Question

On a coordinate plane, a triangle has points J (negative 3, 9), L (negative 3, 2), and K (5, 2).
Consider reflections of ΔJKL.

What line of reflection maps point K to point K' at (–5, 2)?





What line of reflection maps point L to point L' at (–2, 3)?

Answer 1

1) The y-axis, x = 0.
K(5,2) and K'(−5,2) have midpoint (0,2); the perpendicular bisector is the vertical line x = 0, which reflects (5,2) to (−5,2).

2) The line y = −x.
L(−3,2) and L'(−2,3) have midpoint (−2.5,2.5). The segment has slope 1, so the perpendicular bisector has slope −1 and through (−2.5,2.5) gives y = −x (also (x,y) → (−y,−x) maps (−3,2) to (−2,3)).