When I reflect across the x-axis, why does the y-coordinate change sign instead of the x-coordinate?

Because the motion is perpendicular to the mirror. The x-axis is horizontal, so points move straight up or down to reach their images; only the vertical position changes. A point 3 units above the axis lands 3 units below it, at the same horizontal position. The name of the line tells you what stays fixed, not what changes.

How do I reflect across a line like x = -4 or y = 6?

Use (x,y)→(2a-x,\;y) for the vertical line x=a and (x,y)→(x,\;2b-y) for the horizontal line y=b. Across x=-4, the point (1,5) maps to (-8-1,5)=(-9,5). You can also just count: (1,5) is 5 units right of x=-4, so its image is 5 units left, at x=-9. The two methods should always agree.

Is a reflection the same thing as a rotation of 180 degrees?

No. Both can make a figure look upside down, but a 180^ rotation preserves orientation (counterclockwise stays counterclockwise) and has exactly one fixed point, the center. A reflection reverses orientation and fixes an entire line of points. Comparing the vertex order of the image is the quickest way to tell them apart.

How do I find the line of reflection if I am only given a figure and its image?

Pick any vertex and its image, then find the perpendicular bisector of the segment joining them. Compute the midpoint and use the opposite reciprocal of that segment's slope. If the two points share a y-coordinate, the mirror is the vertical line through the average of the x-values; if they share an x-coordinate, it is the horizontal line through the average of the y-values. Always test a second vertex to confirm.

GEOM-4.2

Reflections

Learn reflections in Geometry: the line of reflection as perpendicular bisector, rules across the axes, y = x, and lines x = a or y = b, plus why orientation flips.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Reflections, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Reflections free

What this lesson covers

Hold a triangle up to a mirror and something strange happens: every side keeps its length, every angle keeps its measure, but the figure now reads backwards. That is a reflection — a rigid motion that preserves distance and angle measure while reversing orientation. In this lesson you will learn the precise geometric definition (it is all about perpendicular bisectors, not about "flipping"), the coordinate rules for reflecting across the x-axis, the y-axis, the lines

y=x

and

y=-x

, and any horizontal or vertical line. You will also learn how to work backwards: given a figure and its image, find the mirror line. These skills feed directly into the next lessons, where reflections combine with translations and rotations to build every rigid motion and to define what congruence really means.

The Perpendicular Bisector Definition

A reflection across line $\ell$ is the transformation that sends each point

P

to the point

P'

such that

\ell

is the perpendicular bisector of segment

\overline{PP'}

. If

P

happens to lie on

\ell

, then

P'=P

; points on the mirror line are fixed.

Unpack that definition, because it carries two separate conditions. Perpendicular means

\overline{PP'}

meets

\ell

at a right angle. Bisector means

\ell

crosses

\overline{PP'}

at its midpoint, so

P

and

P'

are the same distance from the mirror. Miss either condition and the transformation is not a reflection: a point moved straight across but too far is not a reflection, and a point moved the right distance but along a slanted path is not one either.

This definition is more useful than the informal "flip it over" picture because it gives you a construction and a test. To construct

P'

by hand, drop a perpendicular from

P

\ell

, measure the distance

d

from

P

to the foot of that perpendicular, and continue the same distance

d

past

\ell

along the same line. To test whether a given pair of points is related by a reflection across

\ell

, check that the midpoint of

\overline{PP'}

lies on

\ell

and that the slope of

\overline{PP'}

is the opposite reciprocal of the slope of

\ell

.

One consequence worth noticing now: since every point of

\ell

stays put, a reflection has infinitely many fixed points, unlike a translation (no fixed points, unless it is the identity) or a rotation (exactly one, the center).

Notation:

r_\ell(P)=P'

, or

r_{y\text{-axis}}(\triangle ABC)=\triangle A'B'C'

.

A common mistake is reflecting each vertex a different distance because you eyeballed it. Always count squares perpendicular to the mirror, not diagonally.

The image of a segment is a segment, and the image of a triangle is a triangle of the same size.

