1. (17 points) Find the area of the triangle cut off by the line $y=3 x+1$ from the figure defined by the inequality $|x-1|+|y-2| \leq 2$.
Answer: 2.

Solution. The figure defined by the given inequality is a square.

![](https://cdn.mathpix.com/cropped/2024_05_06_0785b6f179aec901e215g-1.jpg?height=459&width=549&top_left_y=665&top_left_x=845)

The side of the square is $2 \sqrt{2}$ (this can be found using the Pythagorean theorem). The given line passes through the vertex of the square and cuts off a triangle with the smaller leg equal to $\sqrt{2}$, and the larger leg - the side of the square. Therefore, the area of the cut-off right triangle is $S=\frac{1}{2} \cdot \sqrt{2} \cdot 2 \sqrt{2}=2$ square units.