If area of the triangle formed by `(0, 0), (a^(x^2), 0), (0, a^(6x))` is `1/(2a^5)` sq. units then `x=`

Options

A. 1,5
B. `-1,5`
C. `1,-5`
D. `-1,-5`

Explanation

To solve the problem, we need to find the value of \( x \) given that the area of the triangle formed by the points \( (0, 0) \), \( (a^{x^2}, 0) \), and \( (0, a^{6x}) \) is \( \frac{1}{2a^5} \) square units.

### Step-by-Step Solution:

1. **Identify the Points**:
The points given are:
- \( A(0, 0) \)
- \( B(a^{x^2}, 0) \)
- \( C(0, a^{6x}) \)

2. **Calculate the Area of the Triangle**:
The area \( A \) of a triangle formed by points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our points:
\[ \text{Area} = \frac{1}{2} \left| 0(a^{6x} - 0) + a^{x^2}(0 - 0) + 0(0 - a^{6x}) \right| = \frac{1}{2} \left| a^{x^2} \cdot a^{6x} \right| \]
Since the area is a right triangle, we can also use the base and height:
- Base = \( a^{x^2} \)
- Height = \( a^{6x} \)
Thus,
\[ \text{Area} = \frac{1}{2} \cdot a^{x^2} \cdot a^{6x} = \frac{1}{2} a^{x^2 + 6x} \]

3. **Set the Area Equal to the Given Value**:
We know from the problem statement that:
\[ \frac{1}{2} a^{x^2 + 6x} = \frac{1}{2} a^{-5} \]
By multiplying both sides by 2, we simplify to:
\[ a^{x^2 + 6x} = a^{-5} \]

4. **Equate the Exponents**:
Since the bases are the same, we can equate the exponents:
\[ x^2 + 6x = -5 \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ x^2 + 6x + 5 = 0 \]

6. **Factoring the Quadratic**:
We can factor the quadratic equation:
\[ (x + 5)(x + 1) = 0 \]

7. **Finding the Roots**:
Setting each factor to zero gives:
\[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

8. **Conclusion**:
The values of \( x \) that satisfy the equation are \( x = -5 \) and \( x = -1 \).

### Final Answer:
Thus, the possible values for \( x \) are \( -5 \) and \( -1 \).