Problem Statement
Find the value of $k$ if the area of triangle with vertices $(k, 0), (4, 0)$ and $(0, 2)$ is $4\text{ square units}$.
Verified Solution & Marking Scheme
Set Up Determinant Formula for Area of Triangle
$$\text{Area} = \frac{1}{2} \left| \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} \right| = 4$$
$$\left| \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} \right| = 8$$
Expand Determinant Along Column 2
$$\begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = -2(k(1) - 4(1)) = -2(k - 4) = 8 - 2k$$
Solve Absolute Value Equation
$$|8 - 2k| = 8$$
Case 1: $8 - 2k = 8 \implies -2k = 0 \implies k = 0$.
Case 2: $8 - 2k = -8 \implies -2k = -16 \implies k = 8$.