Solution:
To determine the value of ( k ) for which the given points are collinear, we can use the concept that three points are collinear if the area of the triangle formed by them is zero.
Given points:
– ( A(-k + 1, 2k) )
– ( B(k, 2 – 2k) )
– ( C(-4 – k, 6 – 2k) )
The area of the triangle formed by points ( A(x_1, y_1) ), ( B(x_2, y_2) ), and ( C(x_3, y_3) ) is given by:
[
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|
]
Substituting the coordinates:
[
text{Area} = frac{1}{2} left| (-k + 1)((2 – 2k) – (6 – 2k)) + k((6 – 2k) – 2k) + (-4 – k)(2k – (2 – 2k)) right|
]
Simplify the expressions inside the absolute value:
[
= frac{1}{2} left| (-k + 1)(-4) + k(6 – 4k) + (-4 – k)(4k – 2) right|
]
[
= frac{1}{2} left| 4k – 4 + 6k – 4k^2 – 16k + 8 + 4k^2 + 2k right|
]
[
= frac{1}{2} left| -4 – 8k + 8 right|
]
[
= frac{1}{2} left| 4 – 8k right|
]
Condition for Collinearity:
For the points to be collinear, the area must be zero:
[
frac{1}{2} left| 4 – 8k right| = 0
]
[
left| 4 – 8k right| = 0
]
[
4 – 8k = 0
]
[
8k = 4
]
[
k = frac{1}{2}
]
Conclusion:
The points are collinear when ( k = frac{1}{2} ).
