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AABC is reflected about the line y=−x to give AA'B'C with vertices A(-1, 1), B'(-2,-1), C(-1, 0). What are the vertices of AABC?

User Rgoal
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1 Answer

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Final answer:

The original vertices of ▲ABC can be found by swapping and changing the signs of the coordinates of the reflected triangle ▲A'B'C'. The vertices are A(1, -1), B(1, 2), and C(-1, 0).

Step-by-step explanation:

You have been given the coordinates of one vertex of ▲ABC after reflection and two vertices of the reflected triangle ▲A'B'C'. The reflection is about the line y = -x. When a point is reflected across the line y = -x, the x-coordinate and y-coordinate of that point are swapped and their signs are changed. Therefore, to find the original vertices of ▲ABC, we need to swap and change the signs of the coordinates of ▲A'B'C'.

Let's find the original coordinates of point B since we are given B'(-2,-1). After swapping and changing the signs, B would be at (1, 2). For C, we are given C(-1, 0), and after the reflection, it does not change its position because it lies on the line y = -x itself. Hence, the original C is at (-1, 0). Now, given A' is at (-1, 1) after reflection, the original A would be at (1, -1).

Therefore, the vertices of ▲ABC are A(1, -1), B(1, 2), and C(-1, 0).

User WPFUser
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