Graph and Describe Congruence Transformation Sequences | Grade 8 Math | LevelKite

A transformation moves every point of a figure to a new point, and the new figure is called the image. Image points get prime marks: point $P$ moves to $P'$ (read P prime), and a second move takes $P'$ to $P''$ (P double prime). A sequence of transformations is a list of moves done in order, each one acting on the image left by the move before.

You'll work on a coordinate plane. Its horizontal number line is the x-axis, its vertical number line is the y-axis, and they cross at the origin, $(0, 0)$ . All lengths are in grid units. The moves in this lesson are translations (slides), reflections (flips across a line) and rotations (turns about a fixed point). They are called congruence transformations because none of them changes a length or an angle measure.

In each rule,

(x, y)

is any point of the figure, and the arrow points to where it goes.

Move	What it does	Coordinate rule
Translation	Slides every point $h$ units right and $k$ units up; a negative $h$ means left, a negative $k$ means down	$(x, y) \to (x + h, y + k)$
Reflection across the x-axis	Moves every point straight across the x-axis to the same distance on the other side	$(x, y) \to (x, -y)$
Reflection across the y-axis	Moves every point straight across the y-axis to the same distance on the other side	$(x, y) \to (-x, y)$
Rotation of $90^\circ$ counterclockwise about the origin	Turns every point a quarter-turn around $(0, 0)$ , opposite to a clock's hands, keeping its distance from $(0, 0)$	$(x, y) \to (-y, x)$

Why do these moves keep lengths and angles? Take any two points of a figure, $(x_1, y_1)$ and $(x_2, y_2)$ . The step from the first to the second is $a = x_2 - x_1$ units across and $b = y_2 - y_1$ units up, where a negative value means left or down.

Translation: $(x_2 + h) - (x_1 + h) = x_2 - x_1 = a$ , and in the same way $b$ stays $b$ . Every step is unchanged.
Reflection across the x-axis: $-y_2 - (-y_1) = -(y_2 - y_1) = -b$ , so the step becomes $a$ across, $-b$ up. Across the y-axis, it becomes $-a$ across, $b$ up.
Rotation: a quarter-turn counterclockwise turns right into up and up into left. The rule agrees: the new across change is $-y_2 - (-y_1) = -b$ , and the new up change is $x_2 - x_1 = a$ .

Each move keeps the sizes $|a|$ and $|b|$ , sometimes swapping them. A side's length depends only on those sizes, so every side keeps its length. All the steps in a figure change in the same way, so the shape is kept and angle measures stay the same. Each reflection or rotation rule holds only for its own mirror line, or its own center, angle and direction.

Key concept

Sequences of congruence transformations

Translations, reflections and rotations keep every length and angle measure, so the image after any sequence of them is congruent to the original figure. To graph a sequence, apply each move, in the order given, to every vertex. To describe a congruence, list moves in order, each stated exactly (how far and which way, which mirror line, or which center, angle and direction), so that every vertex lands on its corresponding vertex.

Worked example

Translate, then reflect

Triangle $PQR$ has vertices $P(-4, 3)$ , $Q(-1, 3)$ and $R(-4, 5)$ . Graph its image after this sequence: translate 5 units right and 2 units down, then reflect across the x-axis.

Write the translation as a rule. Moving 5 units right adds 5 to every x-coordinate, and moving 2 units down subtracts 2 from every y-coordinate, so $h = 5$ and $k = -2$ .
$(x, y) \to (x + 5,\ y - 2)$
Apply the rule to every vertex, because the whole triangle moves. This gives $\triangle P'Q'R'$ .
$P' = (-4 + 5,\ 3 - 2) = (1, 1),\quad Q' = (-1 + 5,\ 3 - 2) = (4, 1),\quad R' = (-4 + 5,\ 5 - 2) = (1, 3)$
Now reflect $\triangle P'Q'R'$ , not the original, across the x-axis. Each point goes straight across the axis to the same distance on the other side, so its x-coordinate stays and its y-coordinate changes sign. Plot the three new points and join them to graph the final image.
$(x, y) \to (x, -y):\quad P'' = (1, -1),\quad Q'' = (4, -1),\quad R'' = (1, -3)$
Check with exact coordinates. $PQ$ and $P''Q''$ are horizontal, since each pair of endpoints shares a y-coordinate, and both are 3 units long. $PR$ and $P''R''$ are vertical, since each pair shares an x-coordinate, and both are 2 units long. A horizontal side meets a vertical side at $P$ , and again at $P''$ , so the angles at $P$ and $P''$ are both right angles.
$PQ = -1 - (-4) = 3,\quad P''Q'' = 4 - 1 = 3,\quad PR = 5 - 3 = 2,\quad P''R'' = -1 - (-3) = 2$

