Question

If the points (k,2-2k), (1-k, 2k) and (-k-4, 6-2k) are collinear, then a possible values of k is .

• A. -$\frac{1}{2}$ or 1
• B. $\frac{1}{2}$ or -1
• C. $1$ or -1
• D. $-2$ or -1

Answer 1

The correct option is B $\frac{1}{2}$ or -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
$\Rightarrow \frac{2k-(2-2k)}{1-k-k}=\frac{6-2k-2k}{-k-4-1+k}$
$\Rightarrow \frac{4k-2}{1-2k}=\frac{6-4k}{-5}$
$\Rightarrow 120k+10=6-4k-12k+8k^2$
$\Rightarrow 8k^2+4k-4=0$
$\Rightarrow 2k^2+k-1=0$
$\Rightarrow 2k^2(k+1)-1(k+1)=0$
$\Rightarrow (2k-1)(k+1)=0$
$\Rightarrow 2k-1=0ork+1=0$
$\Rightarrow k=\frac{1}{2}ork=-1$