Triangle `OA_(1)B_(1)` is the reflection of triangle OAB in origin, where `A_(1) (4, -5)` is the image of A and `B_(1) (-7, 0)` is the image of B.
Give a special name to the qudrilateral `ABA_(1)B_(1)`. Give reason.

Answer

To solve the problem, we need to analyze the given points and determine the properties of the quadrilateral \( ABA_1B_1 \).

### Step-by-Step Solution:

1. **Identify the Points:**
- Given points are:
- \( A_1(4, -5) \)
- \( B_1(-7, 0) \)
- Since \( A_1 \) and \( B_1 \) are reflections of points \( A \) and \( B \) in the origin, we can find \( A \) and \( B \) as follows:
- \( A(-4, 5) \) (since reflection in the origin changes the sign of both coordinates)
- \( B(7, 0) \)

2. **Plot the Points:**
- Plot the points \( A(-4, 5) \), \( B(7, 0) \), \( A_1(4, -5) \), and \( B_1(-7, 0) \) on a Cartesian plane.

3. **Connect the Points to Form the Quadrilateral:**
- Connect the points \( A \), \( B \), \( A_1 \), and \( B_1 \) to form the quadrilateral \( ABA_1B_1 \).

4. **Determine the Lengths of the Sides:**
- Use the distance formula to find the lengths of the sides:
- Distance \( AB \):
\[ AB = \sqrt{(7 - (-4))^2 + (0 - 5)^2} = \sqrt{(7 + 4)^2 + (0 - 5)^2} = \sqrt{11^2 + (-5)^2} = \sqrt{121 + 25} = \sqrt{146} \]
- Distance \( A_1B_1 \):
\[ A_1B_1 = \sqrt{(-7 - 4)^2 + (0 - (-5))^2} = \sqrt{(-11)^2 + (5)^2} = \sqrt{121 + 25} = \sqrt{146} \]
- Distance \( AB_1 \):
\[ AB_1 = \sqrt{(-7 - (-4))^2 + (0 - 5)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \]
- Distance \( A_1B \):
\[ A_1B = \sqrt{(7 - 4)^2 + (0 - (-5))^2} = \sqrt{(3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34} \]

5. **Check for Parallelogram Properties:**
- We observe that:
- \( AB = A_1B_1 \) (both equal to \( \sqrt{146} \))
- \( AB_1 = A_1B \) (both equal to \( \sqrt{34} \))
- Since both pairs of opposite sides are equal, \( ABA_1B_1 \) is a parallelogram.

6. **Conclusion:**
- The quadrilateral \( ABA_1B_1 \) is a **parallelogram** because opposite sides are equal in length.