Reflections in the Axes
A reflection sends every point to the opposite side of a mirror line, the same distance away. When the mirror is an axis, that description collapses into a single sign change, and which coordinate changes is the only thing to keep straight.
Plot $(3, 4)$. Now fold the page along the x-axis. Where does the point land? Fold along the y-axis instead. Where now? Which coordinate changed in each case?
Reflecting in the x-axis flips a point vertically, so the y-coordinate changes sign. Reflecting in the y-axis flips it horizontally, so the x-coordinate changes sign. The mirror is the axis whose coordinate stays.
$$\text{x-axis}: (x, y) \to (x, -y) \qquad \text{y-axis}: (x, y) \to (-x, y)$$
The rule to remember is which coordinate survives. Reflecting in the x-axis, points slide up and down but not sideways, so $x$ is untouched. Reflecting in the y-axis, points slide sideways only, so $y$ is untouched.
Know
- that reflection in the x-axis sends $P(x, y)$ to $P'(x, -y)$
- that reflection in the y-axis sends $P(x, y)$ to $P'(-x, y)$
- that reflecting in the x-axis and then the y-axis gives $(-x, -y)$, which is a half turn about the origin
Understand
- why the coordinate belonging to the mirror axis is the one left unchanged
- why a point lying on the mirror line is an invariant point
- why a reflection reverses orientation while a translation does not
Can Do
- find the image of a point or a shape reflected in either axis
- identify which reflection maps a given point onto a given image
- describe the single transformation equivalent to two successive reflections in the axes
A reflection is defined by a mirror line. Each point is sent to the point on the opposite side of the line, at the same perpendicular distance from it.
Two consequences follow immediately. Lengths and angles are unchanged, so the image is congruent. But orientation reverses: a triangle whose vertices read $A$, $B$, $C$ clockwise has an image whose vertices read $A'$, $B'$, $C'$ anticlockwise.
That reversal is the practical signature of a reflection. Translations and rotations preserve orientation; reflections do not, which is why a reflected letter looks wrong in a way that a rotated one does not.
Points on the mirror line do not move at all, since their distance from the line is zero. A reflection therefore always has invariant points, unlike a translation.
The x-axis is horizontal, so reflecting in it moves points straight up or straight down. Nothing moves sideways.
A point at height $y$ above the axis goes to height $y$ below it, which is the point at $-y$. The x-coordinate is untouched. So
$P(x, y) \to P'(x, -y)$
The point $(3, 4)$ maps to $(3, -4)$, and the point $(-2, -5)$ maps to $(-2, 5)$, since the negative of $-5$ is $5$.
The invariant points are those with $y = 0$, which is exactly the x-axis itself, as expected.
The y-axis is vertical, so reflecting in it moves points sideways only. Nothing moves up or down.
A point at horizontal position $x$ to the right goes to the same distance to the left, which is $-x$. The y-coordinate is untouched. So
$P(x, y) \to P'(-x, y)$
The point $(3, 4)$ maps to $(-3, 4)$, and the point $(-2, -5)$ maps to $(2, -5)$.
Comparing the two rules, the coordinate that survives is always the one named by the mirror. Mirror in the x-axis, $x$ survives; mirror in the y-axis, $y$ survives. Stating it in that form is the reliable way to avoid swapping them.
Apply the rule to each vertex and join the images in the same order.
Reflecting triangle $A(1, 2)$, $B(5, 3)$, $C(2, 6)$ in the y-axis gives $A'(-1, 2)$, $B'(-5, 3)$, $C'(-2, 6)$.
Any vertex sitting on the mirror line stays where it is. If a vertex were at $(0, 4)$, its image under reflection in the y-axis would be $(0, 4)$ again, and the original and image shapes would touch at that point.
Check the orientation as a final test. If $ABC$ ran clockwise and your $A'B'C'$ also runs clockwise, something has gone wrong, because a reflection must reverse it.
Reflect in the x-axis and then in the y-axis:
$(x, y) \to (x, -y) \to (-x, -y)$
The result is $(-x, -y)$, which negates both coordinates. That is a rotation of $180°$ about the origin, as the next lesson confirms.
