To be honest I don't understand why there are so many down votes. This is a deep question and doesn't have an easy answer.
In very general terms what you are asking about is metric spaces. A metric space is a certain object we study in mathematics that allows us to define explicitly some notion of length. You'll find that there are in fact many different notions of length you can choose for certain metric spaces and each have their own interesting properties.
The Pythagorean theorem gives us one kind of metric. This is called the standard Euclidean Metric $a^2+b^2=c^2$, or $c = \sqrt{a^2+b^2}$. One of your questions was why don't we take the negative root. The reason is because mathematically $\sqrt{x}$ is defined to be the positive root, if we allowed for both the positive and negative roots to be used then $\sqrt{x}$ would cease to be a function. That is it would fail the vertical line test (think of a sideways parabola).
Now more generally we want a metric space. We can define what a metric space is by the following:
A metric space is a set $M$ (like for instance the set of numbers on the plane $(x,y)\in\mathbb{R}^2)$ on which we can apply a function $d$ which is said to be a distance function between two points in your set $M$. This function $d$ must satisfy certain properties to be considered a metric these are:
- $d(x,y) \geq 0$
- $d(x,y)=0$ if and only if $x=y$
- $d(x,y)=d(y,x)$
- $d(x,z) = d(x,y) + d(y,z)$
If your set $M$ satisfies all of these properties then it is called a metric space. In particular notice the first one, here we explicitly define the metric by the fact that the length between two points in your set can't be negative.
But the above definition is a little notation heavy so let's use the example from before on the set $M=\mathbb{R}^2$ with the Euclidean norm. Let us check that this satisfies all of the conditions required to be a metric space.
First let $(a,b),(c,d)\in\mathbb{R}^2$ be two points in $M$ Then the way we define the Euclidean metric is $d((a,b),(c,d)) = \sqrt{(a-c)^2 + (b-d)^2}$. Because we always take the positive root, and we know that any number squared is non-negative we know that $d((a,b),(c,d))\geq 0$.
If $d((a,b),(c,d))=0$ then we must have that $\sqrt{(a-c)^2+(b-d)^2}=0$. So it follows that $a=c$, and $b=d$ which is enough to say that $(a,b)=(c,d)$.
The remaining conditions follow in the same way. Now I elluded to the fact that the Euclidean metric was not the only one we could have chosen. In fact any function that satisfied the conditions above will be a metric on $M$. One such metric is called the "Taxicab Metric". See if you can play around with the definition and come up with your own metric!
