Question

If three points $(k,2k),(2k,3k),(3,1)$ are collinear, then $k$ is equal to:

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

Answer 1

The correct option is A $-2$

If given points are collinear, then

$\left|\begin{array}{ccc}k& 2k& 1\\ 2k& 3k& 1\\ 3& 1& 1\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}k& 2k& 1\\ k& k& 0\\ 3& 1& 1\end{array}\right|=0,R_2\to R_2-R_1$

$\Rightarrow 1(k-3k)+1(k^2-2k^2)=0$

$\Rightarrow k^2+2k=0$

$\Rightarrow k=0,-2$