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Reflection — chapter notes

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1. Reflection in a mirror line

  • A reflection in a mirror line ll sends each point PP to its image P′P' on the other side of ll, at the same distance from it.
  • The mirror line is the perpendicular bisector of PP′PP': PP′PP' meets ll at right angles, and ll passes through the midpoint of PP′PP'.
  • A point and its image are 2k2k units apart when the point is kk units from the mirror.
  • A point on the mirror is its own image (it is invariant).

Example: P(4,3)P(4, 3) in the line y=1y = 1: PP is 22 units above the mirror, so P′P' is (4,−1)(4, -1), 22 units below it.

  • PP′
The mirror is the perpendicular bisector of $PP'$: $PM = MP' = 2$.
  • triangle ABC
  • image A′B′C′
Each vertex of the image is as far behind the mirror as the vertex is in front of it.

Key terms

Image

The point to which a reflection sends a given point.

Mirror line

The line of reflection; it is the perpendicular bisector of every segment PP′PP'.

2. Reflection in the axes

  • In the xx-axis (y=0y = 0, written MxM_x): keep xx, change the sign of yy.
  • In the yy-axis (x=0x = 0, written MyM_y): change the sign of xx, keep yy.
Mx:(x,y)→(x,−y)My:(x,y)→(−x,y)M_x : (x, y) \to (x, -y) \qquad M_y : (x, y) \to (-x, y)

Example: A(−5,2)A(-5, 2) goes to (−5,−2)(-5, -2) in the xx-axis and to (5,2)(5, 2) in the yy-axis; B(0,−3)B(0, -3) lies on the yy-axis, so it is invariant under MyM_y.

  • AA′
  • BB′
  • CC′
In the $x$-axis: same $x$-coordinate, opposite $y$-coordinate.
  • AA′
  • BB′
  • CC′
In the $y$-axis: opposite $x$-coordinate, same $y$-coordinate.
A point on the mirror does not move.
PointImage in the xx-axisImage in the yy-axis
A(−5,2)A(-5, 2)(−5,−2)(-5, -2)(5,2)(5, 2)
B(0,−3)B(0, -3)(0,3)(0, 3)(0,−3)(0, -3) (invariant)

Formulas

  • Reflection in the x-axis
    Mx:(x,y)→(x,−y)M_x : (x, y) \to (x, -y)
  • Reflection in the y-axis
    My:(x,y)→(−x,y)M_y : (x, y) \to (-x, y)

Key terms

MxM_x

Reflection in the xx-axis: (x,y)→(x,−y)(x, y) \to (x, -y).

MyM_y

Reflection in the yy-axis: (x,y)→(−x,y)(x, y) \to (-x, y).

3. Reflection in the origin

  • In the origin (written MOM_O): change the signs of both coordinates. The origin is the midpoint of PP′PP', and PP, OO, P′P' lie on one straight line.
MO:(x,y)→(−x,−y)M_O : (x, y) \to (-x, -y)
  • Reflection in the xx-axis followed by reflection in the yy-axis (in either order) has the same effect as MOM_O.
  • Only the origin is invariant under MOM_O.

Example: the image of (3,2)(3, 2) in the origin is (−3,−2)(-3, -2); and if the image of PP in the yy-axis is (−6,−1)(-6, -1), then PP is (6,−1)(6, -1) and its image in the origin is (−6,1)(-6, 1).

  • PP′ through O
The origin is the midpoint of $PP'$.
  • P′P
  • PQ through O
Reflect the image back to find the point; then change both signs.

Formulas

  • Reflection in the origin
    MO:(x,y)→(−x,−y)M_O : (x, y) \to (-x, -y)

Key terms

MOM_O

Reflection in the origin: (x,y)→(−x,−y)(x, y) \to (-x, -y).

