2. Given the parabola $y=\sqrt{2} x^{2}$ intersects with the lines $y=1, y=2, y=3$ to form three line segments. The area of the triangle formed by these three line segments is $\qquad$ .
2. 2 .

The intersection points of the lines $y=1, y=2, y=3$ with the parabola $y=\sqrt{2} x^{2}$ are all symmetric about the $y$-axis, so the lengths of the three line segments are $2 x_{1}, 2 x_{2}, 2 x_{3}$, where $x_{1}, x_{2}, x_{3}$ are the positive roots of the equations $\sqrt{2} x_{1}^{2}=1, \sqrt{2} x_{2}^{2}=2, \sqrt{2} x_{3}^{2}=3$ respectively.
Therefore, $x_{1}^{2}=\frac{\sqrt{2}}{2}, x_{2}^{2}=\sqrt{2}, x_{3}^{2}=\frac{3 \sqrt{2}}{2}$.
The squares of the lengths of the three sides are
$$
\left(2 x_{1}\right)^{2}=2 \sqrt{2},\left(2 x_{2}\right)^{2}=4 \sqrt{2},\left(2 x_{3}\right)^{2}=6 \sqrt{2} \text {, }
$$

Thus, the triangle formed by the three line segments of lengths $2 x_{1}, 2 x_{2}, 2 x_{3}$ is a right triangle.
Let the area of this right triangle be $S$. Then
$$
\begin{array}{l}
S^{2}=\left(\frac{1}{2} \times 2 x_{1} \times 2 x_{2}\right)^{2}=\frac{2 \sqrt{2} \times 4 \sqrt{2}}{4}=4 \\
\Rightarrow S=2 .
\end{array}
$$