Find all possible values of a for which the distance between the points A(a, -1) and B(5, 3) is 5 unit.

Answer

To find all possible values of \( a \) for which the distance between the points \( A(a, -1) \) and \( B(5, 3) \) is 5 units, we will use the distance formula.

### Step-by-Step Solution:

1. **Recall the Distance Formula**:
The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

2. **Identify the Points**:
Here, the points are \( A(a, -1) \) and \( B(5, 3) \). So, we have:
- \( x_1 = a \)
- \( y_1 = -1 \)
- \( x_2 = 5 \)
- \( y_2 = 3 \)

3. **Set Up the Distance Equation**:
We know the distance between points \( A \) and \( B \) is 5 units. Therefore, we can set up the equation:
\[ \sqrt{(5 - a)^2 + (3 - (-1))^2} = 5 \]

4. **Simplify the Equation**:
First, simplify \( (3 - (-1))^2 \):
\[ 3 - (-1) = 3 + 1 = 4 \quad \Rightarrow \quad (3 - (-1))^2 = 4^2 = 16 \]
Now substitute this back into the equation:
\[ \sqrt{(5 - a)^2 + 16} = 5 \]

5. **Square Both Sides**:
To eliminate the square root, square both sides:
\[ (5 - a)^2 + 16 = 25 \]

6. **Isolate the Squared Term**:
Subtract 16 from both sides:
\[ (5 - a)^2 = 25 - 16 \]
\[ (5 - a)^2 = 9 \]

7. **Take the Square Root**:
Now take the square root of both sides:
\[ 5 - a = 3 \quad \text{or} \quad 5 - a = -3 \]

8. **Solve for \( a \)**:
For the first equation:
\[ 5 - a = 3 \quad \Rightarrow \quad a = 5 - 3 = 2 \]
For the second equation:
\[ 5 - a = -3 \quad \Rightarrow \quad a = 5 + 3 = 8 \]

9. **Conclusion**:
The possible values of \( a \) are:
\[ a = 2 \quad \text{and} \quad a = 8 \]

### Final Answer:
The values of \( a \) for which the distance between the points \( A(a, -1) \) and \( B(5, 3) \) is 5 units are \( a = 2 \) and \( a = 8 \).