Reflection

Reflections in coordinate axes and diagonals

Reflecting $P(3,-2)$ in the $y$-axis gives $(-3,-2)$. The horizontal coordinate changes sign because the perpendicular movement is horizontal. The vertical coordinate remains unchanged.

Reflecting the same point in $y=x$ gives $(-2,3)$. Its midpoint with the original is $(1/2,1/2)$, which lies on the mirror line. The joining segment has slope $-1$, perpendicular to the mirror’s slope $1$.

Mirrors that are not coordinate axes

For a vertical mirror $x=h$, preserve the vertical coordinate and place the horizontal coordinate symmetrically:

For a horizontal mirror $y=k$, the rule is $x’=x$, $y’=2k-y$.

Reflect $P(5,-1)$ in $x=2$. The point is $3$ units to the right, so the image is $3$ units to the left at $(-1,-1)$. Averaging their horizontal coordinates gives $2$, confirming the correct midpoint.

Reflect $Q(-3,7)$ in $y=4$. Its image is $(-3,1)$. Changing $7$ to $-7$ would reflect in the $x$-axis, not the stated line.

An oblique mirror: construct the perpendicular foot

Reflect $P(3,2)$ in the line $y=x+2$. The perpendicular through $P$ has slope $-1$ and equation $y=-x+5$. Intersect the two lines:

The perpendicular foot is $H(3/2,7/2)$. Because $H$ must be the midpoint of $PP’$, recover the image using $P’=2H-P$:

Both $PH$ and $HP’$ have length $3/\sqrt2$. This is the same perpendicular-foot calculation used in Distance from a Point to a Line, now extended to the opposite side of the mirror.

A general formula

For a line $Ax+By+C=0$ with $A^2+B^2>0$, set

The reflection of $(x_0,y_0)$ is

The vector $(A,B)$ is normal to the line, so the formula moves along the perpendicular by twice the signed displacement to the line. Here $d$ is an algebraic coefficient, not itself the Euclidean distance unless the normal vector is a unit vector.

What reflection preserves and reverses

Reflection preserves lengths, angles, perimeter, area, collinearity and parallelism. It reverses clockwise versus counterclockwise orientation. A reflected triangle is therefore congruent to its original, even though the vertex order around the boundary reverses.

Reflecting twice in the same line returns every point to its starting position. For the $y$-axis rule, $(-(-x),y)=(x,y)$. Reflection is its own inverse.

The mirror line is fixed point by point. Other lines may be carried onto themselves as sets without having all their points fixed; a line perpendicular to the mirror provides an example.

Two reflections can produce another transformation

Reflect in $x=0$, then in $x=3$. The first rule sends $(x,y)$ to $(-x,y)$; the second sends that image to $(6+x,y)$. The combination is a translation $6$ units right, twice the distance between the parallel mirrors.

Reversing the order gives a translation $6$ units left. For intersecting mirror lines, two reflections give a rotation about their intersection by twice the directed angle from the first mirror to the second. Reflecting first in the $x$-axis and then in $y=x$ sends $(x,y)$ to $(-y,x)$, a counterclockwise quarter-turn.

These combinations explain why repeated reflections can create structured patterns without distorting the original tile.

Finding a reflection line

For a non-fixed point-image pair, construct its perpendicular bisector. A second pair must give the same mirror line. If the second pair does not fit, the proposed whole-figure mapping is not that reflection.

Do not confuse a reflection between two figures with reflection symmetry of one figure. Symmetry requires the whole original set to coincide with its own reflected image. A scalene triangle has reflected copies but no reflection symmetry.

Related topics

Geometric Constructions develops perpendicular-bisector methods. Symmetry applies reflection to a figure as a whole, while Transformations in Geometry compares the operations.