Question

If $(k,2-2k),(-k+1,2k),(-4-k,6-2k)$ are collinear, then $k=$

• A. $+1$
• B. $-1$
• C. $-2$
• D. $2$

Answer 1

Answer: (B)

$-1$

$\begin{array}{c}wheremorethanthreepointsarecollinearslopeofthelinearesame\\ So,\\ slope:\left[\left(k,2-2k\right),\left(-k+1,2k\right)\right]=\left[\left(-k+1,2k\right),\left(-4-k,6-2k\right)\right]\\ \Rightarrow \frac{2-2k-2k}{k+k-1}=\frac{2k-6+2k}{-k+1+4+k}\\ \Rightarrow \frac{2-4k}{2k-1}=\frac{4x-6}{5}\\ \Rightarrow \left(1-2k\right)5=\left(2k-3\right)\left(2k-1\right)\end{array}$

$5-10k=4k^2-2k-6k+3$

$4k^2+2k-2=0$

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

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

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

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

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

Hence,option $B$ is the correct answer.