Answer

The image is $\triangle P''Q''R''$ with vertices $P''(1, -1)$ , $Q''(4, -1)$ and $R''(1, -3)$ , and it is congruent to $\triangle PQR$ .

\triangle PQR

is at upper left,

\triangle P'Q'R'

(after the translation) is at upper right, and

\triangle P''Q''R''

is below the x-axis. The labels follow

P

:

(-4, 3)

, then

(1, 1)

, then

(1, -1)

.

Doing the moves in another order can change the image. Suppose you reflect $P(-4, 3)$ across the x-axis first, getting $(-4, -3)$ , and then translate, getting $(-4 + 5, -3 - 2) = (1, -5)$ instead of $P''(1, -1)$ . Here is why: the reflection changes the sign of whatever y-coordinate the point has when you reflect it. Reflecting second also flips the translation's 2 units down into 2 units up: $-(3 - 2) = -3 + 2 = -1$ . Reflecting first leaves those 2 units down for afterward: $-3 - 2 = -5$ . The wrong-order triangle is still congruent to $\triangle PQR$ , but it is not the image this sequence produces. So follow the order given, and state the order whenever you describe a sequence.

Worked example

Rotate, then translate

Triangle $JKL$ has vertices $J(1, 1)$ , $K(4, 1)$ and $L(4, 3)$ . Graph its image after a rotation of $90^\circ$ counterclockwise about the origin, followed by a translation 2 units left and 5 units down.

Use the rotation rule, which comes from turning a step: the point $(x, y)$ is $x$ across and $y$ up from the origin, and the quarter-turn makes that step $-y$ across and $x$ up. The rule works for any signs of $x$ and $y$ , but only for a $90^\circ$ counterclockwise turn about $(0, 0)$ .
$(x, y) \to (-y, x)$
Apply the rule to every vertex so the whole triangle turns: the new x-coordinate is the opposite of the old y-coordinate, and the new y-coordinate is the old x-coordinate.
$J(1, 1) \to J'(-1, 1),\quad K(4, 1) \to K'(-1, 4),\quad L(4, 3) \to L'(-3, 4)$
Translate $\triangle J'K'L'$ by subtracting 2 from each x-coordinate and 5 from each y-coordinate, because left and down are the negative directions.
$J'' = (-1 - 2,\ 1 - 5) = (-3, -4),\quad K'' = (-1 - 2,\ 4 - 5) = (-3, -1),\quad L'' = (-3 - 2,\ 4 - 5) = (-5, -1)$
Check lengths. $JK$ is horizontal and $J''K''$ is vertical, yet both are 3 units long; $KL$ is vertical and $K''L''$ is horizontal, yet both are 2 units long. The turn swapped directions but kept sizes, and a horizontal side still meets a vertical side, so the right angle at $K$ is a right angle at $K''$ too.
$JK = 4 - 1 = 3,\quad J''K'' = -1 - (-4) = 3,\quad KL = 3 - 1 = 2,\quad K''L'' = -3 - (-5) = 2$

Answer

The image is $\triangle J''K''L''$ with vertices $J''(-3, -4)$ , $K''(-3, -1)$ and $L''(-5, -1)$ , and it is congruent to $\triangle JKL$ .

