If the area of a triangle with vertices (- 3, 0), (3, 0) and (0, k) is 9 sq unit, then what is the value of k?
(a)3
(b)6
(c)9
(d)12

Options

A. 3
B. 6
C. 9
D. 12

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 its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 1: Assign the coordinates
Let:
- \((x_1, y_1) = (-3, 0)\)
- \((x_2, y_2) = (3, 0)\)
- \((x_3, y_3) = (0, k)\)

### Step 2: Substitute the coordinates into the area formula
Substituting the coordinates into the area formula gives:

\[ \text{Area} = \frac{1}{2} \left| -3(0 - k) + 3(k - 0) + 0(0 - 0) \right| \]

### Step 3: Simplify the expression
This simplifies to:

\[ \text{Area} = \frac{1}{2} \left| -3(-k) + 3k \right| = \frac{1}{2} \left| 3k + 3k \right| = \frac{1}{2} \left| 6k \right| = 3|k| \]

### Step 4: Set the area equal to 9
We know the area is 9 square units, so we set up the equation:

\[ 3|k| = 9 \]

### Step 5: Solve for \( k \)
Dividing both sides by 3 gives:

\[ |k| = 3 \]

This means \( k \) can be either 3 or -3. However, since the question asks for the value of \( k \) in the context of the triangle's height, we take the positive value:

\[ k = 3 \]

### Final Answer
Thus, the value of \( k \) is \( \boxed{3} \).