Using determinats, find the values of k, if the area of triangle with vertices (-2, 0), (0, 4) and (0, k) is 4 square units.

Answer

To find the values of \( k \) such that the area of the triangle with vertices \((-2, 0)\), \((0, 4)\), and \((0, k)\) is 4 square units, we can use the formula for the area of a triangle given its vertices.

### Step-by-Step Solution

1. **Identify the vertices**:
The vertices of the triangle are:
- \( A(-2, 0) \)
- \( B(0, 4) \)
- \( C(0, k) \)

2. **Area formula using determinants**:
The area \( \Delta \) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:
\[ \Delta = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

3. **Substituting the vertices into the formula**:
Substituting the coordinates of the vertices into the formula:
\[ \Delta = \frac{1}{2} \left| -2(4 - k) + 0(k - 0) + 0(0 - 4) \right| \]
This simplifies to:
\[ \Delta = \frac{1}{2} \left| -2(4 - k) \right| \]
\[ \Delta = \frac{1}{2} \left| -8 + 2k \right| = \frac{1}{2} \left| 2k - 8 \right| \]

4. **Setting the area equal to 4**:
We know the area is 4 square units, so we set up the equation:
\[ \frac{1}{2} \left| 2k - 8 \right| = 4 \]

5. **Solving the equation**:
Multiply both sides by 2:
\[ \left| 2k - 8 \right| = 8 \]
This gives us two cases to consider:
- Case 1: \( 2k - 8 = 8 \)
- Case 2: \( 2k - 8 = -8 \)

6. **Solving Case 1**:
\[ 2k - 8 = 8 \]
\[ 2k = 16 \]
\[ k = 8 \]

7. **Solving Case 2**:
\[ 2k - 8 = -8 \]
\[ 2k = 0 \]
\[ k = 0 \]

8. **Final values of \( k \)**:
The values of \( k \) that satisfy the condition are:
\[ k = 8 \quad \text{or} \quad k = 0 \]

### Summary of the Solution
The values of \( k \) for which the area of the triangle is 4 square units are \( k = 8 \) and \( k = 0 \).