Position of an object relative to another in a 2D plane

Question

I have two objects, A and B, each of them has a set of coordinates (x,y). How do I understand where is A in relation to B's position?

If I where to just check each coordinates I could do it like this:

B(x:5,y:5), A(x:4,y:5); x = 5-4 = 1; y = 5-5 = 0; A it's on the left

B(x:5,y:5), A(x:6,y:5); x = 5-6 = -1; y = 5-5 = 0; A it's on the right

B(x:5,y:5), A(x:5,y:6); x = 5-5 = 0; y = 5-6 = -1; A it's on top of B

B(x:5,y:5), A(x:6,y:5); x = 5-5 = 0; y = 5-4 = 1; A it's under B

But I was wondering, since I'm using a standard Vector2 data structure (from LibGDX), with the ability to normalize, compute distances, etc.. If I could do it using those methods instead of subtracting the coordinates each time.

If it can help, I need to know the position of A relative to B because A is supposed to push B, so if A it's on B's left, B needs to move right, etc...

Answer 1

If you want to get position A relative to B, you just need to substract them, like you were doing. If you want to make use of the methods, you can use the method "sub": https://libgdx.badlogicgames.com/nightlies/docs/api/com/badlogic/gdx/math/Vector2.html#sub-com.badlogic.gdx.math.Vector2-

Vector2 relative = A.sub(B); //contains the position of A relative to B
Vector2 direction = B.sub(A); //direction that B will move after being hit by A

Answer 2

Coordinates of A xa,ya and B xb,yb and vector AB X=xa-xb ,Y=ya-yb

I give a formula to give point A in relation to point B in 8 directions.

For E=0,NE=1,N=2,NW=3,W=4,SW=5,S=6 and SE=7

F(X,Y)=MOD((4-2*(1+SIGN(X))* (1-SIGN(Y^2))-(2+SIGN(X))*SIGN(Y)

  -(1+SIGN(ABS(SIGN(X*Y)*ATAN((ABS(X)-ABS(Y))/(ABS(X)+ABS(Y))))

  -PI()/(8+10^-15)))/2*SIGN((X^2-Y^2)*(X*Y))),8)