For what value of k are the points `(k ,2-2k)(-k+1,2k)a n d(-4-k ,6-2k)` are collinear?

Explanation

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

### Step 1: Set up the area formula
The area \( A \) of a triangle formed by three points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula:

\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
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