Axial Symmetry

Let $\ell$ be a fixed line in the plane. The reflection across $\ell$ (or axial symmetry with axis $\ell$) is the transformation that maps each point $A$ to the point $A'$ such that:

• If $A$ lies on $\ell$, then $A' = A$ (points on $\ell$ are fixed).
• If $A$ does not lie on $\ell$, then $\ell$ is the perpendicular bisector of segment $AA'$, i.e.\ $\ell \perp AA'$ and the foot of the perpendicular from $A$ to $\ell$ is the midpoint of $AA'$.

Let $\sigma_\ell$ denote the reflection across line $\ell$. Then:

1. Isometry: $|A'B'| = |AB|$ for all $A, B$.
2. Opposite isometry: $\sigma_\ell$ reverses orientation.
3. Involution: $\sigma_\ell \circ \sigma_\ell = \text{id}$ (applying the reflection twice returns every point to itself).
4. Lines perpendicular to $\ell$ map to themselves; lines parallel to $\ell$ map to other lines parallel to $\ell$.
5. Composition of two reflections: if $\ell_1 \parallel \ell_2$ at distance $d$, then $\sigma_{\ell_2} \circ \sigma_{\ell_1}$ is a translation by $2d$; if $\ell_1$ and $\ell_2$ intersect at angle $\varphi$, then $\sigma_{\ell_2} \circ \sigma_{\ell_1}$ is a rotation by $2\varphi$ about their intersection.

Proof of (1). Let $A(x_1,y_1)$, $B(x_2,y_2)$, reflected across the $x$-axis: $A'(x_1,-y_1)$, $B'(x_2,-y_2)$. $|A'B'| = \sqrt{(x_2-x_1)^2+(-y_2+y_1)^2} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} = |AB|. \quad \blacktriangleleft$

A line $\ell$ is called an axis of symmetry of a figure $F$ if the reflection of $F$ across $\ell$ coincides with $F$ itself: $\sigma_\ell(F) = F$.

Examples:

• A rectangle (non-square) has exactly 2 axes of symmetry (through midpoints of opposite sides).
• A square has 4 axes of symmetry.
• A regular $n$-gon has $n$ axes of symmetry.
• A circle has infinitely many axes of symmetry (any diameter).
• A scalene triangle has no axis of symmetry.

Find the image of $A(1, 4)$ under reflection across the line $\ell: y = x + 1$.

Solution. The axis $\ell$ has slope $1$, so the perpendicular from $A$ has slope $-1$ and passes through $A(1,4)$: $y - 4 = -1(x - 1) \implies y = -x + 5.$

Find the foot $F$ (intersection of $\ell$ and the perpendicular): $x + 1 = -x + 5 \implies 2x = 4 \implies x = 2,\; y = 3. \quad F = (2, 3).$

Since $F$ is the midpoint of $AA'$: $A' = (2\cdot 2 - 1,\; 2\cdot 3 - 4) = (3,\; 2).$

Answer: $A'(3, 2)$. $\blacktriangleleft$

Points $P(1, 3)$ and $P'(5, 1)$ are symmetric about a line $\ell$. Find the equation of $\ell$.

Solution. Line $\ell$ is the perpendicular bisector of $PP'$.

Midpoint of $PP'$: $M = \left(\dfrac{1+5}{2},\, \dfrac{3+1}{2}\right) = (3, 2)$.

Slope of $PP'$: $k = \dfrac{1-3}{5-1} = -\dfrac{1}{2}$.

Slope of $\ell$ (perpendicular): $k_\ell = 2$.

Equation of $\ell$ through $M(3,2)$ with slope $2$: $y - 2 = 2(x - 3) \implies y = 2x - 4.$

Answer: The axis of symmetry is $y = 2x - 4$. $\blacktriangleleft$

Find the images of the vertices of triangle $ABC$ with $A(2,1)$, $B(5,1)$, $C(5,4)$ under reflection across each axis:

(a) the $x$-axis; (b) the $y$-axis; (c) the line $y = x$.

In each case state whether the image triangle has the same or opposite orientation to the original.

A figure $F$ is symmetric about both coordinate axes. Prove that $F$ is also symmetric about the origin (i.e.\ $F$ has a centre of symmetry at the origin).