Let the parallelogram be ABCD, having co-ordinates of A, Band C as (2,3), (-5-7) and (2,-4) respectively and let (x,y) be the co-ordinates of fourth vertex D. We know that the diagonals of parallelogram bisect each other (say at point P)
Therefore P is the mid point of AC as well as that of BD.
or, -5+x = 4 and -7+y = -1
or, x = 9 and y = 6
so, Co-ordinates of Dare (9,6)