9514 1404 393
Answer:
C. (10, 5), (5, 0), (0, 5)
Step-by-step explanation:
In order for the square to have an area of 50 square units, its side lengths must be √50 = 5√2 units long. The only way to get such a length with integer coordinates is for the sides of the square to be themselves the diagonals of a square 5 units on a side.
That is, two of the other vertices of the square must be among ...
(5±5, 10±5) ⇒ (10, 15), (0, 15), (10, 5), (0, 5)
Only choice C contains two of the vertices on this list.
