Rigid Motions and Translations on the Coordinate Plane | Grade 8 Math | LevelKite

A transformation is a rule that tells every point of a figure where to go. The figure you start with is the original, and the figure you end with is its image. If a vertex is named $A$ , its image is named $A'$ (read "A prime"). $A$ and $A'$ are corresponding vertices.

This lesson covers three transformations: a translation (a slide), a reflection (a flip) and a rotation (a turn). All three are rigid motions: they can change where a figure is and which way it faces, but they keep every side length and every angle measure the same.

You will tell them apart from the exact coordinates of corresponding vertices, never from how a drawing happens to look. Every coordinate and length in this lesson is exact, and every length is counted in grid units. In each figure, the horizontal axis is the $x$ -axis and the vertical axis is the $y$ -axis.

A translation slides a figure. Every point moves the same distance in the same direction, so nothing flips and nothing turns.

The figure below shows one. Triangle $ABC$ has Point A at $(2, 1)$ , Point B at $(5, 1)$ and Point C at $(5, 3)$ . Its image has $A'$ at $(8, 4)$ , $B'$ at $(11, 4)$ and $C'$ at $(11, 6)$ ; the image vertices are labeled with their coordinates.

To check for a translation, subtract for each pair of corresponding vertices: image coordinate minus original coordinate.

$A$ to $A'$ : $x$ changes by $8 - 2 = 6$ , $y$ by $4 - 1 = 3$
$B$ to $B'$ : $x$ changes by $11 - 5 = 6$ , $y$ by $4 - 1 = 3$
$C$ to $C'$ : $x$ changes by $11 - 5 = 6$ , $y$ by $6 - 3 = 3$

Every vertex moved 6 units right and 3 units up. When the stated coordinates give the same change in $x$ and the same change in $y$ for every pair of corresponding vertices of a polygon, the motion is a translation. If even one pair gives a different change, it is not. Lengths are kept too: horizontal side $AB$ spans $x = 2$ to $x = 5$ and its image $A'B'$ spans $x = 8$ to $x = 11$ , so each is 3 units long.

Triangle

ABC

and its image: Point A at

(2, 1)

goes to

(8, 4)

, which is 6 units right and 3 units up.

A reflection flips a figure across a line called the mirror line. Each point goes straight across the mirror line, at a right angle to it, and lands the same distance away on the other side. A point on the mirror line stays where it is.

In the figure below, the mirror line is the $y$ -axis. Triangle $ABC$ is the same triangle as before. Its image has $A'$ at $(-2, 1)$ , $B'$ at $(-5, 1)$ and $C'$ at $(-5, 3)$ , labeled with their coordinates.

Check the definition with the coordinates:

$A(2, 1)$ is 2 units right of the $y$ -axis; $A'(-2, 1)$ is 2 units left of it.
$B(5, 1)$ is 5 units right; $B'(-5, 1)$ is 5 units left.
$C(5, 3)$ is 5 units right; $C'(-5, 3)$ is 5 units left.

Each vertex kept its $y$ -coordinate, so it traveled along a horizontal line, which meets the vertical $y$ -axis at a right angle. So this is a reflection across the $y$ -axis. It is not a translation: $A$ moved 4 units left, but $B$ moved 10 units left. Lengths are still kept: horizontal side $A'B'$ spans $x = -5$ to $x = -2$ , which is 3 units, the same as side $AB$ .

The

y

-axis is the mirror line: Point A at

(2, 1)

and its image at

(-2, 1)

are each 2 units from it.

A rotation turns a figure around a fixed point called the center. Every point stays the same distance from the center and turns through the same angle in the same direction. The center does not move, and the angle can be any size.

In the figure below, the center is the origin, labeled $(0, 0)$ , and triangle $ABC$ is the same triangle again. Its image has $A'$ at $(-2, -1)$ , $B'$ at $(-5, -1)$ and $C'$ at $(-5, -3)$ .

You can confirm the turn from coordinates, without measuring an angle. From $A'(-2, -1)$ to the origin you go 2 units right and 1 unit up. From the origin to $A(2, 1)$ you go 2 units right and 1 unit up again. Two identical steps in a row make one straight path with the origin exactly halfway along it. So $A'$ is straight across the origin from $A$ , the same distance away. That is a half turn: a straight angle, exactly $180^\circ$ .

