Describe, Verify and Correct Rigid Motions | Grade 8 Math | LevelKite

You already know how to write the rule for a transformation. Now you will work the other way. You start with a figure, which is the original shape, and its image, which is the shape after a transformation. You will find the motion, check what it keeps and fix graphs that go wrong. Always reason from exact coordinates, not from how a drawing looks.

Each vertex of the figure has a corresponding vertex in the image, which is the point it lands on. In this lesson a rigid motion is a translation, a reflection or a rotation. Each one moves every point by a single rule:

A translation slides every point the same distance in the same direction.
A reflection flips every point across a mirror line. Each point moves along a path at a right angle to the line and lands the same distance from the line on the other side.
A rotation turns every point through the same angle, in the same direction, around one fixed point called the centre.

To describe a transformation from a figure and its image, find one rule that sends every vertex to its corresponding vertex. Then state all of its details:

for a translation, the shift, such as 2 units left and 5 units up;
for a reflection, the exact mirror line;
for a rotation, the centre, the angle and the direction, clockwise or counterclockwise.

Test the rule on every vertex. One vertex can fit more than one rule, so a rule that works for one vertex is only a guess until it works for all of them.

Worked example

Example 1: Describe a translation

Triangle ABC has vertices $A(-5, 1)$ , $B(-2, 1)$ and $C(-2, 3)$ . Its image is triangle DEF with $D(1, -2)$ , $E(4, -2)$ and $F(4, 0)$ . D, E and F correspond to A, B and C. Describe the transformation.

Start with A and its corresponding vertex D. Subtract the coordinates of A from those of D to see how far A moved across and up.
$x:\ 1 - (-5) = 6, \qquad y:\ -2 - 1 = -3$
A translation moves every point by the same shift. So the only translation that fits A is 6 units right and 3 units down. Test that rule on B.
$B(-2, 1) \to (-2 + 6,\ 1 - 3) = (4, -2) = E$
Test the rule on C too, because one rule must work for every vertex. C lands on F, so the rule fits all three vertices.
$C(-2, 3) \to (-2 + 6,\ 3 - 3) = (4, 0) = F$

Answer

Triangle DEF is the image of triangle ABC under a translation of 6 units right and 3 units down, with rule $(x, y) \to (x + 6, y - 3)$ .

Triangle ABC has

A(1, 1)

,

B(4, 1)

and

C(4, 3)

. Its image, triangle DEF, has

D(-1, 1)

,

E(-1, 4)

and

F(-3, 4)

.

Worked example

Example 2: Describe a rotation

In the figure, triangle DEF is the image of triangle ABC. D, E and F correspond to A, B and C. The vertices are $A(1, 1)$ , $B(4, 1)$ , $C(4, 3)$ , $D(-1, 1)$ , $E(-1, 4)$ and $F(-3, 4)$ . Describe the transformation.

Test for a translation. A translation shifts every vertex by the same amount. Here A moves by (−2, 0) but B moves by (−5, 3). The shifts differ, so this is not a translation.
$A \to D:\ (-1 - 1,\ 1 - 1) = (-2, 0), \qquad B \to E:\ (-1 - 4,\ 4 - 1) = (-5, 3)$
Vertex A alone fits two rules you know. One is the reflection across the y-axis, (x, y) → (−x, y). The other is the 90° counterclockwise rotation about the origin, (x, y) → (−y, x). With x = 1 and y = 1, both rules give (−1, 1), which is D. One vertex cannot decide between them.
$A(1, 1):\quad (-x,\ y) = (-1, 1), \quad (-y,\ x) = (-1, 1)$
Test both rules on B(4, 1). The reflection gives (−4, 1), which is not E, so it is ruled out. The rotation gives (−1, 4), which is E.
$B(4, 1):\quad (-x,\ y) = (-4, 1) \ne E, \quad (-y,\ x) = (-1, 4) = E$
Test the rotation rule on the last vertex, C(4, 3). It gives (−3, 4), which is F, so the rule works for every vertex.
$C(4, 3):\quad (-y,\ x) = (-3, 4) = F$
Check the meaning. From the origin, B is 4 right and 1 up, and E is 4 up and 1 left. A counterclockwise quarter turn (90°) sends right to up and up to left. So the rule matches a 90° counterclockwise turn about the origin.

