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 Opposite Vertices
Let the given opposite vertices of the square be:
- A(2, 6)
- C(0, -2)

### Step 2: Calculate the Midpoint of the Diagonal
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:
\[ M = \left( \frac{2 + 0}{2}, \frac{6 + (-2)}{2} \right) = \left( \frac{2}{2}, \frac{4}{2} \right) = (1, 2) \]

### Step 3: Calculate the Length of the Diagonal
The length of the diagonal AC can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of A and C:
\[ d = \sqrt{(0 - 2)^2 + (-2 - 6)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17} \]

### Step 4: Calculate the Length of Each Side
Since the diagonal of a square is \( s\sqrt{2} \) where \( s \) is the length of a side, we can find the side length as follows:
\[ s = \frac{d}{\sqrt{2}} = \frac{2\sqrt{17}}{\sqrt{2}} = \sqrt{34} \]

### Step 5: Find the Other Vertices
Let the other two vertices be B(x1, y1) and D(x2, y2). The vectors from M to B and D must be perpendicular to the vector AC. The direction vector of AC is:
\[ \text{AC} = (0 - 2, -2 - 6) = (-2, -8) \]
The perpendicular vector can be obtained by rotating AC by 90 degrees:
\[ \text{Perpendicular vector} = (8, -2) \]

Now, we can find the coordinates of B and D by moving from M in the direction of the perpendicular vector scaled by \( \frac{s}{2} \):
\[ B = M + \left( \frac{s}{2} \cdot \frac{8}{\sqrt{68}}, \frac{s}{2} \cdot \frac{-2}{\sqrt{68}} \right) \]
\[ D = M - \left( \frac{s}{2} \cdot \frac{8}{\sqrt{68}}, \frac{s}{2} \cdot \frac{-2}{\sqrt{68}} \right) \]

Calculating the coordinates:
1. Length of half the side:
\[ \frac{s}{2} = \frac{\sqrt{34}}{2} \]
2. Coordinates of B:
\[ B = \left( 1 + \frac{\sqrt{34}}{2} \cdot \frac{8}{\sqrt{68}}, 2 - \frac{\sqrt{34}}{2} \cdot \frac{2}{\sqrt{68}} \right) \]
Simplifying:
\[ B = \left( 1 + \frac{8\sqrt{34}}{2\sqrt{68}}, 2 - \frac{2\sqrt{34}}{2\sqrt{68}} \right) = \left( 1 + \frac{4\sqrt{34}}{\sqrt{17}}, 2 - \frac{\sqrt{34}}{\sqrt{17}} \right) \]
\[ B = \left( 1 + 4, 2 - 1 \right) = (5, 1) \]
3. Coordinates of D:
\[ D = \left( 1 - 4, 2 + 1 \right) = (-3, 3) \]

### Final Result
The coordinates of the other vertices B and D are:
- B(5, 1)
- D(-3, 3)