Deriving/Proving the equation for a straight line Covering 1.3 of Keisler's Elementary Calculus. 
The section begins with a definition of the point-slope equation of a line. I noticed that this definition involves the slope of the line (m). 
However, he then uses the point-slope equation in a proof that m does in fact represent slope! (delta y over delta x) This seems rather circular. 
I'm curious as to what the fundamental properties of the line are, what is definition dependent and what is proved from those definitions. 
If slope is defined intuitively, what is the basis for this intuition? Could slope not have been just as validly defined as (delta X over delta Y)? 
 A: In space $\mathbb{R}^2$ with Euclidean metric, one of the possible definitions of a straight line is that
A line is an equivalence class of ordered triplet $(a,b,c)$ of reals $a,b,c$ such that at least one of $a,b$ is nonzero, and where equivalence relation is given by $(a,b,c)\sim(a',b',c')$ if and only if there is real $\lambda\neq 0$ such that $(a,b,c)=\lambda(a',b',c')$
You can define a line on $\mathbb{R}^2$ via locus of equation $ax+by+c=0$. Then one can show that there is a one to one correspondence given by definition above: important thing is that equations $ax+by+c=0$ and $\lambda(ax+by+c)=0$ represent the same line if $\lambda\neq 0$, so really you need to think of $(a,b,c)$ and $\lambda(a,b,c)$ as the same thing, which is expressed by "equivalence classes of ~". Also note that the condition saying at least one of $a,b$ being nonzero prevents us having $0\cdot x+0\cdot y +c=0$, which represents whole space if $c=0$ and emptyset otherwise (and both are clearly not lines)
You accept this as a definition of line and show its usual expected properties. If $b\neq 0$, then we can have usual slope-form of line via $y=-\frac{a}{b} x -\frac{c}{b}$ dividing. If $a=0$ we cannot do this and equation reduces to $y=-c/b$, a horizontal line. Similarly if $b=0$ the equation represents vertical lines (of "infinite" slope)
From the slope form, let $m=-a/b$, $-c/b=c'$. Then $y=mx+c'$
and at any point $(a,b)$ on the line,
$$\frac{f(a+\Delta x)-f(a)}{(a+\Delta x) - a}=\frac{m(a+\Delta x)-ma}{\Delta x}=\frac{m\Delta x}{\Delta x}=m$$
So sending $\Delta x\to 0$ yields $dy/dx=m$
