Question

Miguel draws a square on a coordinate plane. One vertex is located at (5,4). The length of each side is 3 units. Circle the letter by all the ordered pairs that could be another vertex. A. (5,1) B. (5,7) C. (7,8) D. (2,6) E. (2,1)
I will give brainliest if you answer both questions!

Answer 1

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.