Ok, I'll call "A" the 1st image and "B" the 2nd one.
If the images will be always on the same place you can just create the object at the X and Y that is between those images.
If the images change places try this;
Create object at
x = A.x < B.x ? A.x + (B.x - A.x)/2 : B.x + (A.x - B.x)/2
y = A.y < B.y ? A.y + (B.y - A.y)/2 : B.y + (A.y - B.y)/2
What this is doing is:
(A.x < B.x ?) > Is A.x smaller than B.x? In other words is image A on the left side of image B?
If it is then X will be the part of the code between the "?" and the ":"
So "x = (A.x + (B.x - A.x)/2)" Which is the same distance from the origin as the image A plus half the distance between A and B.
If it is not then X will be the part of the code after the ":"
So "X= B.x + (A.x - B.x)/2" Which is the same thing as before but considering B is on the left side of A
The code for Y is exactly the same.
Of course this codes will create the image always in the exact center between images A or B. To create objects anywhere between the A and B the code should be different.
I'm not sure if this is the best way to do it, but it's how I would do it. Hope it helps