One side of a square is inclined to the x-axis at an angle `alpha` and one of its extremities is at the origin. If the side of the square is `4`, find the equations of the diagonals of the square.

Answer

To find the equations of the diagonals of a square with one vertex at the origin and one side inclined at an angle \( \alpha \) to the x-axis, we can follow these steps:

### Step 1: Determine the coordinates of the vertices of the square

Given:
- One vertex \( O \) at the origin \( (0, 0) \).
- Side length \( s = 4 \).
- The angle \( \alpha \) is the angle between one side of the square and the x-axis.

Let’s denote the vertices of the square as follows:
- \( O(0, 0) \)
- \( A \) (the vertex along the direction of \( \alpha \))
- \( B \) (the vertex adjacent to \( A \))
- \( C \) (the vertex opposite to \( O \))

The coordinates of vertex \( A \) can be calculated as:
- \( A = (s \cos \alpha, s \sin \alpha) = (4 \cos \alpha, 4 \sin \alpha) \)

### Step 2: Find the coordinates of vertex \( B \)

Since \( B \) is perpendicular to \( OA \), we can find its coordinates using the rotation of \( A \) by \( 90^\circ \):
- The coordinates of \( B \) can be derived as:
\[ B = (4 \cos \alpha - 4 \sin \alpha, 4 \sin \alpha + 4 \cos \alpha) \]

### Step 3: Find the coordinates of vertex \( C \)

Vertex \( C \) is opposite to \( O \) and can be found by rotating \( A \) by \( 180^\circ \):
- The coordinates of \( C \) are:
\[ C = (-4 \cos \alpha, -4 \sin \alpha) \]

### Step 4: Find the equations of the diagonals \( OB \) and \( AC \)

#### Diagonal \( OB \)

1. **Slope of line \( OB \)**:
- The slope \( m_{OB} \) can be calculated using the coordinates of \( O \) and \( B \):
\[ m_{OB} = \frac{y_B - y_O}{x_B - x_O} = \frac{(4 \sin \alpha + 4 \cos \alpha) - 0}{(4 \cos \alpha - 4 \sin \alpha) - 0} = \frac{4 (\sin \alpha + \cos \alpha)}{4 (\cos \alpha - \sin \alpha)} = \frac{\sin \alpha + \cos \alpha}{\cos \alpha - \sin \alpha} \]

2. **Equation of line \( OB \)**:
- Using point-slope form \( y - y_1 = m(x - x_1) \):
\[ y - 0 = \frac{\sin \alpha + \cos \alpha}{\cos \alpha - \sin \alpha} (x - 0) \]
- Simplifying gives:
\[ y = \frac{\sin \alpha + \cos \alpha}{\cos \alpha - \sin \alpha} x \]

#### Diagonal \( AC \)

1. **Slope of line \( AC \)**:
- The slope \( m_{AC} \) can be calculated using the coordinates of \( A \) and \( C \):
\[ m_{AC} = \frac{y_C - y_A}{x_C - x_A} = \frac{-4 \sin \alpha - 4 \sin \alpha}{-4 \cos \alpha - 4 \cos \alpha} = \frac{-8 \sin \alpha}{-8 \cos \alpha} = \frac{\sin \alpha}{\cos \alpha} = \tan \alpha \]

2. **Equation of line \( AC \)**:
- Using point-slope form:
\[ y - 4 \sin \alpha = \tan \alpha (x - 4 \cos \alpha) \]
- Rearranging gives:
\[ y = \tan \alpha (x - 4 \cos \alpha) + 4 \sin \alpha \]

### Final Equations of the Diagonals

- The equation of diagonal \( OB \):
\[ y = \frac{\sin \alpha + \cos \alpha}{\cos \alpha - \sin \alpha} x \]

- The equation of diagonal \( AC \):
\[ y = \tan \alpha (x - 4 \cos \alpha) + 4 \sin \alpha \]