Crystal is amazed! She graphed ΔABC using the points A(5, –1), B(3, –7), and C(6, –2). Then she rotated ΔABC 90° counterclockwise (↺) about the origin to find ΔA′B′C′. Meanwhile, her teammate took a different triangle (ΔTUV) and rotated it 90° clockwise (↻) about the origin to find ΔT′U′V′. Amazingly, ΔA′B′C′ and ΔT′U′V′ ended up having exactly the same points! Name the coordinates of the vertices of ΔTUV.

On the coordinate plane, consider the point [tex](x,y)[/tex]. To rotate this point by 90° around the origin in counterclockwise direction, you can always swap the x- and y-coordinates and then multiply the new x-coordinate by -1. In a mathematical language this is as follows:

[tex](x,y)\rightarrow(-y,x)[/tex]

Therefore, for ΔA′B′C′ we have:

[tex]\boxed{A'(1, 5)} \\ \\ \boxed{B'(7, 3)} \\ \\ \boxed{C'(2, 6)}[/tex]

Since ΔA′B′C′ and ΔT′U′V′ ended up having exactly the same points, then:

[tex]T'(1, 5) \\ \\U'(7, 3) \\ \\V'(2, 6)[/tex]

On the other hand, in clockwise direction we have the following rule:

[tex](x,y)\rightarrow(y,-x)[/tex]

Therefore. we must find [tex](x,y)[/tex] to get ΔTUV here, so:

[tex]T'(y,-x)=T'(1, 5) \therefore x=-5 \ and \ y=1 \rightarrow T(-5,1)[/tex]

[tex]T'(y,-x)=T'(1, 5) \therefore x=-5 \ and \ y=1 \rightarrow \boxed{T(-5,1)} \\ \\U'(y,-x)=U'(7, 3) \therefore x=-3 \ and \ y=7 \rightarrow \boxed{U(-3,7)} \\ \\V'(y,-x)=V'(2, 6) \therefore x=-6 \ and \ y=2 \rightarrow \boxed{V(-6,2)}[/tex]