That's going to be the midpoint of the two midpoints of the sides;
[tex]P = \frac 1 2 ( \frac 1 2 (L+H) + \frac 1 2 (K+M)}) = \frac 1 4 (K+L+M+N)[/tex]
That's the average of all the coordinates, which is the center of mass of the figure:
[tex] P = \frac 1 4 ( 4m+4q+4p+0,4n+4n+0+0) = (m+q+p,2n)[/tex]
First choice
