 For what value of 'k' the points ( -k + 1, 2k), (k, 2 - 2k) and ( - 4 - k, 6 - 2k) are collinear. 

To find the value of \( k \) that makes the points \( (-k + 1, 2k) \), \( (k, 2 - 2k) \), and \( (-4 - k, 6 - 2k) \) collinear, we need to ensure that the area of the triangle formed by these points is zero. If the area is zero, the points are collinear.

The area of a triangle given three points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated by the determinant formula:

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

If the points are collinear, the area is zero, so:

\[
x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0
\]

Let’s use the coordinates of the given points:

- Point 1: \( (-k + 1, 2k) \)
- Point 2: \( (k, 2 - 2k) \)
- Point 3: \( (-4 - k, 6 - 2k) \)

Substituting into the formula:

\[
(-k + 1)\left((2 - 2k) - (6 - 2k)\right) + k\left((6 - 2k) - 2k\right) + (-4 - k)\left(2k - (2 - 2k)\right) = 0
\]

Now, let's simplify this equation.

1. Simplify the terms inside the parentheses:
   - \( (2 - 2k) - (6 - 2k) = 2 - 2k - 6 + 2k = -4 \)
   - \( (6 - 2k) - 2k = 6 - 4k \)
   - \( 2k - (2 - 2k) = 2k - 2 + 2k = 4k - 2 \)

Now substitute back:

\[
(-k + 1)(-4) + k(6 - 4k) + (-4 - k)(4k - 2) = 0
\]

2. Expand the terms:
   - \( (-k + 1)(-4) = 4k - 4 \)
   - \( k(6 - 4k) = 6k - 4k^2 \)
   - \( (-4 - k)(4k - 2) = (-4)(4k - 2) + (-k)(4k - 2) = -16k + 8 - 4k^2 + 2k \)

Now the equation becomes:

\[
(4k - 4) + (6k - 4k^2) + (-16k + 8 - 4k^2 + 2k) = 0
\]

3. Combine like terms:
   - \( 4k + 6k - 16k + 2k = -4k \)
   - \( -4 + 8 = 4 \)
   - \( -4k^2 - 4k^2 = -8k^2 \)

The equation simplifies to:

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

4. Divide the entire equation by \(-4\):

\[
2k^2 + k - 1 = 0
\]

5. Solve the quadratic equation \( 2k^2 + k - 1 = 0 \) using the quadratic formula:

\[
k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 2 \), \( b = 1 \), and \( c = -1 \). Substituting into the formula:

\[
k = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm \sqrt{9}}{4} = \frac{-1 \pm 3}{4}
\]

So, the two possible values of \( k \) are:

\[
k = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad k = \frac{-1 - 3}{4} = \frac{-4}{4} = -1
\]

Thus, the values of \( k \) that make the points collinear are \( k = \frac{1}{2} \) and \( k = -1 \).

---

Do you want further details or have any questions?

Here are five related questions to explore:
1. How do you find the slope of a line through two points?
2. What are the conditions for three points to be non-collinear?
3. Can you determine the area of a triangle using its vertices?
4. What is the significance of the quadratic formula in solving equations?
5. How does the determinant method apply to geometry problems?

**Tip**: For collinear points, using the slope formula for two pairs of points can also help verify if they lie on a straight line.

For what value of 'k' the points (-k + 1, 2k), (k, 2 - 2k) and (-4 - k, 6 - 2k) are collinear?

Determine the value of 'k' for which the points (-k + 1, 2k), (k, 2 - 2k), and (-4 - k, 6 - 2k) are collinear. Learn to apply the area of a triangle formula and quadratic equation in this step-by-step guide suitable for students in Grades 10-12.

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