Coordinate Rules for Common Mirror Lines

When the mirror is one of a few standard lines, the perpendicular bisector definition collapses into a short algebraic rule. Each rule below is just "keep the coordinate parallel to the mirror, move the other coordinate to the mirror's opposite side."

Line of reflection	Rule	Example: $(5,2)$ maps to
$x$ -axis	$(x,y)\to(x,-y)$	$(5,-2)$
$y$ -axis	$(x,y)\to(-x,y)$	$(-5,2)$
$y=x$	$(x,y)\to(y,x)$	$(2,5)$
$y=-x$	$(x,y)\to(-y,-x)$	$(-2,-5)$
$x=a$ (vertical)	$(x,y)\to(2a-x,\;y)$	across $x=3$ : $(1,2)$
$y=b$ (horizontal)	$(x,y)\to(x,\;2b-y)$	across $y=4$ : $(5,6)$

Why does

x=a

give

2a-x

? The horizontal distance from

x

to the mirror is

a-x

, so the image sits at

a+(a-x)=2a-x

. Many students prefer to count instead of substitute: if a point is 4 units left of the mirror, its image is 4 units right of it, at the same height. Both methods must agree — use one to check the other.

Two frequent errors show up here. First, students memorize "reflect over the x-axis means change the sign of

x

." It is the opposite: reflecting across the

x

-axis moves points vertically, so the

y

-coordinate changes sign. Tie the rule to the motion, not to the letter in the line's name. Second, students apply the

y

-axis rule to the vertical line

x=3

. The rule

(x,y)\to(-x,y)

only works when the mirror is

x=0

; for

x=3

you need

2(3)-x=6-x

.

For

y=x

, notice the swap makes sense: the mirror line consists of all points where the coordinates are equal, and swapping coordinates is exactly what leaves those points fixed.

Why Distance and Angles Survive but Orientation Flips

A reflection is a rigid motion (an isometry): the distance between any two points equals the distance between their images. You can see this on the coordinate plane. Reflect

A(x_1,y_1)

and

B(x_2,y_2)

across the

x

-axis to get

A'(x_1,-y_1)

and

B'(x_2,-y_2)

. Then

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,

since squaring erases the sign change. Because all distances are preserved, triangles map to congruent triangles, so by SSS every angle measure is preserved too. Segments map to segments, parallel lines map to parallel lines, and a figure's perimeter and area are unchanged.

What is not preserved is orientation. If you list the vertices of

\triangle ABC

in counterclockwise order, their images

A'

B'

C'

appear in clockwise order. A single reflection reverses the sense of rotation around the figure, which is why reflected letters look backwards and why your right hand's mirror image looks like a left hand. Reflections are called opposite isometries; translations and rotations are direct isometries because they preserve orientation.

This orientation reversal has a practical payoff. If you are handed two congruent figures and the vertices run counterclockwise in one and clockwise in the other, no translation or rotation alone can match them — an odd number of reflections must be involved. That observation carries straight into the compositions lesson.

A misconception to clear up: orientation reversal is not the same as "the figure looks different." A shape with a line of symmetry, such as an isosceles triangle reflected across its own axis, can land exactly on itself. The orientation of the labeled vertices still reverses (

A

and

B

trade places), even though the outline of the figure is unchanged. Orientation is about the labeled order of points, not the picture's silhouette.

Working Backwards: Finding the Mirror Line

Problems often give you a figure and its image and ask for the line of reflection. Use the definition in reverse: the mirror must be the perpendicular bisector of every segment connecting a point to its image, so you only need one such segment to locate it (a second one is a good check).

For a segment

\overline{PP'}

with

P(x_1,y_1)

and

P'(x_2,y_2)

, compute the midpoint

M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)

and the slope

m=\frac{y_2-y_1}{x_2-x_1}

. The mirror passes through

M

with slope

-\frac{1}{m}

. Two shortcuts cover most classroom cases: if

P

and

P'

have the same

y

-coordinate, the mirror is the vertical line

x=\frac{x_1+x_2}{2}

; if they share an

x

-coordinate, the mirror is the horizontal line

y=\frac{y_1+y_2}{2}

.

