Constant slope in straight line

Question

I'm now reading about the formal definition of line as the graph of the function $f(x)=f(x_0)+m(x-x_0)$ for an arbitrary $x_0$ through which the graph passes, where the constant $m$ is called slope. If we denote $a-a_0$ by $\Delta a$, it can be easily proved that $m$ for some $x_0$ is $\dfrac{\Delta f(x)}{\Delta x}$.

What makes the equation define straight lines is that if and only if the graph is a straight line does $m$ stay constant for any $x_0$.

I don't understand the intuition behind this definition.

Why does $\dfrac{\Delta y}{\Delta x}=const.$? And why only in them?

Answer 1

The difference quotient $m=\frac{f(x)-f(x_0)}{x-x_0}$ denotes the rate of change of your function.

If the rate of change is constant for any input $x$, then the function has to be linear, since a function can only "curve", if the rate of change differs between two arbitrary points $x_i, x_j$.

You can compare that with driving a car: Imagine you are driving a constant speed of $20 km/h$. You would then cover $20 km$ in an hour, $40 km$ in two hours and so on (in a linear manner).

To cover more or less kilometers, you would have to adjust speed, hence adjust the rate of change with respect to the kilometers covered. Therefore, your function would not be linear anymore.

Answer 2

From your first equation you see that $$ m=\frac{f(x)-f(x_0)}{x-x_0} $$ so, in general, the value of $m$ is not a constant, but depends from $x$ and $x_0$.

If it is the same for any couple $x$ and $x_0$, we can chose an $x_1$ (different from $x$ and $x_0$) and we have $$ m=\frac{f(x)-f(x_0)}{x-x_0}=\frac{f(x_1)-f(x_0)}{x_1-x_0} $$

and this is the equation of a straight line passing through the points $(x_0,f(x_0))$ and $(x_1,f(x_1))$.

So, any three points of $y=f(x)$ stay on the same straight line.