Answer:
D. No, because EFGH is a parallelogram but it’s diagonals are not congruent
Step-by-step explanation:
The differences between the end points of the diagonals are ...
F -H = (1, 1) -(2, -5) = (1 -2, 1 -(-5)) = (-1, 6)
G -E = (4, -2) -(-1, -2) = (4 -(-1), -2 -(-2)) = (5, 0)
The length of FH is more than 6, the length of GE is exactly 5. The diagonals are different length, so the figure cannot be a rectangle.
__
The midpoints of the diagonals will be in the same place if the sum of their end points is the same. (Dividing each sum by 2 gives the midpoint of that segment.)
F+H = (1, 1) +(2, -5) = (3, -4)
G+E = (4, -2) +(-1, -2) = (3, -4)
The diagonals bisect each other (have the same midpoint), so the figure is a parallelogram.
EFGH is a parallelogram, but not a rectangle: its diagonals are not congruent.
