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Why are the coordinates of origin $\left(0,0\right)$?

I mean why these are not $(1,1)$;$(1,2)$ or anything else?
What is special about $(0,0)$? Is this is just to ease the calculation?

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  • $\begingroup$ Basically, yes. The origin is completely arbitrary, so if you are allowed to choose, why not choose the easiest to work with? $\endgroup$
    – mdave16
    Commented Jan 12, 2017 at 16:33
  • $\begingroup$ The xy plane has 4 quadrants. By choosing (0,0) as its center, each quadrant becomes unique in terms of what the signs of the x and y coordinates are going to be. But it does not always have to be (0,0). Think of a logarithmic or exponential scale... $\endgroup$
    – imranfat
    Commented Jan 12, 2017 at 16:33
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    $\begingroup$ In many contexts, yes, it's just for ease of calculation. The word origin specifically means "The point with the coordinates $(0,0)$", but it's only convenience that makes the point special. Once you get a bit further on, though, it is the only single point in the plane that is a vector space all by itself, so at that point (heh) it becomes something more. $\endgroup$
    – Arthur
    Commented Jan 12, 2017 at 16:33
  • $\begingroup$ all lines $y=ax$ go through (0,0) $\endgroup$
    – Widawensen
    Commented Jan 12, 2017 at 16:52
  • $\begingroup$ The point with coordinates $(0,0)$ is called the origin of the coordinate plane. There is nothing more to it. $\endgroup$ Commented Jan 12, 2017 at 17:30

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It's useful to have a point of reference in the coordinate plane -- something from which you can say "that point is way off in the $+x$ direction," for instance -- and someone decided to call this "the origin."

But as for choosing its coordinates: any point CAN work as an origin, as you observe.

The advantage to using $(0,0)$ is that it's easy to remember, because $0$ is a special number in the real number system (it's the additive identity, and has the property that $c \cdot 0 = 0$ for any $c$).

When you start to talk about displacements --- the amount you need to add to one point to get to another point --- it turns out that these, too, are represented by pairs of numbers, and there's a natural inclination to treat these number-pairs (called "vectors") as if they were the same as points (which are also represented by pairs of numbers). But for displacements, the displacement by $(0,0)$ is very special: it doesn't move anything at all. So the vector $(0,0)$ is special, and that gives yet another reason for saying that the point $(0,0)$ is special --- it's the one corresponding to the special vector.

If you really want to go a little further in pursuing this question, Hartshorne's Projective Geometry book is a treasure: it starts from geometric axioms for affine and projective planes and goes through the whole process of establishing coordinates on each; seeing that helps you understand what are the arbitrary choices and what are not.

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$\newcommand{\Reals}{\mathbf{R}}$There's "hidden infrastructure" lurking in the question. To flesh out Arthur's comment: A Euclidean plane $E$ has no distinguished point. Cartesian coordinates with origin $O$ furnish a "dictionary" between points of $E$ and ordered pairs of real numbers.

In this dictionary, a point $X$ of $E$ may be viewed as the arrow with its tail at $O$ and its tip at $X$, in which case:

  • The operation of forming the parallelogram with three vertices given by $O$, $X = (x_{1}, x_{2})$, and $Y = (y_{1}, y_{2})$ corresponds to adding Cartesian coordinates, obtaining the fourth vertex $$ X + Y = (x_{1} + y_{1}, x_{2} + y_{2}). $$

  • The operation of homothety by a factor $c$ about the origin $O$ corresponds to scalar multiplication: $$ cX = (cx_{1}, cx_{2}). $$

With these operations of addition and scalar multiplication, the plane becomes a vector space. There is precisely one distinguished point in a vector space, its "zero vector", the additive element for vector addition.

If arithmetic operations on ordered pairs are to have the "expected" geometric meanings, the origin must have coordinates $(0, 0)$.

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It is defined as such. A point from which the axes extend indefinitely in both the directions. You can always shift the origin if you want. Also, $\left(0,0\right)$ has it's own benefits while dealing with many situations like vectors, complex, dilation, etc.

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I presume we're talking about (rectangular) Cartesian coordinates in the plane. Because, say, in polar coordinates it's not necessarily true.

I would say that from the planar geometry point of view, this is by so by definition. As others have said, there's no special point in the plane. But to set up a (rectangular) Cartesian coordinate system, we have to make some choices and some definitions. We pick a point $O$ to serve as the origin. We pick two mutually perpendicular lines through this point to serve as the axes. Then we define the $x$-coordinate of a point $P$ to be the distance from the origin $O$ to the perpendicular projection of $P$ onto the horizontal axis; similar for the $y$-coordinate. Apply this definition to the origin $O$ itself, and you get $(0,0)$.

Of course, other branches of mathematics offer different points of view and other insights...

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