4. Reflection in the lines x = a and y = b

  • In x=ax = a (a vertical mirror): keep yy; a point x−ax - a to the right goes x−ax - a to the left, so the new xx-coordinate is a−(x−a)=2a−xa - (x - a) = 2a - x.
  • In y=by = b (a horizontal mirror): keep xx; the new yy-coordinate is 2b−y2b - y.
(x,y)→(2a−x,y)(x,y)→(x,2b−y)(x, y) \to (2a - x, y) \qquad (x, y) \to (x, 2b - y)
  • With a=0a = 0 or b=0b = 0 these become the rules for the axes.

Example: R(7,−2)R(7, -2) goes to (2×3−7,−2)=(−1,−2)(2 \times 3 - 7, -2) = (-1, -2) in x=3x = 3 and to (7,4)(7, 4) in y=1y = 1.

  • PP′
  • QQ′
Same distance on each side of $x = 2$; the $y$-coordinate is kept.
  • PP′
  • QQ′
Same distance above and below $y = -1$; the $x$-coordinate is kept.
  • RR′
  • RR″
A vertical mirror moves the point sideways; a horizontal mirror moves it up or down.

Formulas

  • Reflection in x = a
    (x,y)→(2a−x,y)(x, y) \to (2a - x, y)
  • Reflection in y = b
    (x,y)→(x,2b−y)(x, y) \to (x, 2b - y)

Key terms

Line parallel to an axis

x=ax = a is vertical (parallel to the yy-axis); y=by = b is horizontal (parallel to the xx-axis).

5. Working backwards

  • The point from its image: a reflection undoes itself, so reflect the image in the same mirror.
  • The mirror from a point and its image: if the xx-coordinates are equal, the mirror is horizontal, y=y1+y22y = \dfrac{y_1 + y_2}{2}; if the yy-coordinates are equal, it is vertical, x=x1+x22x = \dfrac{x_1 + x_2}{2}; if both coordinates change sign, it is the origin.
  • Unknown coordinates: write the image by the rule, then equate xx-coordinates and yy-coordinates separately.

Example: the image of (2a−3,5)(2a - 3, 5) in the yy-axis is (−7,b+2)(-7, b + 2): 3−2a=−73 - 2a = -7, so a=5a = 5; b+2=5b + 2 = 5, so b=3b = 3.

  • PP′
  • QQ′
The mirror passes through the midpoint of the segment, at right angles to it.
Two equal points have equal $x$-coordinates and equal $y$-coordinates.

Formulas

  • Horizontal mirror through the midpoint
    y=y1+y22y = \dfrac{y_1 + y_2}{2}
  • Vertical mirror through the midpoint
    x=x1+x22x = \dfrac{x_1 + x_2}{2}

Key terms

Self-inverse

Reflecting twice in the same mirror gives back the original point.

6. Invariant points

  • A point is invariant under a reflection if its image is the point itself.
  • For a mirror line, the invariant points are exactly the points on that line: (x,0)(x, 0) for the xx-axis, (0,y)(0, y) for the yy-axis, (a,y)(a, y) for x=ax = a, (x,b)(x, b) for y=by = b.
  • For the origin, only O(0,0)O(0, 0) is invariant.
  • A point invariant under two reflections in perpendicular lines is their point of intersection.

Example: D(−2,−2)D(-2, -2) is invariant under reflection in both x=−2x = -2 and y=−2y = -2.

  • WW′
Points on the mirror are invariant; every other point moves.
Five rules, one example point.
MirrorRuleInvariant pointsImage of (4,−3)(4, -3)
xx-axis(x,y)→(x,−y)(x, y) \to (x, -y)every point of the xx-axis(4,3)(4, 3)
yy-axis(x,y)→(−x,y)(x, y) \to (-x, y)every point of the yy-axis(−4,−3)(-4, -3)
origin(x,y)→(−x,−y)(x, y) \to (-x, -y)only the origin(−4,3)(-4, 3)
line x=1x = 1(x,y)→(2a−x,y)(x, y) \to (2a - x, y) with a=1a = 1every point with x=1x = 1(−2,−3)(-2, -3)
line y=2y = 2(x,y)→(x,2b−y)(x, y) \to (x, 2b - y) with b=2b = 2every point with y=2y = 2(4,7)(4, 7)
$D(-2, -2)$ lies on both $x = -2$ and $y = -2$.
MirrorInvariant pointsReason
xx-axisAAyy-coordinate 00
yy-axisBBxx-coordinate 00
line x=−2x = -2CC and DDxx-coordinate −2-2
line y=−2y = -2BB and DDyy-coordinate −2-2

Key terms

Invariant point

A point that is its own image under a transformation.

