Show that it is not possible for $A$ and $B$ to meet at a point 
Two friends $A$ and $B$ are initially at points $(0,0)$ and $(12,7)$ respectively on the infinite grid plane. $A$ and $B$ takes steps of sizes $4$ and $6$ units respectively along the grid lines. Show that it is not possible for them to meet at a point. An exaple step- $A$ takes three steps towards right and one upwards($3+1=4$).
  

I moved $A$ three times towards right on the x axis($12$ steps), so that it is now vertically below $B$ Now I moved $B$ one time($6$ steps) vertically downward towards $A$ Now $B$ is just one unit above $A$. What next? Am I on the right track?
 A: I suspect that your method won't really help you prove it fully.  You seem to be trying to prove that one particular route won't work;  but the question is asking for you to prove that it's impossible, i.e., out of all the possible things A and B could do, none of them will allow A and B to meet up.
Here's a hint on how to prove that it's impossible:  Make a list of a bunch of points that A can get to by taking steps of 4 or 6 units.   Now make a list of a bunch of points that B can get to.  See if there are any points in common between the two lists (if you've done it right, there won't be.)  More importantly, look for patterns & commonalities within each list, and see if you can discern why it's impossible for them to meet up.  If you spot a pattern, try to figure out why it arises.
A: Notice that $X_A$ and $Y_A$ have to be of the same parity because you start at $(0,0)$ and if you move $x$ steps to right or left and $y$ steps to top or bottom by the rule you have $x+y=4$ and the sum of an odd and an even is always odd so it must be the case that $x,y$ are both even or they are both odd. So $X_A+x$ and $Y_A+y$ are of the same parity.
Now similarly $X_B$ and $Y_B$ must have different parities since you start at $(0,1)$ now you have $x+y=6$ which again implies that $x,y$ have the same parities which implies that $X_B+x$ and $Y_B+y$ have different parities.
