A(-2, 4) and B(-4, 2) are reflected in the y-axis. If A and B' are images of A and B respectively.
Assign a special name to quadrilateral AA' B'B

Answer

To solve the problem step by step, we will reflect the points A and B across the y-axis and then analyze the resulting quadrilateral.

### Step 1: Identify the coordinates of points A and B
- Point A is given as A(-2, 4).
- Point B is given as B(-4, 2).

### Step 2: Reflect point A across the y-axis
- The reflection of a point (x, y) across the y-axis is given by (-x, y).
- Therefore, the reflection of point A(-2, 4) will be:
\[ A' = (2, 4) \]

### Step 3: Reflect point B across the y-axis
- Similarly, reflecting point B(-4, 2) across the y-axis gives:
\[ B' = (4, 2) \]

### Step 4: Identify the coordinates of the reflected points
- Now we have:
- A' = (2, 4)
- B' = (4, 2)

### Step 5: Determine the vertices of quadrilateral AAB'B
- The vertices of quadrilateral AAB'B are:
- A(-2, 4)
- A'(2, 4)
- B'(4, 2)
- B(-4, 2)

### Step 6: Analyze the shape of quadrilateral AAB'B
- To determine the type of quadrilateral formed, we can analyze the lengths of the sides and the slopes.
- The lengths of sides AB and A'B' can be calculated using the distance formula:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
- Calculate the lengths:
- Length of AB:
\[ AB = \sqrt{((-4) - (-2))^2 + (2 - 4)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]
- Length of A'B':
\[ A'B' = \sqrt{(4 - 2)^2 + (2 - 4)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 7: Check the slopes of the sides
- The slopes of AB and A'B' are:
- Slope of AB:
\[ \text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 4}{-4 - (-2)} = \frac{-2}{-2} = 1 \]
- Slope of A'B':
\[ \text{Slope} = \frac{2 - 4}{4 - 2} = \frac{-2}{2} = -1 \]

### Step 8: Conclusion
- Since AB is parallel to A'B' and both pairs of opposite sides are equal in length, quadrilateral AAB'B is an **isosceles trapezium**.

### Final Answer
The special name assigned to quadrilateral AAB'B is **isosceles trapezium**.