7. Combined reflections and figures on graph paper

  • Two reflections: xx-axis then yy-axis (either order) == reflection in the origin; the same mirror twice == no change.
  • A figure: reflect each vertex and join the images in the same order. The image is congruent (same lengths, angles, area) but the order of the vertices is reversed.
  • Naming the figure: horizontal and vertical sides give parallel sides and right angles at a glance; then compare side lengths (rectangle, square, isosceles trapezium, kite, right-angled or isosceles triangle).
  • Area: use horizontal or vertical sides as bases and heights; trapezium 12(a+b)h\dfrac{1}{2}(a + b)h; kite 12d1d2\dfrac{1}{2} d_1 d_2.

Example: A(5,2)A(5, 2) with its images in the yy-axis, the origin and the xx-axis makes a rectangle 1010 by 44: area 4040 square units, perimeter 2828 units.

  • ABCD
$ABCD$ is a rectangle: $10 \times 4 = 40$ square units.
  • in the x-axis, then in the y-axis
  • in the origin
Two axis reflections, one after the other, equal one reflection in the origin.
  • ABB′A′
Parallel sides $8$ and $4$, height $3$: area $18$ square units.
  • ABCD
  • AC
Diagonals $AC = 6$ and $BD = 7$ cross at right angles: area $21$ square units.
Horizontal and vertical sides on graph paper make parallel sides and right angles easy to see.
The image is congruent to the figure: same size and shape, facing the other way.
FactExample
A point kk units from the mirror is 2k2k units from its image(1,6)(1, 6) is 44 units from y=2y = 2 and 88 units from its image (1,−2)(1, -2)
A reflection keeps lengthsO(0,0)O(0, 0) to (3,4)(3, 4) and its image (4,0)(4, 0) to (1,4)(1, 4) in x=2x = 2: both 55 units
A reflection keeps angles and areasa right triangle of area 66 and its image in the xx-axis: both area 66
The order of the vertices is reversedanticlockwise ABCABC becomes clockwise A′B′C′A'B'C'

Formulas

  • Area of a trapezium
    12(a+b)h\dfrac{1}{2}(a + b)h
  • Area of a kite
    12d1d2\dfrac{1}{2} d_1 d_2

Key terms

Congruent

Same size and shape; a reflected figure is congruent to the original.

Isosceles trapezium

A trapezium whose non-parallel sides are equal; a segment and its image often make one.

Exam tips

How marks are awarded

  • Image of a point (1 mark): correct coordinate pair.
  • Mirror line from a point and its image (1–2 marks): horizontal or vertical stated 1; equation 1.
  • Unknown coordinates (2–3 marks): image by the rule 1; each unknown 1.
  • Invariant points (1–2 marks): the points on the mirror, with a reason.
  • 10-mark graph question: parts (a), (b), (c), usually 3 + 3 + 4: plotting and labelling, images, naming the figure with a reason, area or perimeter; show all working.

Common mistakes

  1. Changing the wrong coordinate: reflection in the xx-axis changes the yy-coordinate.
  2. Using a−xa - x or x−2ax - 2a instead of 2a−x2a - x for the line x=ax = a.
  3. Confusing the line x=3x = 3 (vertical) with the line y=3y = 3 (horizontal).
  4. Saying a point on the xx-axis is invariant under reflection in the yy-axis (only the origin is).
  5. Joining the vertices of a figure in the wrong order, or naming it without a reason.
  6. Forgetting the half in the area of a triangle, trapezium or kite.