Question

If the points $(k,2-2k)$, $(1-k,2k)$ and $(-k-4,6-2k)$ be collinear, the number of possible values of $k$ are

• A. $4$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

$(k,2-2k),(1-k,2k)$ & $(-k-4,6-2k)$

if collinear the $m_1=m_2$

$\frac{2k-2+2k}{1-k-k}=\frac{6-2k-2k}{-k-4-1+k}$

$\frac{4k-2}{1-2k}=\frac{6-4k}{-5}$

$\frac{2k-1}{1-2k}=\frac{3-2k}{-5}$

$5=3-2k$

$2k=-2$

$k=-1$

Only $1$ value possible.