $\Delta y$ is in the denominator, which contradicts the geometry, let alone our common sense.

Thus, steepness depends on the angle $\alpha$, and the $\alpha$ ultimately depends on the change in the dependent variable $\Delta y$, since $\alpha = arctan(\frac{\Delta y}{\Delta x})$.

From our above discussion, our understanding of slope in equation $(1)$ may transition to other interpretations of slope, such as rates of change.

Algebraic Answer (Less Insightful)

Suppose we have two points $(x_1,y_1)$ and $(x_2,y_2)$ such that $x_1 < x_2$. Then let us define $y_1$ and $y_2$ as functions of $x_1$ and $x_2$, respectively, as

$$y_1 = mx_1 + b \tag{1}$$ $$y_2 = mx_2 + b \tag{2}$$

where m and b are constants. Then if we subtract equation $(2)$ from equation $(1)$, we have

$$y_2-y_1 = m(x_2-x_1) \tag{3}$$

or

$$m = \frac{y_2-y_1}{x_2-x_1} \tag{4}$$

To be explicit, we may invoke function notation as follows

$$m = \frac{f(x_2)-f(x_2)}{x_2-x_1} \tag{5}$$

~ Fin