Question

If the points (k,2-2k), (1-k, 2k) and (-k-4, 6-2k) lie on a line, then find the possible values of k.$\left[3Marks\right]$

Answer 1

Given: 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$\left[0.5Mark\right]$
$\Rightarrow \frac{2k-(2-2k)}{1-k-k}=\frac{6-2k-2k}{-k-4-1+k}$
$\Rightarrow \frac{4k-2}{1-2k}=\frac{6-4k}{-5}$
$\Rightarrow 120k+10=6-4k-12k+8k^2$
$\Rightarrow 8k^2+4k-4=0$
$\Rightarrow 2k^2+k-1=0$
$\Rightarrow 2k^2(k+1)-1(k+1)=0$
$\Rightarrow (2k-1)(k+1)=0$
$\Rightarrow 2k-1=0andk+1=0$
$\Rightarrow k=\frac{1}{2}andk=-1$$\left[1Mark\right]$

We can also say, slope of AB = slope of AC$\left[0.5Mark\right]$

$\Rightarrow \frac{2k-(2-2k)}{1-k-k}=\frac{6-2k-(2-2k)}{-k-4-k}$
$\Rightarrow \frac{4k-2}{1-2k}=\frac{4}{-2k-4}$
$\Rightarrow -8k^2+4k-16k+8=4-8k$
$\Rightarrow 8k^2+4k-4=0$
$\Rightarrow 2k^2+k-1=0$
we got the same equation of previous case
So,

$k=\frac{1}{2}$ and $k=-1$ are the only possible values of $k$

when given points lie on a line.$\left[1Mark\right]$