Question

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

Answer 1

Consider the given points.

$(-k+1,2k),(k,2-2k),(-4-k,6-2k)$

Since, these points are collinear.

Therefore the area of triangle formed by the triangle formed by the points will be zero.

Therefore,

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

$(-k+1)(-4)+k(6-4k)+(-4-k)(4k-2)=0$

$4k-4+6k-4k^2-16k+8-4k^2+2k=0$

$-2k^2-k+1=0$

$2k^2+k-1=0$

$2k^2+2k-k-1=0$

$2k(k+1)-1(k+1)=0$

$(k+1)(2k-1)=0$

$k=-1$ or $k=\frac{1}{2}$

Hence, this is the answer.