For example, if

A(-1,5)

maps to

A'(7,5)

, the mirror is

x=\frac{-1+7}{2}=3

. If

B(4,1)

maps to

B'(4,-9)

, the mirror is

y=\frac{1+(-9)}{2}=-4

Situation	What to do
Same $y$ -values	Mirror is $x=$ average of the $x$ -values
Same $x$ -values	Mirror is $y=$ average of the $y$ -values
Coordinates swapped, as $(a,b)\to(b,a)$	Mirror is $y=x$
Neither	Find midpoint, then use the opposite reciprocal slope

Where students go wrong: they find the midpoint but forget the perpendicular condition and draw a line through the midpoint in the wrong direction, or they average only one pair of coordinates and report a point instead of an equation. Always answer with an equation of a line. And verify by reflecting a second vertex across your proposed mirror — if it does not land on the stated image, the mirror is wrong.

Key terms

Reflection.: A transformation across a line $\ell$ sending each point $P$ to $P'$ so that $\ell$ is the perpendicular bisector of $\overline{PP'}$ ; points on $\ell$ are fixed.
Line of reflection.: The mirror line $\ell$ ; it is the perpendicular bisector of every segment joining a point to its image and consists entirely of fixed points.
Perpendicular bisector.: A line that passes through the midpoint of a segment and meets it at a $90^\circ$ angle.
Rigid motion (isometry).: A transformation that preserves distance between points, and therefore preserves angle measure, segment length, perimeter, and area.
Orientation.: The clockwise or counterclockwise order in which labeled vertices of a figure appear; a single reflection reverses it.
Image and preimage.: The preimage is the original figure; the image is the result after the transformation, usually labeled with primes such as $A'$ .
Fixed point.: A point that maps to itself. For a reflection, the fixed points are exactly the points on the line of reflection.
Line of symmetry.: A line across which a figure reflects onto itself, so the image coincides with the original figure.

Worked example

Triangle

ABC

has vertices

A(-2,1)

B(1,4)

, and

C(3,0)

. Reflect it across the vertical line

x=2

to get

\triangle A'B'C'

. Then verify that

A'B'=AB

, state whether the orientation reversed, and confirm that

x=2

is the perpendicular bisector of

\overline{AA'}

Step 1: Apply the rule for a vertical mirror. For

x=a

, the rule is

(x,y)\to(2a-x,\;y)

. With

a=2

, that is

(x,y)\to(4-x,\;y)

A(-2,1)\to A'(4-(-2),\,1)=A'(6,1)

B(1,4)\to B'(4-1,\,4)=B'(3,4)

C(3,0)\to C'(4-3,\,0)=C'(1,0)

.

Step 2: Check with counting.

A

4

units left of

x=2

, so

A'

must be

4

units right, at

x=6

— matches.

C

1

unit right of the mirror, so

C'

1

unit left, at

x=1

— matches.

Step 3: Verify distance is preserved.

AB=\sqrt{(1-(-2))^2+(4-1)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}

. And

A'B'=\sqrt{(3-6)^2+(4-1)^2}=\sqrt{9+9}=3\sqrt{2}

. Equal, as required for a rigid motion.

Step 4: Check orientation. Going

A(-2,1)\to B(1,4)\to C(3,0)

traces the triangle clockwise. Going

A'(6,1)\to B'(3,4)\to C'(1,0)

traces it counterclockwise. The orientation reversed, which is exactly what a single reflection does.

Step 5: Confirm the perpendicular bisector property for

\overline{AA'}

. The midpoint of

A(-2,1)

and

A'(6,1)

\left(\frac{-2+6}{2},\frac{1+1}{2}\right)=(2,1)

, which lies on

x=2

. The segment

\overline{AA'}

is horizontal, and

x=2

is vertical, so they are perpendicular. Both conditions hold, so

x=2

is the perpendicular bisector of

\overline{AA'}

— the definition of a reflection is satisfied.

