Without loss of generality we can assume that the positions of all hyperbolae, lines, and points look like in the figure (otherwise we can rotate the plane by the angle which is a multiple of $90^\circ$, and rename the points).

Let $A(a; 1/a)$, $B(b; 1/b)$, $C(c; -1/c)$, $D(d; -1/d)$. Note that all numbers $a$, $b$, $c$, $d$ are pairwise distinct and $c < 0$, $d > a > b > 0$. We write the equations of the line $l_{AB}$ and $l_{CD}$ passing through the pairs of points $A, B$ and $C, D$:
$$l_{AB}: \quad \frac{x-a}{b-a} = \frac{y-1/a}{1/b-1/a} \Leftrightarrow y = -\frac{1}{ab}x + \frac{1}{a} + \frac{1}{b},$$

$$l_{CD}: \frac{x-c}{d-c} = \frac{y+1/c}{-1/d+1/c} \Leftrightarrow y = \frac{1}{cd}x - \frac{1}{c} - \frac{1}{d}$$
Since these lines coincide, $l = l_{AB} = l_{CD}$, we have
$$ab = -cd, \quad \frac{1}{a} + \frac{1}{b} = -\frac{1}{c} - \frac{1}{d} \Rightarrow a+b=c+d. \qquad (1)$$
The abscissa of the midpoint $M$ of the segment $AB$ is equal to $x_M = (a+b)/2$, and from (1) it follows that it is equal to the abscissa $(c+d)/2$ of the midpoint of the segment $CD$. Since the points $A$, $B$, $C$, $D$ lie on the same line, we see that $M$ is the midpoint of the segments $AB$ and $CD$. So $AC = CM + MA = MD + MB = BD$. Since the lengths of the altitudes of the triangles $OAC$ and $OBD$ from the vertex $O$ are equal, we see that the area of these triangles are equal.