Answer:
A, B, and E.
Step-by-step explanation:
We know that one vertex is at (5, 4), and each side of our square is 3 units long.
Then the distance between the known vertex and another vertex is 3 units (if those vertexes are connected by a side of the square) or (√2)*3 units (if those vertexes are connected by the diagonal of the square).
Also remember that the distance between two points (a, b) and (b, c) is:
distance = √( (a - c)^2 + (b - d)^2)
So we need to find the distance between our point and all the ones given in the options:
A) the distance between (5, 4) and (5, 1) is:
distance = √( (5 - 5)^2 + (4 - 1)^2) = 3
Then point (5, 1) can be a vertex.
B) The distance between (5, 4) and (5, 7) is:
distance = √( (5 - 5)^2 + (4 - 7)^2) = 3
Then (5, 7) can be a vertex.
C) The distance between (5, 4) and (7, 8) is:
distance = √( (5 - 7)^2 + (4 - 8)^2) = √( 2^2 + 4^2) = √20
Point (7, 8) can not be a vertex.
D) The distance between (5, 4) and (2, 6) is:
distance = √( (5 - 2)^2 + (4 - 6)^2) = √( 3^2 + 2^2) = √13
Point (2, 6) can not be a vertex.
E) The distance between (5, 4) and (2, 1) is:
distance = √( (5 - 2)^2 + (4 - 1)^2) = √( 3^2 + 3^2) = √18 = √(2*9) = √2*√9 = √2*3
Then point (2, 1) can be a vertex.
