mysterious in slope-intercept form of the equation of the line I found something very mysterious for me in the slope-intercept form of the equation of the line.
First let's start with the slope equation. Assume I have 2 points (2,4) and (1,3) and I want to find slope between two points which is
$$ m = \frac{y - y_1}{x - x_1} = \frac{2 - 1}{4 - 3} = 1  $$
after we construct the slope then we continue to put it into the point-slope form:
which is $$ y-y_1 = m (x-x_1)$$ you can notice that this equation derive from the slope equation above, in the slope equation if the two points we are measuring have the same x value then the equation will be undefined. Now keep that fact in mind and move on to put value we have got into the point-slope equation:
$$ y-4 = 1 (x-2)$$
$$ y = x+ 2$$
(I use the point (2,4) in the equation) Now come to what I call mysterious part for me. if I put (2,4) in to the equation as a domain of a function then I get back $4 = 2+ 2$ this is very strange for me as a newbie because the point (2,4) was used to construct the equation which means if we reverse to the slope form we will get:
$$ 1 = \frac{y - y_1}{x - x_1} = \frac{y - 2}{x - 4} = 1  $$ which means in this case we can not put another (2,4) as a domain of the function then we would get : $$\frac{2 - 2}{4 - 4} = 1 $$ which is so wrong but why when it is in the form of $$y = x+2 $$ (which is already been construct with point (2,4)) we can use (2,4) (Again) as domain of the function?
I do not want someone to just come and show off calculation. I want to get deep to the philosophy and I want crystal clear step by step answer.
 A: Your notation causes the confusion.
You have a straight line by two points, of equation
$$y=\frac{y_1-y_0}{x_1-x_0}(x-x_0)+y_0.$$
Its slope is indeed 
$$m=\frac{y_1-y_0}{x_1-x_0}.$$
Now you can substitute any coordinate pair $(x,y)$ into the equation.
In particular,
$$(x,y)=(x_0,y_0)\to y_0=\frac{y_1-y_0}{x_1-x_0}(x_0-x_0)+y_0,$$
and
$$(x,y)=(x_1,y_1)\to y_1=\frac{y_1-y_0}{x_1-x_0}(x_1-x_0)+y_0,$$
which are two true identities.
Your misconception comes form the fact that you are trying to change the coordinates in the slope formula, though these are fixed.

If you want to "reconstruct" the line by keeping one of the points, let $(x_0,y_0)$ and replacing the other by some $(x_2,y_2)$ drawn form the line equation, you get
$$y_2=\frac{y_1-y_0}{x_1-x_0}(x_2-x_0)+y_0,$$ and the new equation is
$$y=\frac{\frac{y_1-y_0}{x_1-x_0}(x_2-x_0)+y_0-y_0}{x_2-x_0}(x-x_0)+y_0,$$
which simplifies to
$$y=\frac{y_1-y_0}{x_1-x_0}(x-x_0)+y_0,$$
unless $x_2=x_0$. (When this is the case, we no more have two distinct points to define the line.)
Similarly, replacing $(x_0,y_0)$ by $(x_2,y_2)$,
$$y=\frac{y_1-\frac{y_1-y_0}{x_1-x_0}(x_2-x_0)-y_0}{x_1-x_2}(x-x_2)+\frac{y_1-y_0}{x_1-x_0}(x_2-x_0)+y_0$$
also simplifies to
$$y=\frac{y_1-y_0}{x_1-x_0}(x-x_0)+y_0$$
provided $x_1\ne x_2$, after a little more effort.
A: The slope formula is used to find the slope of the line determined by two distinct points. You can't put the same point in twice, which is what you are trying to do--you are using $(2,4)$ and $(2,4)$. Those are the same single point, and a single point does not determine a line.
Because your line is not vertical, there is only one point on the line whose $x$-coordinate is $2$ (namely, the point $(2,4)$, of course).
When you are dealing with vertical lines, however, the slope formula can't be used. Vertical lines have undefined slope, and every point on such a line has the same $x$-coordinate.
A: To begin with, you did one very confusing thing, which was you wrote
$(2,4)$ and $(1,3)$ but when you substituted these into
$m = \frac{y - y_1}{x - x_1}$ you wrote $\frac{2 - 1}{4 - 3}$,
that is, you substituted $x$-coordinates for the $y$-variables and
vice-versa. But you were inconsistent about this: in one place
you wrote $ y-4 = 1 (x-2)$ but in another you wrote 
$\frac{y - 2}{x - 4} = 1  $.
To make things consistent, we have to rewrite some of what you wrote.
Below, I chose to change the original two points
to $(4,2)$ and $(3,1)$ in order to
make the equation $\frac{y - 2}{x - 4} = 1  $ relevant, since that
was the paradoxical equation for you.

This is not the equation of a line:
$$
\frac{y - 2}{x - 4} = 1. \tag1 
$$
Instead, it is the equation of a figure you get by deleting one point
from a line. The point $(4,2)$ is not part of the figure.
If you add the point $(4,2)$ to that figure you get back a line
which has (among other representations) the equation
$$
y - 2 = x - 4. \tag2
$$
The relation between equations $(1)$ and $(2)$ is that
$(1)$ implies $(2)$
(every point on the "line minus a point" is on the line)
but $(2)$ does not imply $(1)$
(not every point on the line is part of the "line minus a point"--in
particular, the point $(4,2)$ is in the first set but not
the second set).
You can conclude $(1)$ if you are given $(2)$ and the fact that $x\neq 4$.
In the case where $x = 4$ and $y = 2$, you have no justification to
say that $(1)$ is true, and indeed it is not.
An equation of the form $(1)$ is indeed useful for finding the
slope of a line, despite the fact that it is not completely equivalent
to $(2)$.
It is useful because in order to find a slope, you must have two points
with two different $x$-coordinates.
In effect this prevents you from writing something like
$\frac{2 - 2}{4 - 4}$, in which you would have substituted the same number,
$4$, for both $x$ and $x_1$ in the formula
$ m = \frac{y - y_1}{x - x_1}$.