Doing them in the other order gives $(x, y) \to (-x, y) \to (-x, -y)$, the same result. So for these two particular reflections the order does not matter, because the two mirrors are perpendicular.
Two reflections in the same line return every point to where it started, since each point crosses the mirror and crosses back. Reflecting twice in one line is doing nothing at all, which is worth knowing when a question chains transformations together.
Note also that orientation reverses twice and so ends up unchanged, which fits: the composite is a rotation, and rotations preserve orientation.
Watch Me Solve It · 3 examples
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1Reflect in the x-axis$(x, y) \to (x, -y)$$A(3, -5) \to (3, 5)$The mirror is the x-axis, so $x$ survives and the y-coordinate changes sign. The negative of $-5$ is $5$.
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2Reflect in the y-axis$(x, y) \to (-x, y)$$A(3, -5) \to (-3, -5)$The mirror is the y-axis, so $y$ survives and the x-coordinate changes sign.
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3Check against a sketchThe original is in the fourth quadrant. The first image is in the first quadrant, directly above it; the second is in the third quadrant, directly left of it. Both are consistent with the direction each mirror flips in.
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1Apply the y-axis rule to each vertex$P(0, 4) \to P'(0, 4)$$Q(5, 1) \to Q'(-5, 1)$$R(3, 7) \to R'(-3, 7)$Negate the x-coordinate each time. The negative of $0$ is $0$.
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2Identify the invariant point$P$ lies on the y-axis, which is the mirror line, so it does not move. It is an invariant point, and $P'$ coincides with $P$.
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3Check the orientation reversedReading $P$, $Q$, $R$ in order traces one rotational sense, and $P'$, $Q'$, $R'$ traces the opposite one, as a reflection requires.
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1First reflection$B(-2, 6) \to B'(-2, -6)$Mirror is the x-axis, so the y-coordinate changes sign.
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2Second reflection$B'(-2, -6) \to B''(2, -6)$Mirror is the y-axis, so the x-coordinate changes sign.
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3Describe the composite in general$(x, y) \to (x, -y) \to (-x, -y)$Both coordinates end up negated, so the composite sends $(x, y)$ to $(-x, -y)$.
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4Name itNegating both coordinates is a rotation of $180°$ about the origin. Checking on the numbers: $(-2, 6)$ becomes $(2, -6)$, which is the point diametrically opposite through the origin.
Brain Trainer · 5 problems
Five items. Name the surviving coordinate before you write anything down.
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1 Reflect $(7, 2)$ in the x-axis.
Mirror is the x-axis, so $x$ survives.$(7, -2)$ -
2 Reflect $(7, 2)$ in the y-axis.
Mirror is the y-axis, so $y$ survives.$(-7, 2)$ -
3 Reflect $(-4, -9)$ in the x-axis.
Negate the y-coordinate, and the negative of $-9$ is $9$.$(-4, 9)$ -
4 $(6, 3)$ maps to $(-6, 3)$. Which reflection was it?
The y-coordinate survived, so the mirror is the y-axis.In the y-axis -
5 Which points are invariant under reflection in the x-axis?
Those whose flipping coordinate is already zero.All points on the x-axis
Multiple Choice · 5 questions
Reflected in the x-axis, the point $(-5, 8)$ maps to:
A reflection maps $(4, -3)$ to $(-4, -3)$. The mirror line is:
Which property distinguishes a reflection from a translation?
A point is reflected in the x-axis and then in the y-axis. The combined effect on $(x, y)$ is:
A triangle has a vertex at $(0, -7)$. Under reflection in the y-axis, this vertex:
Short Answer · 3 questions
The mirror names the survivor
Reflecting in the x-axis gives $(x, -y)$; reflecting in the y-axis gives $(-x, y)$. The coordinate matching the mirror's name is the one that stays.
Points on the mirror do not move
A reflection always has invariant points, namely the whole mirror line. A translation, by contrast, moves every point without exception.
Orientation reverses
A clockwise-labelled triangle comes back anticlockwise. This is the property that distinguishes a reflection from a translation or a rotation.
Two axis reflections make a half turn
$(x, y) \to (x, -y) \to (-x, -y)$, in either order, which is a rotation of $180°$ about the origin. Reflecting twice in the same line changes nothing at all.
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