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**Hint:**Orthocenter of a triangle is defined as the point where perpendiculars drawn from the vertices to the opposite side will intersect each other. For a right-angled triangle, the orthocenter lies at the vertex containing the right angle.

**Step By Step Solution:**

Let us assume the points as

A = $\left( {{x_1},{y_1}} \right)$ = $\left( {2,\dfrac{{\sqrt 3 - 1}}{2}} \right)$

B = $\left( {{x_2},{y_2}} \right)$ = $\left( {\dfrac{1}{2},\dfrac{{ - 1}}{2}} \right)$

C = $\left( {{x_3},{y_3}} \right)$ = $\left( {2,\dfrac{{ - 1}}{2}} \right)$

Firstly, we calculate the slope of the line AC using the formula,

Slope of AC = $\dfrac{{{y_3} - {y_1}}}{{{x_3} - {x_1}}}$

$\Rightarrow$ Slope of AC = $\dfrac{{\dfrac{{ - 1}}{2} - \left( {\dfrac{{\sqrt 3 - 1}}{2}} \right)}}{{2 - 2}}$

$\Rightarrow$ Slope of AC = $\infty$

As slope of the line is infinity, the line AC will be perpendicular to the X-axis

Then we calculate the slope of the line BC using the formula

Slope of BC = $\dfrac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}}$

$\Rightarrow$ Slope of BC = $\dfrac{{\dfrac{{ - 1}}{2} - \left( {\dfrac{{ - 1}}{2}} \right)}}{{2 - \dfrac{1}{2}}}$

$\Rightarrow$ Slope of BC = 0

As the slope of the line is equal to zero, the line BC will be parallel to X-axis.

Therefore $\Delta ABC$ will be a right angle triangle with $C = 90^\circ$

For a right-angled triangle, the orthocenter will be the vertex having an angle of 90 degrees.

$\therefore$ The point $\left( {2,\dfrac{{ - 1}}{2}} \right)$ will be the orthocenter of the

triangle formed by the given vertices.

**Therefore the correct answer is Option (B)**

**Note:**

After finding the lines with slope equal to infinity and slope equal to zero, the common point from both the lines should be selected as the vertex containing the right angle. In the above calculation, the slope of AC is infinite and the slope of BC is zero. So point C is selected as a vertex having the right angle.

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