2. In $\triangle A B C$, it is known that the angle bisector of $\angle B$ intersects $A C$ at point $K$. If $B C=2, C K=1, B K=\frac{3 \sqrt{2}}{2}$, then the area of $\triangle A B C$ is
2. $\frac{15 \sqrt{7}}{16}$.

As shown in Figure 1, let $A C = b, A B = c$. Then, by the cosine rule, we have
$$
\begin{array}{l}
8 + 2 b^{2} - 2 c^{2} \\
= b .
\end{array}
$$

Also, $\frac{A K}{C K} = \frac{A B}{B C}$, so
$$
\frac{b-1}{1} = \frac{c}{2} \text{. }
$$

From equations (1) and (2), we know $b = \frac{5}{2}, c = 3$.
Given $\cos C = \frac{1}{8}$, we get $\sin C = \frac{3 \sqrt{7}}{8}$.
Therefore, $S_{\triangle A B C} = \frac{1}{2} a b \sin C = \frac{15 \sqrt{7}}{16}$.