To determine the relationship between the points \((-2, -1)\), \((1, 0)\), \((4, 3)\), and \((1, 2)\), we will check if they are collinear and if they form a parallelogram.
### Step-by-step Solution:
1. **Identify the Points**:
We have the points:
- \(A(-2, -1)\)
- \(B(1, 0)\)
- \(C(4, 3)\)
- \(D(1, 2)\)
2. **Check for Collinearity**:
To check if the points are collinear, we can use the area method. The area of the triangle formed by three points \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
If the area is zero, the points are collinear.
Let's calculate the area for points \(A\), \(B\), and \(C\):
\[
\text{Area} = \frac{1}{2} \left| -2(0 - 3) + 1(3 - (-1)) + 4(-1 - 0) \right|
\]
\[
= \frac{1}{2} \left| -2(-3) + 1(4) + 4(-1) \right|
\]
\[
= \frac{1}{2} \left| 6 + 4 - 4 \right| = \frac{1}{2} \left| 6 \right| = 3
\]
Since the area is not zero, points \(A\), \(B\), and \(C\) are not collinear.
3. **Check if they form a Parallelogram**:
To check if the points form a parallelogram, we need to find the midpoints of the diagonals \(AC\) and \(BD\) and see if they are the same.
- **Midpoint of \(AC\)**:
\[
M_{AC} = \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right) = \left( \frac{-2 + 4}{2}, \frac{-1 + 3}{2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1)
\]
- **Midpoint of \(BD\)**:
\[
M_{BD} = \left( \frac{x_2 + x_4}{2}, \frac{y_2 + y_4}{2} \right) = \left( \frac{1 + 1}{2}, \frac{0 + 2}{2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1)
\]
4. **Conclusion**:
Since the midpoints of the diagonals \(AC\) and \(BD\) are the same, the points form a parallelogram.
### Final Answer:
The points \((-2, -1)\), \((1, 0)\), \((4, 3)\), and \((1, 2)\) are the vertices of a parallelogram.
