 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 \( k \) such that the points \((-k + 1, 2k)\), \((k, 2 - 2k)\), and \((-4 - k, 6 - 2k)\) are collinear, we can use the concept that three points are collinear if the area of the triangle formed by these points is zero.

The area of a triangle formed by three points \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is given by the 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. Let’s plug in the coordinates of the points into the formula:

- Point \( A \): \((-k + 1, 2k)\)
- Point \( B \): \((k, 2 - 2k)\)
- Point \( C \): \((-4 - k, 6 - 2k)\)

### Substituting these into the formula:

\[
\text{Area} = \frac{1}{2} \left| (-k+1)((2 - 2k) - (6 - 2k)) + k((6 - 2k) - 2k) + (-4 - k)(2k - (2 - 2k)) \right| = 0
\]

Simplify the terms:

1. \( (2 - 2k) - (6 - 2k) = -4 \)
2. \( (6 - 2k) - 2k = 6 - 4k \)
3. \( 2k - (2 - 2k) = 2k - 2 + 2k = 4k - 2 \)

Now substituting these into the equation:

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

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

Expanding all the terms:

1. \( 4(k - 1) = 4k - 4 \)
2. \( k(6 - 4k) = 6k - 4k^2 \)
3. \( (-4 - k)(4k - 2) = -4(4k - 2) - k(4k - 2) = -16k + 8 - 4k^2 + 2k \)

Now, substitute these expanded terms back:

\[
\frac{1}{2} \left| 4k - 4 + 6k - 4k^2 - 16k + 8 - 4k^2 + 2k \right| = 0
\]

Simplify the terms:

\[
\frac{1}{2} \left| -8k - 4 - 8k^2 + 8 \right| = 0
\]

\[
\frac{1}{2} \left| -8k^2 - 8k + 4 \right| = 0
\]

Since we want the absolute value to be zero, we can ignore the \(\frac{1}{2}\) and solve:

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

Divide through by \(-4\) to simplify:

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

This is a quadratic equation. We can solve it using the quadratic formula:

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

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

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

\[
k = \frac{-2 \pm \sqrt{4 + 8}}{4}
\]

\[
k = \frac{-2 \pm \sqrt{12}}{4}
\]

\[
k = \frac{-2 \pm 2\sqrt{3}}{4}
\]

\[
k = \frac{-1 \pm \sqrt{3}}{2}
\]

Thus, the two possible values for \(k\) are:

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

These are the values of \( k \) for which the points are collinear.

Would you like further details or explanations on any part of the solution? Here are five related questions to explore:

1. What is the geometric interpretation of collinearity in 2D space?
2. How do we derive the area of a triangle from three given points?
3. What is the significance of using the determinant method to check for collinearity?
4. How would the equation change if the points were in 3D space instead of 2D?
5. Can the quadratic formula be used for non-real roots, and how does that affect geometry?

**Tip:** When solving quadratic equations, always simplify the coefficients as much as possible before applying the quadratic formula to avoid unnecessary complexity.

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

Find the value of k that makes the points (-k + 1, 2k), (k, 2 - 2k), and (-4 - k, 6 - 2k) collinear by solving a quadratic equation. This problem covers coordinate geometry concepts, including the area of a triangle and the quadratic formula. Suitable for students in Grades 9-12.

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