Q.Find the value(s) of $k$ if the area of the triangle with vertices $(k,0),(4,0),(0,2)$ is $4$ square units.

[!TLDR] Set the area determinant equal to $\pm8$ (twice the given area) and solve the resulting linea …

Area $=\tfrac12\begin{vmatrix}k&0&1\\4&0&1\\0&2&1\end{vmatrix}=4$, so $\begin{vmatrix}k&0&1\\4&0&1\\0&2&1\end{vmatrix}=\pm8$. Expanding along row 1: $k\begin{vmatrix}0&1\\2&1\end{vmatrix}-0+1\begin{vmatrix}4&0\\0&2\end{vmatrix}=k(0-2)+1(8-0)=-2k+8$. Setting $-2k+8=8$ gives $k=0$; setting $- …

Expand the area determinant in terms of $k$, set it equal to $\pm$(twice the given area) to account for the absolute value, a …

Solving only one of the two sign cases and missing the second valid value of $k$ -- the absolute value in the area formu …