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

Options

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

Explanation

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 its vertices.

### Step-by-Step Solution:

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

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

3. **Substitute the coordinates into the formula**:
Here, we have:
- \( x_1 = -3, y_1 = 0 \)
- \( x_2 = 3, y_2 = 0 \)
- \( x_3 = 0, y_3 = k \)

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

4. **Simplify the expression**:
This simplifies to:
\[ A = \frac{1}{2} \left| -3(-k) + 3k \right| = \frac{1}{2} \left| 3k + 3k \right| = \frac{1}{2} \left| 6k \right| = 3|k| \]

5. **Set the area equal to 9**:
We know the area is given as 9 square units:
\[ 3|k| = 9 \]

6. **Solve for \( k \)**:
Dividing both sides by 3 gives:
\[ |k| = 3 \]
This means \( k \) can be either 3 or -3.

### Conclusion:
Thus, the possible values of \( k \) are:
\[ k = 3 \quad \text{or} \quad k = -3 \]