The points (2 , -2) , (8 , 4) , (4 , 6) and (-1 , 1) in order are the vertices of which one of the following quadrilaterals ?

Options

A. Square
B. Rhombus
C. Rectangle (but not square)
D. Trapezium

Answer

To determine the type of quadrilateral formed by the points (2, -2), (8, 4), (4, 6), and (-1, 1), we will follow these steps:

### Step 1: Plot the Points
First, we need to plot the given points on a coordinate plane.

- Point A (2, -2)
- Point B (8, 4)
- Point C (4, 6)
- Point D (-1, 1)

### Step 2: Connect the Points
Next, we connect the points in the order they are given: A → B → C → D → A.

### Step 3: Calculate the Lengths of the Sides
We will calculate the lengths of the sides of the quadrilateral using the distance formula, which is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

1. **Length of AB**:
\[ AB = \sqrt{(8 - 2)^2 + (4 - (-2))^2} = \sqrt{(6)^2 + (6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \]

2. **Length of BC**:
\[ BC = \sqrt{(4 - 8)^2 + (6 - 4)^2} = \sqrt{(-4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]

3. **Length of CD**:
\[ CD = \sqrt{(-1 - 4)^2 + (1 - 6)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \]

4. **Length of DA**:
\[ DA = \sqrt{(2 - (-1))^2 + (-2 - 1)^2} = \sqrt{(3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \]

### Step 4: Analyze the Sides
Now we have the lengths of the sides:
- AB = \(6\sqrt{2}\)
- BC = \(2\sqrt{5}\)
- CD = \(5\sqrt{2}\)
- DA = \(3\sqrt{2}\)

### Step 5: Determine the Type of Quadrilateral
To classify the quadrilateral, we can observe the lengths of the sides and the angles formed. Since opposite sides are not equal and the lengths vary significantly, we can conclude that the quadrilateral is a trapezium (trapezoid).

### Conclusion
The points (2, -2), (8, 4), (4, 6), and (-1, 1) form a trapezium.

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