The same check works for $B$ and $B'$ (5 right and 1 up, twice) and for $C$ and $C'$ (5 right and 3 up, twice). Every vertex made the same $180^\circ$ turn about the origin, so this is a rotation. Horizontal side $A'B'$ spans $x = -5$ to $x = -2$ : still 3 units.

The origin

(0, 0)

is the center: Point A at

(2, 1)

and its image at

(-2, -1)

are straight across it from each other.

Worked example

Which motion is it?

Rectangle $PQRS$ has vertices $P(1, 1)$ , $Q(4, 1)$ , $R(4, 3)$ and $S(1, 3)$ . Its image has $P'(-1, 1)$ , $Q'(-4, 1)$ , $R'(-4, 3)$ and $S'(-1, 3)$ . Use the coordinates of corresponding vertices to decide whether the motion is a translation, a reflection or a rotation.

Test for a translation first, because it is the quickest check. For $P$ and $P'$ , subtract image coordinate minus original coordinate.
$-1 - 1 = -2, \qquad 1 - 1 = 0$
Do the same for $Q$ and $Q'$ . A translation needs the same change for every vertex, but $P$ moved 2 units left and $Q$ moved 8 units left. So this is not a translation.
$-4 - 4 = -8, \qquad 1 - 1 = 0$
Look at what the coordinates did. Every $y$ -coordinate stayed the same, and every $x$ -coordinate became its opposite (shown for $P$ , $Q$ , $R$ , $S$ in order). That suggests the $y$ -axis as a mirror line, so check the definition of a reflection.
$1 \to -1, \quad 4 \to -4, \quad 4 \to -4, \quad 1 \to -1$
Compare distances from the $y$ -axis. $P$ and $S$ are 1 unit to its right, and $P'$ and $S'$ are 1 unit to its left. $Q$ and $R$ are 4 units to its right, and $Q'$ and $R'$ are 4 units to its left. Each vertex kept its $y$ -coordinate, so it traveled horizontally, at a right angle to the vertical $y$ -axis.
Every vertex went straight across the $y$ -axis to the same distance on the other side. That matches the definition of a reflection exactly.

Answer

The motion is a reflection across the $y$ -axis. The shape of the image alone could not settle this, because sliding the rectangle 5 units left would land on the same region; the stated corresponding vertices decide it.

Now look more closely at translations. In the first figure, every vertex of triangle $ABC$ moved 6 units right and 3 units up. Moving right makes an $x$ -coordinate larger, so each $x$ gained 6. Moving up makes a $y$ -coordinate larger, so each $y$ gained 3.

You can write that whole slide as one short rule. Let $(x, y)$ stand for any point of the triangle. Then $(x, y) \to (x + 6, y + 3)$ , where the arrow is read "goes to". Test it on Point A: $(2, 1) \to (2 + 6, 1 + 3) = (8, 4)$ , which is exactly $A'$ .

A slide to the left makes $x$ smaller, and a slide down makes $y$ smaller, so those shifts are subtracted. For example, 4 units left and 2 units down is $(x, y) \to (x - 4, y - 2)$ .

Key concept

A translation adds the same shift to every point

Let $(x, y)$ be any point of a figure. Let $a$ be the horizontal shift and $b$ the vertical shift, both real numbers in grid units. A translation sends every point to $(x + a, y + b)$ .

$a > 0$ moves right, $a < 0$ moves left, $a = 0$ gives no sideways move.
$b > 0$ moves up, $b < 0$ moves down, $b = 0$ gives no up-or-down move.
If $a = 0$ and $b = 0$ , no point moves.

Find $a$ and $b$ from the exact coordinates of corresponding points (image minus original), never from how a drawing looks. The sign of a shift gives its direction; the distance moved is never negative. Every point gets the same $a$ and $b$ , so the horizontal and vertical gaps between any two points stay the same. That is why side lengths and angle measures do not change.

