Question

If the points (k, 2-2k), (1-k, 2k) and (-k-4, 6-2k) are collinear, find the possible values of k.
$\left[4Marks\right]$

Answer 1

Points A(k, 2-2k), B (1-k, 2k) and C(-k-4, 6-2k) are collinear.
Then, slope of line AB = slope of line BC
$\frac{2k-(2-2k)}{1-k-k}=\frac{6-2k-2k}{-k-4-1+k}$
$\left[1Mark\right]$
$\frac{4k-2}{1-2k}=\frac{6-4k}{-5}$
$-20k+10=6-4k-12k+8k^2$
$8k^2+4k-4=0$
$2k^2+k-1=0$
$\left[1Mark\right]$
$2k(k+1)-1(k+1)=0$
$(2k-1)(k+1)=0$
2k-1=0 and k+1=0
$k=\frac{1}{2}andk=-1$
$\left[1Mark\right]$
we can also say, slope of AB = slope of AC $\frac{2k-(2-2k)}{1-k-k}=\frac{6-2k-(2-2k)}{-k-4-k}$
$\frac{4k-2}{1-2k}=\frac{4}{-2k-4}$

$-8k^2+4k-16k+8=4-8k$
$8k^2+4k-4=0$
$2k^2+k-1=0$
we got the same equation of previous case
So $k=\frac{1}{2}$ and k=-1 are the only possible values of k when the given points are collinear.
$\left[1Mark\right]$