Grid system conversion I am attempting to convert between two different two dimensional grid coordinate systems. That are defined as follows.
System 1
The x and y axis start at point 0,0. Y increases going up and decreases going down. X increases going right and decreases going left.
System 2
The x and y axis start at point 0,0. Y increases going up and left and decreases going down and right. X increases going up and right and decreases going down and left.
I would like to be able to make a formula for translating from system 1 to 2 and vice versa.
For example
x:50 y:0 system 1 would be system 2 x:100 y:-100
x70.71 y:70.71 system 1 would be x:100 y:0
I can calculate a few points but a generic formula for translation is beyond my skill... I'm not even 100% certain I even got the above correct
 A: Good question. This is called a change of basis. You notice how in the standard grid system, you need one number (the $x$ coordinate) to tell you how far to the left or right to go, and another (the $y$ coordinate) to tell you how far up and down you need to go.
In your system, we want one coordinate  (call it the $x'$ coordinate) to tell your how far in the up right diagonal to go, and the other coordinate (call it the $y'$ coordinate) to tell you how far in the up left direction you need to go. 
So let's think about how to express going to the coordinate $(x,y) = (1,0)$ in the $(x', y')$ system. You see there's a right triangle formed - you need to go $a$ units in the $x'$ directions and then $a$ units in the $y'$ direction. Solving using the Pythagorean theorem gives us $a = \frac{1}{\sqrt 2}$.
Repeating this to try to figure out how to get to $(x,y) = (0,1)$, we see we have to move $a$ units in the $x'$ direction but now $-a$ units in the $y'$ direction. Doing the calculation again reveals that $a$ is the same value as last time. 
So now we know if you want to go to the point $(1,0)$ in the standard system, you need to go $(a,a)$ units in the new sysyem. If you want to go to $(n,0)$, you this need to go $(a,a)$ units $n$ times. This is thus $(na,na)$ in the new system.
Similarly, to get to $(m,0)$ in the new system, this will me $(ma, -ma)$ in the new system. 
Putting this together, suppose you want to get to $(n,m)$ under the standard system. In the new system you can think of this as first getting to $(n,0)$ by taking $na$ steps in the $x'$ direction and $y'$ direction, and then going $m$ units over by going $ma$ steps in the $x'$ direction and $-ma$ units in the $y'$ direction. So in total we have 
$$(na+ma, na-ma)$$
In the new system.
If you're familar with matrix notation, this is equivalent to saying 
$$ \begin{pmatrix} 
a & a \\
a & -a 
\end{pmatrix} $$
is the change of basis matrix between the $(x,y)$ basis and $(x',y')$ basis.
This is in fact a very general phenomenon. If you choose any different directions (they don't need to be perpendicular to each other), you can come up with such a formula for that system as well.
