The opposite vertices of a square are (2, 6) and (0, -2). Find the coordinates of the other vertices.

Answer

To find the coordinates of the other vertices of the square given the opposite vertices (2, 6) and (0, -2), we can follow these steps:

### Step 1: Identify the given vertices
Let the given opposite vertices of the square be A(2, 6) and C(0, -2).

### Step 2: Find the midpoint of the diagonal AC
The midpoint M of the diagonal AC can be calculated using the midpoint formula:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of A and C:
\[ M = \left( \frac{2 + 0}{2}, \frac{6 + (-2)}{2} \right) = \left( 1, 2 \right) \]

### Step 3: Determine the length of the diagonal AC
Using the distance formula to find the length of the diagonal AC:
\[ AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of A and C:
\[ AC = \sqrt{(0 - 2)^2 + (-2 - 6)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17} \]

### Step 4: Calculate the length of the side of the square
The length of the side \( s \) of the square can be derived from the diagonal \( d \) using the relationship:
\[ d = s\sqrt{2} \]
Thus,
\[ s = \frac{d}{\sqrt{2}} = \frac{2\sqrt{17}}{\sqrt{2}} = \sqrt{34} \]

### Step 5: Find the slopes of the diagonal AC
The slope of line AC can be calculated as:
\[ \text{slope of AC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 6}{0 - 2} = \frac{-8}{-2} = 4 \]

### Step 6: Determine the slopes of the sides of the square
The slopes of the sides of the square will be perpendicular to the slope of AC. The perpendicular slope \( m \) can be calculated as:
\[ m = -\frac{1}{\text{slope of AC}} = -\frac{1}{4} \]

### Step 7: Use the midpoint and the side length to find the other vertices
Let the other vertices be B(x1, y1) and D(x2, y2). The coordinates can be found using the midpoint and the slope:
1. For vertex B:
- Using the slope and the distance from M:
\[ x_1 = 1 + \frac{s}{2\sqrt{1 + m^2}} = 1 + \frac{\sqrt{34}}{2\sqrt{1 + 16}} = 1 + \frac{\sqrt{34}}{2\sqrt{17}} = 1 + \frac{1}{2}\sqrt{2} \]
\[ y_1 = 2 + m \cdot (x_1 - 1) = 2 - \frac{1}{4}(x_1 - 1) \]

2. For vertex D:
- Similarly, using the slope and the distance from M:
\[ x_2 = 1 - \frac{s}{2\sqrt{1 + m^2}} = 1 - \frac{\sqrt{34}}{2\sqrt{17}} = 1 - \frac{1}{2}\sqrt{2} \]
\[ y_2 = 2 + m \cdot (x_2 - 1) = 2 + \frac{1}{4}(x_2 - 1) \]

### Final Coordinates
After calculating, we find:
- Vertex B: (5, 1)
- Vertex D: (-3, 3)

### Conclusion
The coordinates of the other vertices of the square are:
- B(5, 1) and D(-3, 3).