Name the type of quadrilateral formed, if any, by the following points and give reasona for your Solution :
(-1,-2), (1,0), (-1,2), (-3,0)

Answer

To determine the type of quadrilateral formed by the points (-1, -2), (1, 0), (-1, 2), and (-3, 0), we will follow these steps:

### Step 1: Identify the Points
We have the following points:
- A (-1, -2)
- B (1, 0)
- C (-1, 2)
- D (-3, 0)

### Step 2: Use the Distance Formula
We will use the distance formula to find the lengths of the sides of the quadrilateral. The distance formula between two points (x1, y1) and (x2, y2) is given by:
\[ d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \]

### Step 3: Calculate the Lengths of the Sides
1. **Calculate AB:**
\[ AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{(1 + 1)^2 + (0 + 2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

2. **Calculate BC:**
\[ BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

3. **Calculate CD:**
\[ CD = \sqrt{(-3 - (-1))^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

4. **Calculate DA:**
\[ DA = \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 4: Check the Lengths of the Sides
We find that:
- AB = 2√2
- BC = 2√2
- CD = 2√2
- DA = 2√2

All sides are equal, which suggests that the quadrilateral could be a rhombus or a square.

### Step 5: Calculate the Lengths of the Diagonals
Next, we will calculate the lengths of the diagonals AC and BD.

1. **Calculate AC:**
\[ AC = \sqrt{(-1 - (-1))^2 + (2 - (-2))^2} = \sqrt{(0)^2 + (4)^2} = \sqrt{16} = 4 \]

2. **Calculate BD:**
\[ BD = \sqrt{(1 - (-3))^2 + (0 - 0)^2} = \sqrt{(4)^2 + (0)^2} = \sqrt{16} = 4 \]

### Step 6: Check the Lengths of the Diagonals
We find that:
- AC = 4
- BD = 4

Both diagonals are equal.

### Conclusion
Since all sides are equal and both diagonals are equal, the quadrilateral formed by the points (-1, -2), (1, 0), (-1, 2), and (-3, 0) is a **square**.