1. As shown in Figure 4, in the Cartesian coordinate system, $O$ is the origin, the diagonals of $\square A B O C$ intersect at point $M$, and the hyperbola $y=\frac{k}{x}(x<0)$ passes through points $B$ and $M$. If the area of $\square A B O C$ is 24, then $k=$ . $\qquad$
2, 1. -8.
Let $M\left(\frac{k}{y}, y\right)$. Then $B\left(\frac{k}{2 y}, 2 y\right), C\left(\frac{3 k}{2 y}, 0\right)$.
From $S_{\text {OABOC }}=4 S_{\triangle O C M}=2\left|x_{c} y_{M}\right|$
$\Rightarrow 24=2\left|\frac{3 k}{2}\right| \Rightarrow|k|=8 \Rightarrow k=-8$.