The area of a triangle with vertices `(-3,0)`, `(3,0)` and `(0,k)` is `9` sq. Units. Then , then value of k will be

Options

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

Answer

To find the value of \( k \) such that the area of the triangle with vertices at \((-3, 0)\), \((3, 0)\), and \((0, k)\) is \( 9 \) square units, we can use the formula for the area of a triangle given by its vertices.

### Step-by-Step Solution:

1. **Area Formula**: The area \( A \) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

2. **Substituting the Vertices**: For our triangle, the vertices are:
- \( (x_1, y_1) = (-3, 0) \)
- \( (x_2, y_2) = (3, 0) \)
- \( (x_3, y_3) = (0, k) \)

Plugging these values into the area formula:
\[ A = \frac{1}{2} \left| -3(0 - k) + 3(k - 0) + 0(0 - 0) \right| \]

3. **Simplifying the Expression**:
\[ A = \frac{1}{2} \left| -3(-k) + 3k \right| \]
\[ = \frac{1}{2} \left| 3k + 3k \right| \]
\[ = \frac{1}{2} \left| 6k \right| \]
\[ = 3|k| \]

4. **Setting the Area Equal to 9**: According to the problem, the area is \( 9 \) square units:
\[ 3|k| = 9 \]

5. **Solving for \( k \)**:
\[ |k| = \frac{9}{3} = 3 \]
This gives us two possible values for \( k \):
\[ k = 3 \quad \text{or} \quad k = -3 \]

### Conclusion:
The possible values of \( k \) are \( 3 \) and \( -3 \).