The area of the triangle formed by three point `sqrt3 +i, -1+sqrt3i and (sqrt3-1) +(sqrt3+1)i` is ________

Answer

To find the area of the triangle formed by the points \( A = \sqrt{3} + i \), \( B = -1 + \sqrt{3} i \), and \( C = (\sqrt{3} - 1) + (\sqrt{3} + 1)i \), we can follow these steps:

### Step 1: Identify the coordinates of the points
The points can be expressed in Cartesian coordinates as follows:
- \( A(\sqrt{3}, 1) \)
- \( B(-1, \sqrt{3}) \)
- \( C(\sqrt{3} - 1, \sqrt{3} + 1) \)

### Step 2: Calculate the lengths of the sides of the triangle
We will first find the length of side \( AB \).

Using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
where \( A(x_1, y_1) \) and \( B(x_2, y_2) \).

Substituting the coordinates of points \( A \) and \( B \):
\[ AB = \sqrt{(-1 - \sqrt{3})^2 + (\sqrt{3} - 1)^2} \]
Calculating the squares:
\[ = \sqrt{(-1 - \sqrt{3})^2 + (\sqrt{3} - 1)^2} = \sqrt{(1 + 2\sqrt{3} + 3) + (3 - 2\sqrt{3} + 1)} = \sqrt{4 + 2\sqrt{3} + 4 - 2\sqrt{3}} = \sqrt{8} = 2\sqrt{2} \]

### Step 3: Find the midpoint of \( AB \)
The midpoint \( D \) of \( AB \) is given by:
\[ D\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Substituting the coordinates of points \( A \) and \( B \):
\[ D\left(\frac{\sqrt{3} - 1}{2}, \frac{1 + \sqrt{3}}{2}\right) \]

### Step 4: Calculate the height from point \( C \) to line \( AB \)
To find the height \( CD \), we can use the distance formula from point \( C \) to line \( AB \). The coordinates of point \( C \) are:
\[ C(\sqrt{3} - 1, \sqrt{3} + 1) \]

We need to find the perpendicular distance from point \( C \) to line \( AB \). The line \( AB \) can be represented in the form \( Ax + By + C = 0 \).

First, we find the slope of \( AB \):
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\sqrt{3} - 1}{-1 - \sqrt{3}} \]
The equation of line \( AB \) can be derived from the slope-intercept form.

### Step 5: Calculate the area of the triangle
The area \( A \) of triangle \( ABC \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Substituting the values we found:
\[ \text{Area} = \frac{1}{2} \times AB \times CD \]
where \( AB = 2\sqrt{2} \) and \( CD = \sqrt{3} \).

Thus,
\[ \text{Area} = \frac{1}{2} \times 2\sqrt{2} \times \sqrt{3} = \sqrt{6} \]

### Final Answer
The area of the triangle formed by the points is \( \sqrt{6} \).

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