Denote the abscissas of the points $A$, $B$, $C$ and $D$ by $a$, $b$, $c$ and $d$ respectively. The perpendicularity of the lines $AB$ and $CD$ is equivalent to $acbd = -1/4$ and the fact that the point $F$ belongs to the lines $AC$ and $BD$ is equivalent to the equalities $a+c = 2ac+1$ and $b+d = 2bd+1$. Denote $ac = p$ and $bd = q$, then the numbers $a$ and $c$ are the roots of the quadratic equation $x^2 - (2p+1)x + p = 0$ and the numbers $b$ and $d$ are the roots of the quadratic equation $x^2 - (2q+1)x + q = 0$. Hence $c - a = \sqrt{4p^2 + 1}$ and $d - b = \sqrt{4q^2 + 1}$. Note that
$$\frac{1}{2a} - \frac{1}{2c} = \frac{c-a}{2ac} = \frac{\sqrt{4p^2+1}}{2p} = -2q \cdot \sqrt{\frac{1}{4q^2} + 1} = \sqrt{4q^2+1} = d-b.$$
Similarly, $\frac{1}{2d} - \frac{1}{2b} = c-a$. Thus the projections of the segment $AC$ on the abscissa and ordinate axes are equal, respectively, to the projections of the segment $BD$ on the ordinate and abscissa axes, hence $AC = BD$. Since the diagonals $AC$ and $BD$ of the quadrilateral $ABCD$ are perpendicular, its area $S$ is equal to half the product of the diagonals. Therefore
$$S = \frac{1}{2} \cdot AC \cdot BD = \frac{1}{2} AC^2 = \frac{1}{2}((c-a)^2 + (b-d)^2) = \frac{1}{2}(4p^2 + 4q^2 + 2).$$