Let O be the origin, and P and Q be two points on the rectangular...

The equation of the chord of the hyperbola $xy = 12$ with midpoint $(x_1, y_1)$ is given by $T = S_1$.

$\dfrac{x y_1 + y x_1}{2} - 12 = x_1 y_1 - 12$

$x y_1 + y x_1 = 2 x_1 y_1$

Substituting the midpoint \left(\dfrac{1}{2}, -\dfrac{1}{2}\right)$:<br><br>$x\left(-\dfrac{1}{2}\right) + y\left(\dfrac{1}{2}\right) = 2\left(\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)$<br><br>$-x + y = -1 \Rightarrow y = x - 1$<br><br>To find the coordinates of P and Q, substitute$y = x - 1$into the equation of the hyperbola$xy = 12$:<br><br>$x(x - 1) = 12$<br><br>$x^2 - x - 12 = 0$<br><br>$(x - 4)(x + 3) = 0 \Rightarrow x = 4, -3$<br><br>For$x = 4$,$y = 3$.<br><br>For$x = -3$,$y = -4$.<br><br>Thus, the coordinates of P and Q are$(4, 3)$and (-3, -4)$.

The area of triangle OPQ with vertices $(0,0)$, $(4,3)$, and $(-3,-4)$ is:

$\text{Area} = \dfrac{1}{2} |x_P y_Q - x_Q y_P|$

$\text{Area} = \dfrac{1}{2} |(4)(-4) - (-3)(3)|$

$\text{Area} = \dfrac{1}{2} |-16 + 9| = \dfrac{7}{2}$

Answer: $\dfrac{7}{2}$