Question

If the points $(k,2-2k),(1-k,2k)$ and $(-k-4,6-2k)$ be collinear the possible value(s) of $k$ is/are

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

Answer 1

Answer: (B), (C)

$1$
$\frac{1}{2}$

Let the three points be,

$A=(k,2-2k)$

$B=(1-k,2k)$

$C=(-k-4,6-2k)$

If $A,B,C$ to be collinear, the area of the triangle formed by these three points must be $0$.

For these points to be collinear, we know that

$\left|\begin{array}{ccc}k& 2-2k& 1\\ 1-k& 2k& 1\\ -k& 6-2k& 1\end{array}\right|=0$

$k(2k-6+2k)-(2-2k)\left[\right(1-k)+k]+1\left[\right(1-k\left)\right(6-2k)+2k^2]=0$

$4k^2-6k-2+2k+6-8k+2k^2+2k^2=0$

$8k^2-12k+4=0$

$2k^2-3k+1=0$

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

$k=1,\frac{1}{2}$