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

Explanation

Let the three given points be `A-=(x_1,y_1)-=(k,2-2k),B-=(x_2,y_2)-=(-k+1,2k)`, and `C-=(x_3,y_3)-=(-4-k,6-2k)`.
If the given points are collinear, then `Delta=0`, i.e.,
`x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0`
or `k(2k-6+2k)+(-k+1)(6-2k-2+2k)+ (-4-k)(2-2k) =0`
or `k(4k-6)-4(k-1) +(4+k)(4k-2)=0`
or `4k^2 -6k-4k+4+4k^2+14k-8=0`
or `8k^2+4k-4=0`
or `2k^2+k-1=0`
or `(2k^2-1)(k+1)=0`
i.e., `k=(1)/(2)` or `-1`
But for `t=1//2`, the first two points are coincident. Hence ,`K=-1`.