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

Options

A. `-1`
B. `1/2`
C. `2`
D. Both A and B

Answer

To find 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 determinant method. The points are collinear if the following determinant is equal to zero:

\[ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 \]

### Step 1: Set up the determinant

Substituting the points into the determinant, we have:

\[ \begin{vmatrix} k & 2-2k & 1 \\ 1-k & 2k & 1 \\ -4-k & 6-2k & 1 \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant

To calculate the determinant, we can expand it along the third column:

\[ = 1 \cdot \begin{vmatrix} 2-2k & 1 \\ 2k & 1 \\ 6-2k & 1 \end{vmatrix} \]

### Step 3: Simplify the determinant

Now we can simplify the determinant:

\[ = (2-2k) \cdot \begin{vmatrix} 1-k & 1 \\ -4-k & 1 \end{vmatrix} - (2k) \cdot \begin{vmatrix} k & 1 \\ -4-k & 1 \end{vmatrix} \]

Calculating these 2x2 determinants:

1. For \( \begin{vmatrix} 1-k & 1 \\ -4-k & 1 \end{vmatrix} \):
\[ = (1-k)(1) - (1)(-4-k) = 1-k + 4 + k = 5 \]

2. For \( \begin{vmatrix} k & 1 \\ -4-k & 1 \end{vmatrix} \):
\[ = k(1) - (1)(-4-k) = k + 4 + k = 2k + 4 \]

### Step 4: Substitute back into the determinant

Substituting these back into our determinant expression gives:

\[ (2-2k) \cdot 5 - (2k)(2k + 4) = 0 \]

### Step 5: Expand and simplify

Expanding this:

\[ 10 - 10k - (4k^2 + 8k) = 0 \]

Combining like terms:

\[ -4k^2 - 18k + 10 = 0 \]

### Step 6: Multiply through by -1

To make it easier to work with, multiply the entire equation by -1:

\[ 4k^2 + 18k - 10 = 0 \]

### Step 7: Solve the quadratic equation

Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 4 \), \( b = 18 \), and \( c = -10 \):

\[ b^2 - 4ac = 18^2 - 4(4)(-10) = 324 + 160 = 484 \]

Now, substituting into the quadratic formula:

\[ k = \frac{-18 \pm \sqrt{484}}{2 \cdot 4} = \frac{-18 \pm 22}{8} \]

Calculating the two possible values for \( k \):

1. \( k = \frac{4}{8} = \frac{1}{2} \)
2. \( k = \frac{-40}{8} = -5 \)

### Final Answer

Thus, the values of \( k \) for which the points are collinear are:

\[ k = \frac{1}{2} \quad \text{or} \quad k = -5 \]