Okay so let's repeat a definition here
Slope is calculated by finding the ratio of the "vertical change" to the "horizontal change" between (any) two distinct points on a line.
So first of all, you cannot get the slope of a line by taking two equal points and finding the ratio between their distance (which would be $\frac 00$). With your reasoning, you find the gradient of $y=x$ at the point $(0,0)$ by taking $(0,0)$ as reference point, and then calculating $\frac{0-0}{0-0}$, after which you proceed to argue that that should be equal to $1$. However, the line $y=2x$ for example, has slope $2$ (yes, also in $(0,0)$), yet with your reasoning, it should be $\frac{0-0}{0-0}$, which you said should be $1$.
Second of all, vertical lines have no "horizontal change", and as such, the ratio between the "vertical change" and the "horizontal change" cannot exist, since we'd be dividing by $0$.
Why can we not divide by $0$?
In the real numbers (this works for complex numbers too, but we'll just go with $\mathbb{R}$), we have the following property (for all $b\neq 0$):
$$\frac ab=c\iff a=bc$$
When you say that division by $0$ should be possible, you implicitly say that the above rule shouldn't only work for all $b\neq 0$, but also for $b=0$. This makes no sense however, since with that, we have:
$$\frac a0=c\iff a=0\cdot b$$
But it is known, even without division by $0$, that $0\cdot c=0$ for all $c$, and so that is still true when division by zero is possible; but now we encounter a problem. The property with $b=0$ now states that for all $a,c$, we must have
$$\frac a0=c\iff a=0$$
This is problematic, because, if division by zero was possible, that would mean that if you divide any number $n$ by $0$, that number must've been $0$; moreover, we have $\frac 00=c$ for all $c$. See how that produces problems?
To sum it up, if you allow division by $0$, then all numbers suddenly become equal, and we're dealing with $\{0\}$ instead of the usual, much more useful $\mathbb{R}$. This is why we don't do division by $0$.
As a sidenote, scepticism makes a good mathematician; however, don't underestimate the mathematicians that've existed and thought about all sorts of things over the ages. With such basic results as this, you can assume very talented and respected mathematicians have thought about this, and especially if they all agree, they're probably right.
Edit
It seems like you view real numbers more like quantifiers than a set of mathematical objects; this view is fine, but often doesn't work. You say $0$ is "just a placeholder", but it really doesn't work that way. The set of real numbers, as well as the set of complex numbers, the set of fractions, the integers, and many more, are sets with elements that follow specific rules; those rules work for all numbers in the set, and so $0$ is being treated, as should be treated, as just "one of the elements" of said set. $0$ does not mean "nothing" nor "NaN", even though real-life applications suggest the nothingness of $0$. You have to keep in mind mathematics is purely theoretical, and there's applications in the real world that mathematics wouldn't fully agree with (take for example physics, which does a lot of things like "rounding off", "discarding terms because they're insignificant" which are fine for the real world, and it works, but mathematically speaking, some things would be considered non-rigorous).
It depends on the question. Different questions imply different answers.
Q1: Using the given data points given below, a straight line passing
through the origin is fitted using least squares method. What is the
value of a if the regression function is $y=ax$
Answer: The value of the slope a is equal to $1$
Q2: Using the given data points given below, a straight is fitted using least squares method. What are the
values of a and b if the regression function is $y=ax+b$
Answer: The values of a is $0.6$ and the value of b is $\frac{14}{15}$
These questions can be solved without using matrix algebra, since we have 3 data point only. Just minimize
$\sum_{i=1}^{3} (y_i-ax)^2 =(1.5-a)^2+(2.2-2a)^2+(2.7-3a)^2$ respectively
$\sum_{i=1}^{3} (y_i-ax-b)^2 = (1.5-a-b)^2+(2.2-2a-b)^2+(2.7-3a-b)^2$
Here are the results of Q1 and Q2.
Best Answer
In my experience, when trying to estimate the slope at a point, it is better to use the slope of the line between the preceding and following point.
This is analogous to the fact that $f'(x)$ is more accurately estimated by $(f(x+h)-f(x-h))/(2h)$ (error of order $h^2$) than by $(f(x+h)-f(x))/(h)$ (error of order $h$).