To find the coordinates of the other vertices of the square given that one vertex A(2, 1) and the center of the square is at the origin (0, 0), we can follow these steps:
### Step 1: Identify the given information
We know:
- The center of the square (O) is at (0, 0).
- One vertex (A) is at (2, 1).
### Step 2: Find the coordinates of the opposite vertex (C)
The opposite vertex (C) can be found by reflecting point A across the center O. The coordinates of C can be calculated as:
\[ C(x, y) = (-x, -y) \]
Thus,
\[ C = (-2, -1) \]
### Step 3: Calculate the distance from the center to vertex A
Using the distance formula:
\[ AO = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Where \( O(0, 0) \) and \( A(2, 1) \):
\[ AO = \sqrt{(2 - 0)^2 + (1 - 0)^2} = \sqrt{4 + 1} = \sqrt{5} \]
### Step 4: Find the slope of line AO
The slope (m) of line AO can be calculated as:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates of A and O:
\[ m_{AO} = \frac{1 - 0}{2 - 0} = \frac{1}{2} \]
### Step 5: Determine the slope of line BD
Since AO is perpendicular to BD, the product of their slopes is -1:
\[ m_{AO} \cdot m_{BD} = -1 \]
Thus,
\[ \frac{1}{2} \cdot m_{BD} = -1 \]
This implies:
\[ m_{BD} = -2 \]
### Step 6: Find the coordinates of vertices B and D
Since B and D are at a distance of \( \sqrt{5} \) from the center O (0, 0), we can use the slope to find their coordinates. The coordinates can be expressed as:
\[ B = (0 + \sqrt{5} \cdot \cos(\theta), 0 + \sqrt{5} \cdot \sin(\theta)) \]
\[ D = (0 - \sqrt{5} \cdot \cos(\theta), 0 - \sqrt{5} \cdot \sin(\theta)) \]
Using the slope \( m_{BD} = -2 \), we can find \( \cos(\theta) \) and \( \sin(\theta) \):
- From the slope, we can say:
\[ \tan(\theta) = -2 \]
Thus, we can find:
\[ \cos(\theta) = \frac{-1}{\sqrt{5}}, \quad \sin(\theta) = \frac{2}{\sqrt{5}} \]
### Step 7: Calculate the coordinates of B and D
Substituting the values of \( \cos(\theta) \) and \( \sin(\theta) \):
- For B:
\[ B = (0 + \sqrt{5} \cdot \frac{-1}{\sqrt{5}}, 0 + \sqrt{5} \cdot \frac{2}{\sqrt{5}}) = (-1, 2) \]
- For D:
\[ D = (0 - \sqrt{5} \cdot \frac{-1}{\sqrt{5}}, 0 - \sqrt{5} \cdot \frac{2}{\sqrt{5}}) = (1, -2) \]
### Final Coordinates
The coordinates of the other vertices of the square are:
- B(-1, 2)
- D(1, -2)
