Let $$O$$ be the origin, and $$P$$ and $$Q$$ be two points on the rectangular hyperbola $$xy = 12$$ such that the mid point of the line segment $$PQ$$ is $$\left(\frac{1}{2}, -\frac{1}{2}\right)$$. Then the area of the triangle $$OPQ$$ equals :

The rectangular hyperbola is $$xy = 12$$.
Let $$P(x_1 , y_1)$$ and $$Q(x_2 , y_2)$$ lie on it, so

$$x_1y_1 = 12, \qquad x_2y_2 = 12$$ $$-(1)$$

The midpoint of $$PQ$$ is given as $$\left(\tfrac{1}{2}, -\tfrac{1}{2}\right)$$, hence

$$\frac{x_1+x_2}{2} = \frac12 \Longrightarrow x_1 + x_2 = 1$$
$$\frac{y_1+y_2}{2} = -\frac12 \Longrightarrow y_1 + y_2 = -1$$ $$-(2)$$

The area of $$\triangle OPQ$$ is

$$\text{Area} = \frac12\,\left|x_1y_2 - x_2y_1\right|$$ $$-(3)$$

To obtain $$x_1y_2 - x_2y_1$$, first evaluate the sum $$x_1y_2 + x_2y_1$$. Multiply the two equalities in $$(2)$$:

$$(x_1+x_2)(y_1+y_2)=1\cdot(-1)=-1$$

Expanding the left side:

$$(x_1y_1 + x_1y_2 + x_2y_1 + x_2y_2)= -1$$

Using $$(1)$$, $$x_1y_1 + x_2y_2 = 12 + 12 = 24$$, hence

$$24 + (x_1y_2 + x_2y_1) = -1 \Longrightarrow x_1y_2 + x_2y_1 = -25$$ $$-(4)$$

Now introduce the convenient parametrisation of the hyperbola: take $$P( t,\; \tfrac{12}{t})$$ and $$Q( s,\; \tfrac{12}{s})$$.

From $$(2)$$:

$$t+s = 1$$
$$\frac{12}{t} + \frac{12}{s} = -1 \;\Longrightarrow\; 12\Bigl(\frac{s+t}{st}\Bigr) = -1$$
Since $$s+t = 1$$, this gives $$\displaystyle \frac{12}{st} = -1 \;\Longrightarrow\; st = -12$$ $$-(5)$$

Thus $$t$$ and $$s$$ are the roots of $$u^2 - (s+t)u + st = 0$$:

$$u^2 - u - 12 = 0 \;\Longrightarrow\; (u-4)(u+3)=0$$

Therefore $$\{t,s\} = \{4,\,-3\}$$.

Choose $$P(4,3),\quad Q(-3,-4)$$ (the other assignment only swaps the labels and does not change the area).

Evaluate the determinant in $$(3)$$:

$$x_1y_2 = 4\cdot(-4) = -16,\qquad x_2y_1 = (-3)\cdot3 = -9$$
$$x_1y_2 - x_2y_1 = -16 - (-9) = -7$$

Hence

$$\text{Area} = \frac12 \bigl| -7 \bigr| = \frac{7}{2}$$

Option C which is: $$\frac{7}{2}$$