We are given that one vertex of a square lies at the origin $(0,0)$, and the square has a side length of 2 units. Additionally, one of its diagonals lies along the x-axis. Our goal is to determine the coordinates of the three remaining vertices of the square.

Step-by-Step Breakdown:

1. Diagonal Length: Since the square has a side length of 2 units, the length of its diagonal can be calculated using the Pythagorean theorem. The diagonal $d$ of a square with side $s$ is given by: $d = s\sqrt{2}$ For $s = 2$, the diagonal becomes: $d = 2\sqrt{2}$

2. Position of Vertices:

   • Let’s assume the diagonal lies on the x-axis. This means the diagonal runs from $(0,0)$ to $(2\sqrt{2}, 0)$.
   • Since the square is rotated and its diagonal is along the x-axis, the other two vertices will be positioned symmetrically about the x-axis.

3. Coordinates of the Remaining Vertices:

   • To determine the coordinates of the remaining two vertices, we note that they will be rotated by 45° (or $\frac{\pi}{4}$ radians) from the x-axis. This means they will be equidistant from both ends of the diagonal.
   • Using rotation matrix transformations, we calculate the positions of the remaining vertices as:
     • Vertex 2: $(1, 1)$
     • Vertex 3: $(1, -1)$

Final Answer:

The coordinates of the three remaining vertices are: $(2\sqrt{2}, 0), (1, 1), (1, -1)$

Would you like more details or have any questions?

Here are some related questions for further practice:

1. What is the area of the square based on the given side length?
2. How would you calculate the perimeter of the square?
3. If the square were rotated by another 45°, how would the coordinates of the vertices change?
4. What are the coordinates of the center of the square?
5. How would the problem change if the side length was 3 units?

Tip: For geometric shapes, using symmetry properties often simplifies the coordinate calculations.