Find the value(s) of k so that PQ will be parallel to RS. Given :
`P(3, -1), Q(7, 11), R(-1, -1)` and `S(1, k)`

Answer

To find the value(s) of \( k \) such that line segment \( PQ \) is parallel to line segment \( RS \), we will follow these steps:

### Step 1: Identify the points
We have the points:
- \( P(3, -1) \)
- \( Q(7, 11) \)
- \( R(-1, -1) \)
- \( S(1, k) \)

### Step 2: Calculate the slope of line segment \( PQ \)
The formula for the slope \( m \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For points \( P(3, -1) \) and \( Q(7, 11) \):
- \( x_1 = 3, y_1 = -1 \)
- \( x_2 = 7, y_2 = 11 \)

Now, substituting these values into the slope formula:
\[ m_{PQ} = \frac{11 - (-1)}{7 - 3} = \frac{11 + 1}{7 - 3} = \frac{12}{4} = 3 \]

### Step 3: Calculate the slope of line segment \( RS \)
Using the same slope formula for points \( R(-1, -1) \) and \( S(1, k) \):
- \( x_1 = -1, y_1 = -1 \)
- \( x_2 = 1, y_2 = k \)

Substituting these values into the slope formula:
\[ m_{RS} = \frac{k - (-1)}{1 - (-1)} = \frac{k + 1}{1 + 1} = \frac{k + 1}{2} \]

### Step 4: Set the slopes equal for parallel lines
Since \( PQ \) is parallel to \( RS \), their slopes must be equal:
\[ m_{PQ} = m_{RS} \]
This gives us the equation:
\[ 3 = \frac{k + 1}{2} \]

### Step 5: Solve for \( k \)
To solve for \( k \), we will multiply both sides of the equation by 2:
\[ 2 \cdot 3 = k + 1 \]
\[ 6 = k + 1 \]

Now, subtract 1 from both sides:
\[ k = 6 - 1 \]
\[ k = 5 \]

### Final Answer
The value of \( k \) such that line segment \( PQ \) is parallel to line segment \( RS \) is:
\[ \boxed{5} \]