Question

Find the value of $k$, if the points $(k,3),(6,-2)$ and $(-3,4)$ are collinear.

Answer 1

Points $A(K,3)$; $B(6,-2)$; $C(-3,4)$ are coplanar.

Hence, area $=0$

$\Rightarrow \frac{1}{2}\left|\begin{array}{cc}{x}_1-x_2& y_1-y_2\\ x_1-x_3& y_1-y_3\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{cc}k-6& 3+2\\ k+3& 3-4\end{array}\right|=0$

$\Rightarrow (k-6)(-1)-\left(5\right)(k+3)=0$

$\Rightarrow -k+6-5k-15=0$

$\Rightarrow 6k=-6$

$\Rightarrow k=-\frac{3}{2}$