Math Core

A reflection is the transformation you see in a mirror or a still pond: the figure flips over a line and lands on the other side, the same distance away. On the coordinate plane, reflections over the axes and the lines $y = x$ and $y = -x$ have short coordinate rules, and a reflection over any line can be found with slopes and midpoints.

What a reflection does

Definition

Reflection

A reflection over a line $\ell$ (the line of reflection) maps each point $P$ to a point $P'$ so that $\ell$ is the perpendicular bisector of $\overline{PP'}$ . A point on $\ell$ maps to itself.

That definition packs in two facts you'll use constantly:

The segment $\overline{PP'}$ is perpendicular to the line of reflection.
The midpoint of $\overline{PP'}$ lies on the line of reflection, so $P$ and $P'$ are the same distance from it.

Coordinate rules

For the most common lines, you can apply the definition once and get a rule that works for every point.

Reflection rules

line of reflection	rule	example with $(5, 2)$
x-axis	$(x, y) \rightarrow (x, -y)$	$(5, -2)$
y-axis	$(x, y) \rightarrow (-x, y)$	$(-5, 2)$
$y = x$	$(x, y) \rightarrow (y, x)$	$(2, 5)$
$y = -x$	$(x, y) \rightarrow (-y, -x)$	$(-2, -5)$
$x = h$	$(x, y) \rightarrow (2h - x,\ y)$	over $x = 1$ : $(-3, 2)$
$y = k$	$(x, y) \rightarrow (x,\ 2k - y)$	over $y = 4$ : $(5, 6)$

Where do the last two come from? Reflecting over the vertical line $x = h$ keeps $y$ and moves $x$ to the other side of $h$ . The midpoint of $x$ and the new x-coordinate must be $h$ , so the new x-coordinate is $2h - x$ . The same reasoning over $y = k$ gives $2k - y$ .

Worked example: Reflecting over y = x

Reflect triangle $ABC$ with $A(1, 3)$ , $B(1, 6)$ and $C(3, 5)$ over the line $y = x$ .

The rule $(x, y) \rightarrow (y, x)$ swaps the coordinates:

\begin{aligned} A(1, 3) &\rightarrow A'(3, 1) \\ B(1, 6) &\rightarrow B'(6, 1) \\ C(3, 5) &\rightarrow C'(5, 3) \end{aligned}

Triangle ABC reflected over the dashed line y = x.Open in grapher →

Check one pair against the definition. The midpoint of $A(1, 3)$ and $A'(3, 1)$ is $(2, 2)$ , which is on $y = x$ . The slope of $\overline{AA'}$ is $\dfrac{1 - 3}{3 - 1} = -1$ , perpendicular to the slope $1$ of $y = x$ . ✓

Worked example: A vertical line that isn't an axis

Reflect the point $P(5, -2)$ over the line $x = -1$ .

$P$ is $6$ units to the right of $x = -1$ , so $P'$ is $6$ units to the left: $x = -1 - 6 = -7$ . The y-coordinate doesn't change. So $P' = (-7, -2)$ .

With the rule: $2h - x = 2(-1) - 5 = -7$ . ✓

Common mistake

Students often mix up which coordinate changes. The line $x = h$ is vertical, so reflecting over it moves points left or right: the x-coordinate changes and y stays. The line $y = k$ is horizontal, so only y changes. Sketch the line before you use a rule.

What a reflection preserves

A reflection is a rigid motion: distances and angle measures are preserved, so the image is congruent to the pre-image. But a reflection reverses orientation. In the example above, $A \to B \to C$ runs clockwise, while $A' \to B' \to C'$ runs counterclockwise. That reversal is how you can tell a reflection apart from a translation or a rotation.

Finding the line of reflection

If you know a point and its image, the line of reflection is the perpendicular bisector of the segment joining them: it passes through the midpoint, with the negative reciprocal slope.

Worked example: Finding the line

A reflection maps $P(-2, 3)$ to $P'(4, -1)$ . Find the line of reflection.

Midpoint: $\left(\dfrac{-2 + 4}{2},\ \dfrac{3 + (-1)}{2}\right) = (1, 1)$ .
Slope of $\overline{PP'}$ : $\dfrac{-1 - 3}{4 - (-2)} = \dfrac{-4}{6} = -\dfrac{2}{3}$ .
Perpendicular slope: $\dfrac{3}{2}$ .

Point-slope form through $(1, 1)$ : $y - 1 = \dfrac{3}{2}(x - 1)$ , so the line is $y = \dfrac{3}{2}x - \dfrac{1}{2}$ .

Reflecting over any line

The same two facts let you reflect over a slanted line with no special rule. Draw the perpendicular from the point to the line, find where they meet (that's the midpoint $M$ ), and go the same distance again past $M$ .

Worked example: Reflecting over y = x + 1

Reflect $P(4, 1)$ over the line $y = x + 1$ .

The line has slope $1$ , so the perpendicular through $P$ has slope $-1$ : $y - 1 = -(x - 4)$ , or $y = -x + 5$ .

Find the intersection: $x + 1 = -x + 5$ , so $x = 2$ and $y = 3$ . The midpoint is $M(2, 3)$ .

$M$ is halfway from $P$ to $P'$ , so $P' = (2 \cdot 2 - 4,\ 2 \cdot 3 - 1) = (0, 5)$ .

Check: the midpoint of $(4, 1)$ and $(0, 5)$ is $(2, 3)$ , which is on $y = x + 1$ . ✓

Tip

Whenever you know the midpoint $M$ and one endpoint $P$ , the other endpoint is $P' = 2M - P$ , coordinate by coordinate. This shortcut shows up in reflections, rotations by $180^\circ$ and point symmetry.

Practice

Practice 1

Reflect the point $(5, -2)$ over the line $y = x$ . What is the image?

Enter a point like (2, -3)

Practice 2

Reflect the point $(3, 7)$ over the line $y = -x$ . What is the image?

Enter a point like (2, -3)

Practice 3

Reflect the point $(-4, 6)$ over the line $x = 2$ . What is the image?

Enter a point like (2, -3)

Practice 4

Reflect the point $(1, -3)$ over the line $y = 4$ . What is the image?

Enter a point like (2, -3)

Practice 5

A reflection over a horizontal line $y = k$ maps $(2, 9)$ to $(2, -3)$ . What is $k$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

A reflection maps $P(1, 2)$ to $P'(5, 6)$ . What is the line of reflection?

Practice 7

Triangle $ABC$ is reflected over the x-axis. Which statement is true?

Practice 8

Reflect the point $(6, 1)$ over the line $y = x - 3$ . What is the image?

Enter a point like (2, -3)