Question

The points $(k-1,k+2),(k,k+1),(k+1,k)$ are collinear for

• A. any value of $k$
• B. $k=-\frac{1}{2}$ only
• C. no value of $k$
• D. integral values of $k$ only

Answer 1

The correct option is A any value of $k$

$△=\frac{1}{2}\left|\begin{array}{ccc}k-1& k+2& 1\\ k& k+1& 1\\ k+1& k& 1\end{array}\right|=0$
$\Rightarrow (k-1)(k+1-k)-(k+2)(k-k-1)+1(k^2-(k+1{)}^2)=0$
$\Rightarrow 2k+1+(-1-2k)=0$
$\Rightarrow 0=0$
Hence, The given points are collinear for any value of $k$.