Answer

Triangle DEF is the image of triangle ABC under a rotation of $90^\circ$ counterclockwise about the origin $(0, 0)$ , with rule $(x, y) \to (-y, x)$ .

Key concept

One rule for every vertex

A rigid motion moves every point by one rule. It always keeps every length, every angle measure and every pair of parallel lines. To describe or check a motion, find the rule that sends every vertex to its corresponding vertex and state its details. Matching lengths and angles show only that the image has the same size and shape. Checking the rule vertex by vertex shows which motion it is.

Rigid motions always keep every length, every angle measure and every pair of parallel lines. Parallel lines are lines in the same plane that never meet. When sides are horizontal or vertical, you can verify this exactly. For two vertices $(x_1, y_1)$ and $(x_2, y_2)$ , the step from the first to the second is $(a, b)$ , where $a = x_2 - x_1$ and $b = y_2 - y_1$ . A side joins two different points, so its step is never $(0, 0)$ . Lengths are in grid units.

A horizontal side has $b = 0$ and length $|a|$ . A vertical side has $a = 0$ and length $|b|$ . The absolute value is a number's size without its sign, so a length is never negative.
Where a horizontal side meets a vertical side, they form a right angle, $90^\circ$ , because grid lines cross at right angles.
Two horizontal sides at different heights never meet, so they are parallel. Two vertical sides at different $x$ -values are parallel for the same reason.

What each rule does to the step (a, b) between two vertices

Rigid motion	Rule	New step	Why
Translation $h$ units right and $k$ units up (negative values mean left or down)	$(x, y) \to (x + h,\ y + k)$	$(a, b)$ , unchanged	$(x_2 + h) - (x_1 + h) = a$ and $(y_2 + k) - (y_1 + k) = b$
Reflection across the vertical line $x = c$ (same distance, other side)	$(x, y) \to (2c - x,\ y)$	$(-a, b)$	new $x$ -change: $(2c - x_2) - (2c - x_1) = -a$ ; $y$ -change unchanged
Rotation $90^\circ$ counterclockwise about the origin	$(x, y) \to (-y,\ x)$	$(-b, a)$	new $x$ -change: $(-y_2) - (-y_1) = -b$ ; new $y$ -change: $x_2 - x_1 = a$

Read down the New step column. Each of these rules keeps the sizes $|a|$ and $|b|$ of every step. It only changes their signs or swaps the two parts. For horizontal and vertical sides, that is enough to show all three properties under these rules:

A horizontal or vertical side lands on a horizontal or vertical side of the same length. For example, the rotation sends the step $(4, 0)$ to $(0, 4)$ , so a horizontal side 4 units long becomes a vertical side 4 units long.
Where a horizontal side meets a vertical side, the images are again one horizontal side and one vertical side, so the angle stays $90^\circ$ .
Two horizontal sides at different heights land on two horizontal sides or two vertical sides at different positions, so they stay parallel. The same holds for two vertical sides.

Rectangle ABCD and its image EFGH under

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

. The unlabelled corners are

D(1, 3)

and its image

H(-3, 1)

.

Worked example

Example 3: Verify what a rotation keeps

The figure shows rectangle ABCD with $A(1, 1)$ , $B(5, 1)$ , $C(5, 3)$ and $D(1, 3)$ . The rotation from Example 2, $90^\circ$ counterclockwise about the origin, sends it to EFGH. E, F, G and H correspond to A, B, C and D. Use exact coordinates to verify that the rotation keeps the side lengths, the angle measures and the parallel sides.

