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

Hi there!
Reflections across the line y = -x always go by the rule (-y, -x). We can use this rule to get our answer here. We are given the aftermath of the reflection coordinates, which are A'(-1, 1), B'(-2, -1), and C'(-1, 0). All we have to do now is switch up the coordinate values and multiply them by -1. Here is the work - 
A'(-1, 1) => (1, -1) => x -1 => A(-1, 1)
B'(-2, -1) => (-1, -2) => x -1 => B(1, 2)
C'(-1, 0) => (0, -1) => x -1 => C(0, 1)
Therefore, the coordinates of Triangle ABC are A(-1, 1); B(1, 2); C(0,1). Hope this helped and have a phenomenal day!