The ends of a diagonal of a square have co-ordinates `(-2,p) and (p,2) .` Find p, if the area of the square is 40 sq. units.

Answer

To solve the problem, we need to find the value of \( p \) given the coordinates of the ends of the diagonal of a square and the area of the square.

### Step-by-Step Solution:

1. **Identify the coordinates of the diagonal ends:**
The ends of the diagonal are given as \( (-2, p) \) and \( (p, 2) \).

2. **Use the area of the square to find the side length:**
The area of the square is given as \( 40 \) square units.
\[ \text{Area} = a^2 = 40 \quad \Rightarrow \quad a = \sqrt{40} = 2\sqrt{10} \]

3. **Calculate the length of the diagonal using the side length:**
The length of the diagonal \( D \) of a square is given by the formula:
\[ D = a\sqrt{2} \]
Therefore,
\[ D^2 = (a\sqrt{2})^2 = 2a^2 = 2 \times 40 = 80 \]

4. **Use the distance formula to find \( D^2 \):**
The distance \( D \) between the points \( (-2, p) \) and \( (p, 2) \) can be calculated using the distance formula:
\[ D^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \]
Plugging in the coordinates:
\[ D^2 = (p - (-2))^2 + (2 - p)^2 \]
Simplifying this gives:
\[ D^2 = (p + 2)^2 + (2 - p)^2 \]

5. **Expand the squares:**
Expanding both terms:
\[ D^2 = (p^2 + 4p + 4) + (4 - 4p + p^2) \]
Combining like terms:
\[ D^2 = 2p^2 + 4 + 4 = 2p^2 + 8 \]

6. **Set the two expressions for \( D^2 \) equal:**
We have two expressions for \( D^2 \):
\[ 2p^2 + 8 = 80 \]

7. **Solve for \( p^2 \):**
Rearranging the equation:
\[ 2p^2 = 80 - 8 \]
\[ 2p^2 = 72 \quad \Rightarrow \quad p^2 = \frac{72}{2} = 36 \]

8. **Find \( p \):**
Taking the square root of both sides:
\[ p = \pm 6 \]

### Final Answer:
The possible values of \( p \) are \( 6 \) and \( -6 \).