If the value of K, is the points (k, 2-2k), (1-k, 2k) and (-4-k, 6-2 k) are collinear.

Answer

To determine the value of \( k \) for which the points \( (k, 2-2k) \), \( (1-k, 2k) \), and \( (-4-k, 6-2k) \) are collinear, we can use the concept that the area of the triangle formed by three collinear points is zero.

### Step-by-Step Solution:

1. **Set up the Area Formula**:
The area \( A \) of the triangle formed by the points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the determinant:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our points, we have:
- \( (x_1, y_1) = (k, 2-2k) \)
- \( (x_2, y_2) = (1-k, 2k) \)
- \( (x_3, y_3) = (-4-k, 6-2k) \)

2. **Substituting the Points into the Area Formula**:
Substitute the coordinates into the area formula:
\[ A = \frac{1}{2} \left| k(2k - (6 - 2k)) + (1-k)((6 - 2k) - (2 - 2k)) + (-4-k)((2 - 2k) - 2k) \right| \]

3. **Simplifying the Expression**:
Now simplify each term:
- First term:
\[ k(2k - 6 + 2k) = k(4k - 6) = 4k^2 - 6k \]
- Second term:
\[ (1-k)(6 - 2k - 2 + 2k) = (1-k)(4) = 4 - 4k \]
- Third term:
\[ (-4-k)(2 - 2k - 2k) = (-4-k)(2 - 4k) = -8 + 16k - 2k^2 + 4k = -8 + 20k - 2k^2 \]

4. **Combining All Terms**:
Combine all the terms:
\[ A = \frac{1}{2} \left| (4k^2 - 6k) + (4 - 4k) + (-8 + 20k - 2k^2) \right| \]
Simplifying this gives:
\[ A = \frac{1}{2} \left| (4k^2 - 2k^2) + (-6k - 4k + 20k) + (4 - 8) \right| \]
\[ = \frac{1}{2} \left| 2k^2 + 10k - 4 \right| \]

5. **Setting the Area to Zero**:
Since the points are collinear, set the area to zero:
\[ \left| 2k^2 + 10k - 4 \right| = 0 \]
This implies:
\[ 2k^2 + 10k - 4 = 0 \]

6. **Solving the Quadratic Equation**:
Divide the entire equation by 2:
\[ k^2 + 5k - 2 = 0 \]
Now apply the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ k = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{-5 \pm \sqrt{25 + 8}}{2} = \frac{-5 \pm \sqrt{33}}{2} \]

### Final Values of \( k \):
Thus, the values of \( k \) for which the points are collinear are:
\[ k = \frac{-5 + \sqrt{33}}{2} \quad \text{and} \quad k = \frac{-5 - \sqrt{33}}{2} \]