`(x_(1)-x_(2))^(2)+(y_(1)-y_(2))^(2)=a^(2)`
`(x_(2)-x_(3))^(2)+(y_(2)-y_(3))^(2)=b^(2)` and
`(x_(3)-x_(1))^(2)+(y_(3)-y_(1))^(2)=c^(2)` then
`1/4|(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1)|` is equal to

Options

A. `s(s-a)^(2)`
B. `(s-b)(s-c)^(2)`
C. `s(s-a)(s-b)(s-c)` wher `2s=a+b+c`
D. none

Answer

To solve the problem, we need to find the area of the triangle formed by the points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) given the distances between the points. We can use the formula for the area of a triangle based on its vertices and the lengths of its sides.

### Step-by-Step Solution:

1. **Identify the Given Distances**:
We have three equations representing the distances between the points:
\[ (x_1 - x_2)^2 + (y_1 - y_2)^2 = a^2 \]
\[ (x_2 - x_3)^2 + (y_2 - y_3)^2 = b^2 \]
\[ (x_3 - x_1)^2 + (y_3 - y_1)^2 = c^2 \]

2. **Understanding the Area Formula**:
The area \(A\) of the triangle formed by the points can be calculated using the determinant formula:
\[ A = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \]
This determinant gives twice the area of the triangle.

3. **Using Heron's Formula**:
We can also express the area in terms of the sides \(a\), \(b\), and \(c\) using Heron's formula:
\[ s = \frac{a + b + c}{2} \]
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]

4. **Relating the Two Area Formulas**:
To find the area using the determinant, we can relate it to Heron's formula. The area can also be expressed as:
\[ A = \frac{1}{4} \sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)} \]
However, since we already have the area in terms of the determinant, we can equate the two expressions.

5. **Final Expression**:
Therefore, we can conclude that:
\[ \frac{1}{4} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| = \sqrt{s(s-a)(s-b)(s-c)} \]

6. **Conclusion**:
Since the problem states that the area is equal to this determinant divided by 4, we can conclude that the area of the triangle formed by the points is given by the expression derived above.