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 $k = -1$.

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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.