Apply the rule (x, y) → (−y, x) to each vertex. This way every image point comes from the rule, not from how the graph looks.
$A(1, 1) \to E(-1, 1),\ B(5, 1) \to F(-1, 5),\ C(5, 3) \to G(-3, 5),\ D(1, 3) \to H(-3, 1)$
Find the side lengths of ABCD. Every side is horizontal or vertical, so subtract the coordinates that change. Then take the absolute value, because a length is never negative.
$AB = |5 - 1| = 4,\ BC = |3 - 1| = 2,\ CD = |1 - 5| = 4,\ DA = |1 - 3| = 2$
Find the matching side lengths of EFGH the same way. Each one equals its original side, even though the horizontal side AB became the vertical side EF.
$EF = |5 - 1| = 4,\ FG = |-3 - (-1)| = 2,\ GH = |1 - 5| = 4,\ HE = |-1 - (-3)| = 2$
Check the angles. Each corner of ABCD joins a horizontal side and a vertical side, so it measures 90°. The rotation turned the horizontal sides vertical and the vertical sides horizontal. So each corner of EFGH also joins one side of each kind and still measures 90°.
$\angle A = \angle E,\ \angle B = \angle F,\ \angle C = \angle G,\ \angle D = \angle H,\ \text{each } 90^\circ$
Check the parallel sides. AB and DC lie on the horizontal lines y = 1 and y = 3, which never meet, so they are parallel. Their images EF and HG lie on the vertical lines x = −1 and x = −3, so they are parallel too. Likewise, BC and AD on x = 5 and x = 1 become FG and EH on y = 5 and y = 1.
$AB \parallel DC \to EF \parallel HG, \qquad BC \parallel AD \to FG \parallel EH$

Answer

The rotation keeps every measure. Matching sides are $4$ , $2$ , $4$ and $2$ units long, every angle stays $90^\circ$ , and both pairs of parallel sides stay parallel, even though the horizontal and vertical sides traded directions.

Worked example

Example 4: Find and fix a graphing error

A student was asked to reflect triangle ABC, with $A(3, 1)$ , $B(6, 1)$ and $C(3, 4)$ , across the line $x = 2$ . The student graphed triangle DEF with $D(-3, 1)$ , $E(-6, 1)$ and $F(-3, 4)$ . D, E and F correspond to A, B and C. Find the error and correct the graph.

Write the requested rule. The mirror line x = 2 is vertical, so a path at a right angle to it is horizontal and each point keeps its y-coordinate. A point at x lies x − 2 units right of the line, where a negative value means left. Its image lies the same distance on the other side.
$(x, y) \to (2 - (x - 2),\ y) = (4 - x,\ y)$
Apply the rule to every vertex to build the correct image. Call it triangle PQR, where P, Q and R correspond to A, B and C.
$A \to (4 - 3,\ 1) = P(1, 1),\ B \to (4 - 6,\ 1) = Q(-2, 1),\ C \to (4 - 3,\ 4) = R(1, 4)$
Compare the student's graph with the correct image. None of D, E and F matches P, Q and R. Each student vertex keeps the original y and takes the opposite of the original x. That is the rule (x, y) → (−x, y), a reflection across the y-axis (the line x = 0), not across x = 2.
$A(3, 1) \to D(-3, 1),\ B(6, 1) \to E(-6, 1),\ C(3, 4) \to F(-3, 4)$
Confirm with distances to the mirror line x = 2. A is 1 unit right of the line. The correct P is 1 unit left of it, as a reflection requires. The student's D is 5 units left of it.
$A:\ |3 - 2| = 1, \qquad P:\ |1 - 2| = 1, \qquad D:\ |-3 - 2| = 5$
See why measuring did not catch the error. AB and DE are both 3 units, AC and DF are both 3 units, and the angles at A and D are both 90°. A reflection across the y-axis is also a rigid motion, so it keeps every length and angle. Equal measures cannot tell one rigid motion from another. Only checking the rule vertex by vertex can.
$AB = |6 - 3| = 3 = |-6 - (-3)| = DE, \qquad AC = |4 - 1| = 3 = DF$

Answer

The student reflected across the $y$ -axis instead of across the line $x = 2$ . The corrected image is triangle PQR with $P(1, 1)$ , $Q(-2, 1)$ and $R(1, 4)$ . Each of its vertices is the same distance from $x = 2$ as its original vertex, on the other side.

Right: triangle ABC, with A at

(3, 1)

. Far left: the student's triangle DEF, with D at

(-3, 1)

. Middle: the corrected triangle PQR, with P at

(1, 1)

. The vertical line is

x = 2

.

Takeaway: Work vertex by vertex with exact coordinates. To describe a motion, find the one rule that sends every vertex to its corresponding vertex. Then give its details: the shift, the mirror line, or the centre, angle and direction. Every rigid motion keeps lengths, angle measures and parallel lines, so matching measures confirm only the same size and shape. To confirm or correct a graph, apply the requested rule to every vertex and compare.