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

Answer

To find all possible values of \( x \) for which the distance between the points \( A(x, -1) \) and \( B(5, 3) \) is 5 units, we can follow these steps:

### Step 1: Write down the distance formula
The distance \( d \) between two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

### Step 2: Assign coordinates
For our points:
- Point \( A \) has coordinates \( (x, -1) \) (where \( x_1 = x \) and \( y_1 = -1 \))
- Point \( B \) has coordinates \( (5, 3) \) (where \( x_2 = 5 \) and \( y_2 = 3 \))

### Step 3: Set up the equation
We know the distance \( d \) is 5 units, so we can set up the equation:
\[ 5 = \sqrt{(5 - x)^2 + (3 - (-1))^2} \]

### Step 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:
\[ 5 = \sqrt{(5 - x)^2 + 16} \]

### Step 5: Square both sides
To eliminate the square root, square both sides:
\[ 5^2 = (5 - x)^2 + 16 \]
\[ 25 = (5 - x)^2 + 16 \]

### Step 6: Rearrange the equation
Subtract 16 from both sides:
\[ 25 - 16 = (5 - x)^2 \]
\[ 9 = (5 - x)^2 \]

### Step 7: Take the square root
Now take the square root of both sides:
\[ \sqrt{9} = 5 - x \quad \Rightarrow \quad 3 = 5 - x \quad \text{or} \quad -3 = 5 - x \]

### Step 8: Solve for \( x \)
1. For \( 3 = 5 - x \):
\[ x = 5 - 3 = 2 \]
2. For \( -3 = 5 - x \):
\[ x = 5 + 3 = 8 \]

### Conclusion
The possible values of \( x \) are:
\[ \boxed{2 \text{ and } 8} \]

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