(x, y) \to (x + a, y + b)

Worked example

Translate a triangle with a rule

Triangle $DEF$ has vertices $D(-3, 2)$ , $E(1, 2)$ and $F(1, 5)$ . Translate it 2 units right and 6 units down. Find the vertices of the image, then check that side $DE$ keeps its length.

Turn the words into shifts. Right makes $x$ larger, so $a = 2$ . Down makes $y$ smaller, so $b = -6$ . Write the rule for any point $(x, y)$ of the triangle.
$(x, y) \to (x + 2, y - 6)$
Apply the rule to $D(-3, 2)$ : add 2 to the $x$ -coordinate and subtract 6 from the $y$ -coordinate.
$D' = (-3 + 2, 2 - 6) = (-1, -4)$
Apply the same rule to $E(1, 2)$ , because a translation uses the same shifts for every point.
$E' = (1 + 2, 2 - 6) = (3, -4)$
Apply the same rule to $F(1, 5)$ for the same reason.
$F' = (1 + 2, 5 - 6) = (3, -1)$
Check a length with the coordinates. $D$ and $E$ share $y = 2$ , so side $DE$ is horizontal and its length is the larger $x$ -coordinate minus the smaller one. $D'$ and $E'$ share $y = -4$ , so do the same there. Both are 4 units, because adding 2 to both $x$ -coordinates leaves their difference unchanged.
$1 - (-3) = 4, \qquad 3 - (-1) = 4$

Answer

The image has vertices $D'(-1, -4)$ , $E'(3, -4)$ and $F'(3, -1)$ . Side $D'E'$ is 4 units long, the same as side $DE$ .

Worked example

Find the rule from one pair of vertices

Parallelogram $JKLM$ has vertices $J(1, 2)$ , $K(5, 2)$ , $L(6, 5)$ and $M(2, 5)$ . A translation takes $J$ to $J'(-4, -2)$ . Write the rule for this translation, find $K'$ , $L'$ and $M'$ , and check that sides $J'K'$ and $M'L'$ are still parallel.

Find the shifts from the pair you know. Subtract image minus original, because a shift is what you add to the original coordinate to reach the image coordinate.
$a = -4 - 1 = -5, \qquad b = -2 - 2 = -4$
Read the signs: $a = -5$ means 5 units left and $b = -4$ means 4 units down. Write the rule for any point $(x, y)$ of the parallelogram.
$(x, y) \to (x - 5, y - 4)$
Apply the rule to $K(5, 2)$ . A translation uses the same shifts for every point.
$K' = (5 - 5, 2 - 4) = (0, -2)$
Apply it to $L(6, 5)$ and $M(2, 5)$ , again with the same shifts.
$L' = (6 - 5, 5 - 4) = (1, 1), \qquad M' = (2 - 5, 5 - 4) = (-3, 1)$
Check with the coordinates. $J'$ and $K'$ share $y = -2$ , and $M'$ and $L'$ share $y = 1$ , so both sides are horizontal and therefore parallel, just as $JK$ (on $y = 2$ ) and $ML$ (on $y = 5$ ) were. That had to happen: both ends of a horizontal side get the same $b$ . Each of the two sides is still 4 units long.
$0 - (-4) = 4, \qquad 1 - (-3) = 4$

Answer

The rule is $(x, y) \to (x - 5, y - 4)$ , a slide of 5 units left and 4 units down. The image has $J'(-4, -2)$ , $K'(0, -2)$ , $L'(1, 1)$ and $M'(-3, 1)$ , and sides $J'K'$ and $M'L'$ are still parallel, each 4 units long.

Translations, reflections and rotations are rigid motions: each moves a figure without changing any side length or angle measure. You tell them apart from the exact coordinates of corresponding vertices, not from the look of a drawing. A translation gives every vertex the same change in $x$ and the same change in $y$ . A reflection sends each point straight across a mirror line to the same distance on the other side. A rotation turns every point about a center through the same angle, at an unchanged distance from it.

You can now carry out a translation yourself with $(x, y) \to (x + a, y + b)$ , using the same shifts $a$ and $b$ for every point. Next, you will carry out reflections on the coordinate plane.