The orthocentre of triangle with vertices `(2,(sqrt3-1)/2) , (1/2,-1/2) , (2,,-1/2)`

Options

A. `((3)/(2),(sqrt(3)-2)/(6))`
B. `(2,-(1)/(2))`
C. `((5)/(4),((sqrt(3)-2))/(4))`
D. ((1)/(2),-(1)/(2))`

Answer

To find the orthocenter of a triangle with given vertices, we can follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of the triangle are given as:
- \( A(2, \frac{\sqrt{3}-1}{2}) \)
- \( B(\frac{1}{2}, -\frac{1}{2}) \)
- \( C(2, -\frac{1}{2}) \)

### Step 2: Determine the slopes of the sides of the triangle
To find the orthocenter, we first need to find the slopes of the sides of the triangle.

1. **Slope of line AB**:
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-\frac{1}{2} - \frac{\sqrt{3}-1}{2}}{\frac{1}{2} - 2} = \frac{-\frac{1 + \sqrt{3} - 1}{2}}{-\frac{3}{2}} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{3}{2}} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \]

2. **Slope of line AC**:
\[ m_{AC} = \frac{y_3 - y_1}{x_3 - x_1} = \frac{-\frac{1}{2} - \frac{\sqrt{3}-1}{2}}{2 - 2} = \text{undefined (vertical line)} \]

3. **Slope of line BC**:
\[ m_{BC} = \frac{y_3 - y_2}{x_3 - x_2} = \frac{-\frac{1}{2} - (-\frac{1}{2})}{2 - \frac{1}{2}} = \frac{0}{\frac{3}{2}} = 0 \]

### Step 3: Find the equations of the altitudes
The orthocenter is the intersection of the altitudes.

1. **Altitude from A** (perpendicular to BC):
Since BC is horizontal (slope = 0), the altitude from A will be a vertical line through A:
\[ x = 2 \]

2. **Altitude from B** (perpendicular to AC):
Since AC is vertical (undefined slope), the slope of the altitude from B will be horizontal (slope = 0):
\[ y = -\frac{1}{2} \]

### Step 4: Find the intersection of the altitudes
To find the orthocenter, we need to find the intersection of the lines:
1. From altitude at A: \( x = 2 \)
2. From altitude at B: \( y = -\frac{1}{2} \)

Substituting \( x = 2 \) into the equation for the altitude from B:
\[ y = -\frac{1}{2} \]

Thus, the orthocenter \( H \) is at the point:
\[ H(2, -\frac{1}{2}) \]

### Conclusion
The orthocenter of the triangle with vertices \( (2, \frac{\sqrt{3}-1}{2}), (\frac{1}{2}, -\frac{1}{2}), (2, -\frac{1}{2}) \) is:
\[ \boxed{(2, -\frac{1}{2})} \]