`A(-3,-1)` and `B(2,2)` are two adjacent vertices and `O(3,0)` is the point of intersection of the diagonals of a parallelogram ABCD. Write down the equations of the four sides of the parallelogram.

Answer

To find the equations of the four sides of the parallelogram ABCD given the vertices A(-3, -1), B(2, 2), and the point of intersection of the diagonals O(3, 0), we will follow these steps:

### Step 1: Find the coordinates of points C and D
Since O is the midpoint of both diagonals AC and BD, we can use the midpoint formula to find the coordinates of points C and D.

Let the coordinates of C be (x, y) and D be (h, k).

Using the midpoint formula:
- For diagonal AC:
\[ O = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) = (3, 0) \]
Substituting A(-3, -1):
\[ \frac{-3 + x}{2} = 3 \quad \text{and} \quad \frac{-1 + y}{2} = 0 \]

From the first equation:
\[ -3 + x = 6 \implies x = 9 \]

From the second equation:
\[ -1 + y = 0 \implies y = 1 \]
Thus, point C is (9, 1).

- For diagonal BD:
\[ O = \left( \frac{x_B + x_D}{2}, \frac{y_B + y_D}{2} \right) = (3, 0) \]
Substituting B(2, 2):
\[ \frac{2 + h}{2} = 3 \quad \text{and} \quad \frac{2 + k}{2} = 0 \]

From the first equation:
\[ 2 + h = 6 \implies h = 4 \]

From the second equation:
\[ 2 + k = 0 \implies k = -2 \]
Thus, point D is (4, -2).

### Step 2: Write the equations of the sides of the parallelogram
Now we have all four vertices: A(-3, -1), B(2, 2), C(9, 1), and D(4, -2). We can find the equations of the sides AB, BC, CD, and DA.

#### Equation of AB
Using the two points A(-3, -1) and B(2, 2):
\[ y - y_1 = m(x - x_1) \]
where \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-1)}{2 - (-3)} = \frac{3}{5} \).

Thus,
\[ y + 1 = \frac{3}{5}(x + 3) \]
Multiplying through by 5:
\[ 5y + 5 = 3x + 9 \implies 3x - 5y + 4 = 0 \]

#### Equation of BC
Using points B(2, 2) and C(9, 1):
\[ m = \frac{1 - 2}{9 - 2} = \frac{-1}{7} \]
Thus,
\[ y - 2 = \frac{-1}{7}(x - 2) \]
Multiplying through by 7:
\[ 7y - 14 = -x + 2 \implies x + 7y - 16 = 0 \]

#### Equation of CD
Using points C(9, 1) and D(4, -2):
\[ m = \frac{-2 - 1}{4 - 9} = \frac{-3}{-5} = \frac{3}{5} \]
Thus,
\[ y - 1 = \frac{3}{5}(x - 9) \]
Multiplying through by 5:
\[ 5y - 5 = 3x - 27 \implies 3x - 5y - 22 = 0 \]

#### Equation of DA
Using points D(4, -2) and A(-3, -1):
\[ m = \frac{-1 - (-2)}{-3 - 4} = \frac{1}{-7} \]
Thus,
\[ y + 2 = \frac{1}{-7}(x - 4) \]
Multiplying through by -7:
\[ -7y - 14 = x - 4 \implies x + 7y + 10 = 0 \]

### Summary of the Equations
1. **AB**: \( 3x - 5y + 4 = 0 \)
2. **BC**: \( x + 7y - 16 = 0 \)
3. **CD**: \( 3x - 5y - 22 = 0 \)
4. **DA**: \( x + 7y + 10 = 0 \)