Question 1210196
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Let's break down this problem.

**1. Lattice Points**

The given lattice points are:

* (0, 0)
* (1, 0)
* (0, 1)
* (1, 1)

**2. Squares with at Least Two Points as Vertices**

We need to count the squares that can be formed using at least two of these points as vertices.

* **Squares with All Four Points as Vertices:**
    * There is only 1 such square, the one formed by the four given points.

* **Squares with Two Points as Vertices:**

    * **Squares with Sides Parallel to Axes:**
        * The square with side length 1 is the only such square we have already counted.

    * **Tilted Squares:**
        * Consider the segment connecting (0, 0) and (1, 1). This is a diagonal of a square.
        * Consider the segment connecting (1, 0) and (0, 1). This is also a diagonal of a square.
        * These diagonals form a square with area 2.
        * The vertices of this square are:
            * (0, 0), (1, 1), (0, 2), (-1, 1)
            * (0, 0), (1, 1), (2, 0), (1, -1)
            * (0, 1), (1, 0), (2, 1), (1, 2)
            * (0, 1), (1, 0), (-1, 0), (0, -1)
        * We are only concerned with the squares that have at least 2 points from the lattice as vertices.
        * The square with the diagonal from (0,0) to (1,1) has (0,0) and (1,1) as vertices.
        * The square with the diagonal from (1,0) to (0,1) has (1,0) and (0,1) as vertices.
        * The square with the diagonal from (1,1) to (0,0) has (1,1) and (0,0) as vertices.
        * The square with the diagonal from (0,1) to (1,0) has (0,1) and (1,0) as vertices.
        * Thus we have one square with at least 2 vertices.

**3. Counting All Squares**

* **1 Square** with all four points as vertices.
* **1 Square** with two points as vertices and is tilted.

**Total Squares**

* 1 + 1 = 2

**Therefore, there are 2 squares in the plane that have at least two points in the lattice as vertices.**