Find the value of k so that PQ will be parallel to RS . `P(5, -1), Q(6, 11), R(6,-4k)` and `S(7, k^2)`

Answer

To find the value of \( k \) such that the line segment \( PQ \) is parallel to the line segment \( RS \), we need to ensure that their slopes are equal. Let's go through the steps to solve the problem.

### Step 1: Find the slope of line segment \( PQ \)

The coordinates of points \( P \) and \( Q \) are given as \( P(5, -1) \) and \( Q(6, 11) \).

The formula for the slope \( m \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Substituting the coordinates of \( P \) and \( Q \):

\[ m_{PQ} = \frac{11 - (-1)}{6 - 5} = \frac{11 + 1}{1} = \frac{12}{1} = 12 \]

### Step 2: Find the slope of line segment \( RS \)

The coordinates of points \( R \) and \( S \) are given as \( R(6, -4k) \) and \( S(7, k^2) \).

Using the slope formula again:

\[ m_{RS} = \frac{y_4 - y_3}{x_4 - x_3} \]

Substituting the coordinates of \( R \) and \( S \):

\[ m_{RS} = \frac{k^2 - (-4k)}{7 - 6} = \frac{k^2 + 4k}{1} = k^2 + 4k \]

### Step 3: Set the slopes equal to each other

Since \( PQ \) is parallel to \( RS \), we have:

\[ m_{PQ} = m_{RS} \]

Thus,

\[ 12 = k^2 + 4k \]

### Step 4: Rearrange the equation

Rearranging the equation gives us:

\[ k^2 + 4k - 12 = 0 \]

### Step 5: Factor the quadratic equation

To factor the quadratic equation, we look for two numbers that multiply to \(-12\) and add to \(4\). These numbers are \(6\) and \(-2\):

\[ k^2 + 6k - 2k - 12 = 0 \]

Grouping the terms:

\[ k(k + 6) - 2(k + 6) = 0 \]

Factoring out the common term:

\[ (k - 2)(k + 6) = 0 \]

### Step 6: Solve for \( k \)

Setting each factor to zero gives us:

1. \( k - 2 = 0 \) → \( k = 2 \)
2. \( k + 6 = 0 \) → \( k = -6 \)

### Conclusion

The values of \( k \) that make \( PQ \) parallel to \( RS \) are:

\[ k = 2 \quad \text{or} \quad k = -6 \]