The quarter-turn about

(0, 0)

takes

\triangle JKL

(upper right) to

\triangle J'K'L'

(upper left); the translation then gives

\triangle J''K''L''

(lower left). Labels follow

K

:

(4, 1)

,

(-1, 4)

,

(-3, -1)

.

To describe a sequence that shows congruence, you start with a figure and its image and find moves that carry one exactly onto the other. Rely on coordinates, not on how the figures look.

Use the congruence statement to pair each vertex with its corresponding vertex.
From one vertex, find the steps along two of its sides, and along the matching sides of the image. Use two sides, because one step alone can fit more than one move: a reflection across the x-axis leaves a horizontal step unchanged, just as a translation does.
If the steps changed, find the reflection or rotation that changes both steps that same way, and apply it to every vertex. If no single move fits, you need more than one.
Once the steps match, a translation finishes the job. Find it from one vertex and its partner, then check every vertex.

Other correct sequences may exist. Any sequence that sends every vertex to its partner shows the congruence.

\triangle ABC

has

A(1, 1)

,

B(4, 1)

and

C(2, 3)

; its congruent partner

\triangle DEF

has

D(-2, -3)

,

E(-5, -3)

and

F(-3, -1)

.

Worked example

Describe a sequence from coordinates

In the figure above, $\triangle ABC \cong \triangle DEF$ , with $A(1, 1)$ , $B(4, 1)$ , $C(2, 3)$ and $D(-2, -3)$ , $E(-5, -3)$ , $F(-3, -1)$ . Describe a sequence of transformations that maps $\triangle ABC$ onto $\triangle DEF$ .

The statement $\triangle ABC \cong \triangle DEF$ pairs $A$ with $D$ , $B$ with $E$ and $C$ with $F$ . Subtract coordinates to find the step along side $AB$ and along its partner, $DE$ .
$AB:\ 4 - 1 = 3 \text{ across},\ 1 - 1 = 0 \text{ up};\qquad DE:\ -5 - (-2) = -3 \text{ across},\ -3 - (-3) = 0 \text{ up}$
Do the same for side $AC$ and its partner, $DF$ . In both pairs the across change flipped sign while the up change stayed, and the sizes match (3 and 0, then 1 and 2), so the side lengths agree. Of the single moves in this lesson, only a reflection across the y-axis changes steps this way.
$AC:\ 2 - 1 = 1 \text{ across},\ 3 - 1 = 2 \text{ up};\qquad DF:\ -3 - (-2) = -1 \text{ across},\ -1 - (-3) = 2 \text{ up}$
Reflect $\triangle ABC$ across the y-axis, which keeps each y-coordinate and changes the sign of each x-coordinate.
$(x, y) \to (-x, y):\quad A' = (-1, 1),\quad B' = (-4, 1),\quad C' = (-2, 3)$
Now every step matches, so a translation finishes the job. Moving $A'(-1, 1)$ to its partner $D(-2, -3)$ takes $-2 - (-1) = -1$ across and $-3 - 1 = -4$ up, which is 1 unit left and 4 units down.
$(x, y) \to (x - 1,\ y - 4)$
Check every vertex, not just one. The translation sends $A'$ to $D$ , $B'$ to $E$ and $C'$ to $F$ , so the sequence carries the whole triangle onto $\triangle DEF$ .
$A' \to (-1 - 1,\ 1 - 4) = (-2, -3) = D,\quad B' \to (-4 - 1,\ 1 - 4) = (-5, -3) = E,\quad C' \to (-2 - 1,\ 3 - 4) = (-3, -1) = F$

Answer

Reflect $\triangle ABC$ across the y-axis, then translate the result 1 unit left and 4 units down. This sequence maps $\triangle ABC$ onto $\triangle DEF$ , which shows the triangles are congruent.

Takeaway: A sequence of congruence transformations moves a figure without changing its size or shape. To graph a sequence, send every vertex through each move in order. To describe one, name exact moves in order and use coordinates to check that every vertex reaches its partner. Next, you'll use these sequences to justify whether two figures are congruent.