The area of triangle formed by the vertices `(a, 1/a), (b, 1/b), (c, 1/c)` is (a) `|((a+b)(b+c)(c+a))/(2abc)|` (b) `|((a-b)(b-c)(c-a))/(2abc)|` (c) `|((a-(1)/(a))(b-(1)/(b))(c-(1)/(c )))/(2)|` (d) `|((a-1)(b-1)(c-1))/(2abc)|`

Options

A. (a) `|((a+b)(b+c)(c+a))/(2abc)|`
B. (b) `|((a-b)(b-c)(c-a))/(2abc)|`
C. (c) `|((a-(1)/(a))(b-(1)/(b))(c-(1)/(c )))/(2)|`
D. (d) `|((a-1)(b-1)(c-1))/(2abc)|`

Explanation

To find the area of the triangle formed by the vertices \((a, \frac{1}{a})\), \((b, \frac{1}{b})\), and \((c, \frac{1}{c})\), we can use the formula for the area of a triangle given by three 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:
- \((x_1, y_1) = (a, \frac{1}{a})\)
- \((x_2, y_2) = (b, \frac{1}{b})\)
- \((x_3, y_3) = (c, \frac{1}{c})\)

### Step 2: Substitute into the area formula
Now substitute these coordinates into the area formula:

\[ \text{Area} = \frac{1}{2} \left| a\left(\frac{1}{b} - \frac{1}{c}\right) + b\left(\frac{1}{c} - \frac{1}{a}\right) + c\left(\frac{1}{a} - \frac{1}{b}\right) \right| \]

### Step 3: Simplify each term
Calculating each term separately:
1. For the first term:
\[ a\left(\frac{1}{b} - \frac{1}{c}\right) = a\left(\frac{c-b}{bc}\right) = \frac{a(c-b)}{bc} \]

2. For the second term:
\[ b\left(\frac{1}{c} - \frac{1}{a}\right) = b\left(\frac{a-c}{ac}\right) = \frac{b(a-c)}{ac} \]

3. For the third term:
\[ c\left(\frac{1}{a} - \frac{1}{b}\right) = c\left(\frac{b-a}{ab}\right) = \frac{c(b-a)}{ab} \]

### Step 4: Combine the terms
Now combine these terms:
\[ \text{Area} = \frac{1}{2} \left| \frac{a(c-b)}{bc} + \frac{b(a-c)}{ac} + \frac{c(b-a)}{ab} \right| \]

### Step 5: Find a common denominator
The common denominator for these fractions is \(abc\):
\[ \text{Area} = \frac{1}{2} \left| \frac{a^2c(a-b) + b^2a(b-c) + c^2b(c-a)}{abc} \right| \]

### Step 6: Factor the numerator
The numerator can be factored as:
\[ (a-b)(b-c)(c-a) \]
Thus, we have:
\[ \text{Area} = \frac{1}{2} \left| \frac{(a-b)(b-c)(c-a)}{abc} \right| \]

### Final Result
Therefore, the area of the triangle is:
\[ \text{Area} = \left| \frac{(a-b)(b-c)(c-a)}{2abc} \right| \]

### Conclusion
The correct answer is option (b):
\[ \text{Area} = \left| \frac{(a-b)(b-c)(c-a)}{2abc} \right| \]