Practice questions

Point

P(-3,7)

is reflected across the line

y=x

. What are the coordinates of

P'

$(3,-7)$
$(7,-3)$
$(-7,3)$
$(-3,-7)$

Answer: $(7,-3)$

Reflecting across

y=x

swaps the coordinates:

(x,y)\to(y,x)

, so

(-3,7)\to(7,-3)

. Check it against the definition: the midpoint of

\overline{PP'}

\left(\frac{-3+7}{2},\frac{7+(-3)}{2}\right)=(2,2)

, which lies on

y=x

, and the slope of

\overline{PP'}

\frac{-3-7}{7-(-3)}=-1

, the opposite reciprocal of the mirror's slope

1

. The choice

(-7,3)

comes from applying the

y=-x

rule by mistake, and

(3,-7)

comes from negating both coordinates instead of swapping them.

Quadrilateral

WXYZ

is reflected so that

W(2,-1)

maps to

W'(2,9)

. Find the equation of the line of reflection, justify your answer using the definition of a reflection, and state where

Y(-5,3)

maps.

Answer: The line of reflection is $y=4$ , and $Y'(-5,5)$ .

Since

W

and

W'

share the

x

-coordinate

2

, the segment

\overline{WW'}

is vertical, so its perpendicular bisector is horizontal. The midpoint is

\left(2,\frac{-1+9}{2}\right)=(2,4)

, so the mirror is the horizontal line

y=4

. It passes through the midpoint of

\overline{WW'}

and is perpendicular to it, which is exactly the definition of the line of reflection. Applying the rule for

y=b

, namely

(x,y)\to(x,2b-y)

with

b=4

Y(-5,3)\to(-5,8-3)=(-5,5)

. Counting confirms it —

Y

1

unit below

y=4

, so

Y'

1

unit above.

Explain why reflecting a triangle across a line cannot change its area, but can change whether its vertices read clockwise or counterclockwise.

Answer: A reflection preserves all distances, so the image triangle has congruent sides and therefore equal area; orientation is a separate property about vertex order, and a reflection always reverses it.

Distance preservation follows from the coordinate computation: reflecting across the

x

-axis changes

y_2-y_1

-(y_2-y_1)

, and squaring in the distance formula erases the sign, so lengths are unchanged. Equal corresponding side lengths give congruent triangles by SSS, and congruent triangles have equal area and equal angle measures. Orientation, however, is not a measurement — it records the rotational order of the labeled vertices. Because a reflection sends every point to the opposite side of the mirror along a perpendicular, that circular order reverses. So a reflection is an isometry that is not orientation-preserving, unlike a translation or rotation.

FAQ

When I reflect across the x-axis, why does the y-coordinate change sign instead of the x-coordinate?: Because the motion is perpendicular to the mirror. The $x$ -axis is horizontal, so points move straight up or down to reach their images; only the vertical position changes. A point $3$ units above the axis lands $3$ units below it, at the same horizontal position. The name of the line tells you what stays fixed, not what changes.
How do I reflect across a line like x = -4 or y = 6?: Use $(x,y)\to(2a-x,\;y)$ for the vertical line $x=a$ and $(x,y)\to(x,\;2b-y)$ for the horizontal line $y=b$ . Across $x=-4$ , the point $(1,5)$ maps to $(-8-1,5)=(-9,5)$ . You can also just count: $(1,5)$ is $5$ units right of $x=-4$ , so its image is $5$ units left, at $x=-9$ . The two methods should always agree.
Is a reflection the same thing as a rotation of 180 degrees?: No. Both can make a figure look upside down, but a $180^\circ$ rotation preserves orientation (counterclockwise stays counterclockwise) and has exactly one fixed point, the center. A reflection reverses orientation and fixes an entire line of points. Comparing the vertex order of the image is the quickest way to tell them apart.
How do I find the line of reflection if I am only given a figure and its image?: Pick any vertex and its image, then find the perpendicular bisector of the segment joining them. Compute the midpoint and use the opposite reciprocal of that segment's slope. If the two points share a $y$ -coordinate, the mirror is the vertical line through the average of the $x$ -values; if they share an $x$ -coordinate, it is the horizontal line through the average of the $y$ -values. Always test a second vertex to confirm.

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