ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D 2, -4). Find
(i) Coordiantes of A

Answer

To find the coordinates of point A in the parallelogram ABCD where A is (x, y), B is (5, 8), C is (4, 7), and D is (2, -4), we will use the property that the diagonals of a parallelogram bisect each other.

### Step-by-Step Solution:

1. **Identify the Midpoint of Diagonal BD**:
The midpoint \( P \) of diagonal \( BD \) can be calculated using the midpoint formula:
\[ P = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
where \( B(5, 8) \) and \( D(2, -4) \).

- For the x-coordinate:
\[ x_P = \frac{5 + 2}{2} = \frac{7}{2} \]
- For the y-coordinate:
\[ y_P = \frac{8 + (-4)}{2} = \frac{4}{2} = 2 \]
Thus, the coordinates of point \( P \) are \( \left( \frac{7}{2}, 2 \right) \).

**Hint**: Remember that the midpoint formula averages the coordinates of the two points.

2. **Set Up the Midpoint of Diagonal AC**:
Since \( P \) is also the midpoint of diagonal \( AC \), we can set up the equation:
\[ P = \left( \frac{x + 4}{2}, \frac{y + 7}{2} \right) \]
where \( A(x, y) \) and \( C(4, 7) \).

3. **Equate the Midpoints**:
Since both expressions represent the same point \( P \):
\[ \frac{x + 4}{2} = \frac{7}{2} \quad \text{(1)} \]
\[ \frac{y + 7}{2} = 2 \quad \text{(2)} \]

4. **Solve for x**:
From equation (1):
\[ x + 4 = 7 \quad \Rightarrow \quad x = 7 - 4 = 3 \]

**Hint**: To isolate \( x \), subtract 4 from both sides.

5. **Solve for y**:
From equation (2):
\[ y + 7 = 4 \quad \Rightarrow \quad y = 4 - 7 = -3 \]

**Hint**: To isolate \( y \), subtract 7 from both sides.

6. **Conclusion**:
The coordinates of point \( A \) are \( (3, -3) \).

### Final Answer:
The coordinates of point A are \( (3, -3) \).