Formula: 1 Mark
Calculations: 2 Marks
Answer: 1 Mark
For points A,B,C to be collinear, area of triangle ABC formed by these points should be zero.
$\Rightarrow$ The area of  $\Delta ABC=0$
 $\Rightarrow \frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=0$
$\Rightarrow \frac{1}{2}\left[k\right(2k-6+2k)+(–k+1\left)\right(6–2k-2+2k)+(–4–k\left)\right(2–2k-2k\left)\right]=0$
$\Rightarrow \frac{1}{2}\left[k\right(4k-6)+(–k+1\left)\right(4)+(–4–k\left)\right(2–4k\left)\right]=0$
$\Rightarrow \frac{1}{2}[4k^2-6k-4k+4-8+16k-2k+4k^2]=0$
$\Rightarrow 8k^2+4k-4=0$
$\Rightarrow 8k^2+4k-4=0$
$\Rightarrow 2k^2+k-1=0$
$\Rightarrow (k+1)(2k-1)=0$
$\Rightarrow k=-1ork=\frac{1}{2}$