What is the area of the triangle with vertices `(x_(1),(1)/(x_(1))),(x_(2),(1)/(x_(2))),(x_(3),(1)/(x_(3)))`?

Options

A. `|(x_(1)-x_(2))(x_(2)-x_(3))(x_(3)-x_(1))|`
B. 0
C. `|((x_(1)-x_(2))(x_(2)-x_(3))(x_(3)-x_(1)))/(x_(1)x_(2)x_(3))|`
D. `|((x_(1)-x_(2))(x_(2)-x_(3))(x_(3)-x_(1)))/(2x_(1)x_(2)x_(3))|`

Explanation

To find the area of the triangle with vertices \((x_1, \frac{1}{x_1}), (x_2, \frac{1}{x_2}), (x_3, \frac{1}{x_3})\), we can use the formula for the area of a triangle given by the determinant of a matrix formed by the coordinates of the vertices.

### Step-by-Step Solution:

1. **Set Up the Determinant**:
The area \(A\) of the triangle can be calculated using the formula:
\[ A = \frac{1}{2} \left| \begin{vmatrix} x_1 & \frac{1}{x_1} & 1 \\ x_2 & \frac{1}{x_2} & 1 \\ x_3 & \frac{1}{x_3} & 1 \end{vmatrix} \right| \]

2. **Calculate the Determinant**:
We will compute the determinant:
\[ \begin{vmatrix} x_1 & \frac{1}{x_1} & 1 \\ x_2 & \frac{1}{x_2} & 1 \\ x_3 & \frac{1}{x_3} & 1 \end{vmatrix} \]
Using the properties of determinants, we can simplify this by performing row operations. We can subtract the first row from the second and third rows:
\[ = \begin{vmatrix} x_1 & \frac{1}{x_1} & 1 \\ x_2 - x_1 & \frac{1}{x_2} - \frac{1}{x_1} & 0 \\ x_3 - x_1 & \frac{1}{x_3} - \frac{1}{x_1} & 0 \end{vmatrix} \]

3. **Expand the Determinant**:
The determinant simplifies to:
\[ = \begin{vmatrix} x_1 & \frac{1}{x_1} & 1 \\ x_2 - x_1 & \frac{1}{x_2} - \frac{1}{x_1} & 0 \\ x_3 - x_1 & \frac{1}{x_3} - \frac{1}{x_1} & 0 \end{vmatrix} \]
The third column contains zeros, so we can expand along this column:
\[ = 1 \cdot \begin{vmatrix} x_2 - x_1 & \frac{1}{x_2} - \frac{1}{x_1} \\ x_3 - x_1 & \frac{1}{x_3} - \frac{1}{x_1} \end{vmatrix} \]

4. **Calculate the 2x2 Determinant**:
The 2x2 determinant is calculated as:
\[ = (x_2 - x_1) \left(\frac{1}{x_3} - \frac{1}{x_1}\right) - (x_3 - x_1) \left(\frac{1}{x_2} - \frac{1}{x_1}\right) \]
Simplifying this gives:
\[ = \frac{(x_2 - x_1)(x_1 - x_3) - (x_3 - x_1)(x_1 - x_2)}{x_1 x_2 x_3} \]

5. **Final Area Calculation**:
Therefore, the area \(A\) is:
\[ A = \frac{1}{2} \cdot \left| \frac{(x_2 - x_1)(x_1 - x_3) - (x_3 - x_1)(x_1 - x_2)}{x_1 x_2 x_3} \right| \]
After simplifying, we find:
\[ A = \frac{|(x_1 - x_2)(x_2 - x_3)(x_3 - x_1)|}{2x_1 x_2 x_3} \]

### Final Answer:
The area of the triangle is:
\[ A = \frac{|(x_1 - x_2)(x_2 - x_3)(x_3 - x_1)|}{2x_1 x_2 x_3} \]