Find the value of k if the points $\mathrm{A}(2,\mathrm{ }3), \mathrm{B}(4, \mathrm{k})$ and $\mathrm{C}(6,\mathrm{ }-3)$ are collinear.

The given points are,
A (2, 3), B (4, k) and C (6, −3) Now, for the collinear point,

Area of $△\mathrm{ABC}=0$
$\mathbf{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$
We have, ${\mathrm{x}}_1=2, {\mathrm{y}}_1=3$

${\mathrm{x}}_2=4, {\mathrm{y}}_2=\mathrm{k}$

${\mathrm{x}}_3=6, {\mathrm{y}}_3=-3$

So,
$\Rightarrow \frac{1}{2}\left[2\right(\mathrm{k}-(-3))+4(-3-3)+6(3-\mathrm{k}\left)\right)]=0\mathrm{ }$

$\Rightarrow 2\mathrm{k}+6-24+18-6\mathrm{k}=0\mathrm{ }$

$\Rightarrow -4\mathrm{k}=0\mathrm{ }$

$\Rightarrow$k = 0

So, the value of k is zero.