Question

For what value of k are the points (k, 2 – 2k), (-k + 1, 2k), (-4 –k, 6 – 2k) collinear?

Answer 1

$\Delta =\frac{1}{2}\left|\begin{array}{ccc}k& 2-2k& 1\\ -k+1& 2k& 1\\ -4-k& 6-2k& 1\end{array}\right|=0$
$Apply{R}_2-R_{1}and{R}_3-R_1$
$\Delta =\frac{1}{2}\left|\begin{array}{ccc}k& 2-2k& 1\\ -2k+1& 4k-2& 0\\ -4-2k& 4& 0\end{array}\right|=0$
$or\left|\begin{array}{cc}-2k+1& 4k-2\\ -4-2k& 4\end{array}\right|=0$
$or4(1-2k)-(-4-2k)(4k-2)=0$
$or(1-2k)+(k+2)(2k-1)=0$
$or1-2k+(2k^2+3k-2)=0$
$or2k^2+k-1=0$
$or(k+1)(2k-1)=0$
$\therefore k=\frac{1}{2},-1$