Reflexion über die Linie y=x

Problem

Reflect triangle $ABC$ across the line $y = x$, where $A = (1,2)$, $B = (4,2)$, and $C = (3,5)$, then identify the single transformation rule that maps $ABC$ to $A'B'C'$.

Step 1: Swap the coordinates

Reflection across the line $y = x$ swaps the $x$- and $y$-coordinates of each vertex. So

$$A(1,2) \to A'(2,1), \quad B(4,2) \to B'(2,4), \quad C(3,5) \to C'(5,3).$$

The reflected triangle has vertices $A' = (2,1)$, $B' = (2,4)$, and $C' = (5,3)$.

Step 2: Check the reflected side lengths

The image matches the original because corresponding side lengths agree. For $AB$,

$$AB = \sqrt{(4-1)^2 + (2-2)^2} = 3,$$

and $A'B'$ is also $3$. Likewise,

$$BC = \sqrt{(3-4)^2 + (5-2)^2} = \sqrt{10},$$

and $B'C' = \sqrt{10}$. Also,

$$CA = \sqrt{(1-3)^2 + (2-5)^2} = \sqrt{13},$$

and $C'A' = \sqrt{13}$.

Step 3: State the transformation rule

Each point and its image lie on opposite sides of $y = x$ at equal distance, and the midpoint of each segment joining a point to its image lies on the line $y = x$. That confirms the rule is reflection across $y = x$, or equivalently, $x$ and $y$ are swapped.

Answer

The reflected vertices are $A' = (2,1)$, $B' = (2,4)$, and $C' = (5,3)$, and the transformation rule is $(x,y) \mapsto (y,x)$.