If the area of a triangle with vertices `(-3,0),(3,0)` and `(0,0)` is 9 sq. units. Then the value of `k` will be

Options

A. `9`
B. `3`
C. `-9`
D. `6`

Explanation

To solve the problem, we need to find the value of \( k \) such that the area of the triangle formed by the vertices \((-3, 0)\), \((3, 0)\), and \((0, k)\) is equal to 9 square units.

### Step-by-Step Solution:

1. **Identify the vertices of the triangle**: The vertices are given as \( A(-3, 0) \), \( B(3, 0) \), and \( C(0, k) \).

2. **Use the formula for the area of a triangle using determinants**:
The area \( A \) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the determinant formula:
\[ A = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \]

3. **Substitute the coordinates into the determinant**:
For our triangle, substituting the coordinates:
\[ A = \frac{1}{2} \left| \begin{vmatrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{vmatrix} \right| \]

4. **Calculate the determinant**:
We can expand this determinant. Since the second column has two zeros, we can expand along that column:
\[ = \frac{1}{2} \left( 0 \cdot \begin{vmatrix} 3 & 1 \\ 0 & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} -3 & 1 \\ 0 & 1 \end{vmatrix} + k \cdot \begin{vmatrix} -3 & 1 \\ 3 & 1 \end{vmatrix} \right) \]
The determinant simplifies to:
\[ = \frac{1}{2} \left( k \cdot \left( -3 \cdot 1 - 3 \cdot 1 \right) \right) = \frac{1}{2} \left( k \cdot (-6) \right) = -3k \]

5. **Set the area equal to 9**:
We know that the area is given as 9 square units, so we set up the equation:
\[ \frac{1}{2} \left| -3k \right| = 9 \]
This simplifies to:
\[ \left| -3k \right| = 18 \]

6. **Solve for \( k \)**:
The absolute value gives us two cases:
\[ -3k = 18 \quad \text{or} \quad -3k = -18 \]
Solving these:
- For \( -3k = 18 \):
\[ k = -6 \]
- For \( -3k = -18 \):
\[ k = 6 \]

7. **Conclusion**:
The possible values for \( k \) are \( k = -6 \) and \( k = 6 \). However, since the question states that the area is positive, we take \( k = 6 \).

### Final Answer:
The value of \( k \) is \( 6 \).