Transformation Sequences and Justifying Similarity | Grade 8 Math | LevelKite

Two figures have the same shape when you can move, flip, turn and resize one so it fits exactly on the other. Transformations make that idea exact. A transformation is a rule that sends every point of a figure to a new point, called its image.

A translation slides every point the same distance in the same direction.
A reflection sends every point across a mirror line to the point the same distance from the line on the other side.
A rotation turns every point through the same angle about a fixed point.

These three are rigid motions: they keep every length and every angle the same. A dilation with centre $C$ and scale factor $k$ moves each point along the ray from $C$ through it until its distance from $C$ is $k$ times as large, while $C$ itself stays put. In this lesson $k$ is always positive. On every grid, the horizontal axis is the $x$ -axis and the vertical axis is the $y$ -axis, and lengths are in grid units.

Key concept

Similarity through transformations

Two figures are similar when a sequence of rigid motions and dilations maps one figure exactly onto the other. Rigid motions keep lengths and angles, and a dilation keeps angles while multiplying every length by its scale factor. So in similar figures, corresponding angles are equal and all corresponding side lengths share one ratio. A dilation with centre $(a, b)$ and scale factor $k > 0$ sends any point $(x, y)$ to the image given by this rule:

(x,\ y) \mapsto \big(a + k(x - a),\ b + k(y - b)\big)

Read the dilation rule in three moves. First, $x - a$ and $y - b$ are the point's horizontal and vertical offsets from the centre. Next, multiply both offsets by $k$ . Last, add the centre's coordinates back. For two points with $x$ -coordinates $x_1$ and $x_2$ , the new $x$ -coordinates differ by $\big(a + k(x_1 - a)\big) - \big(a + k(x_2 - a)\big) = k(x_1 - x_2)$ , and the same holds for $y$ , so every horizontal and vertical distance is multiplied by $k$ . Because $k > 0$ , each moved point stays on the same ray from the centre, and every length stays positive.

When the centre is the origin, $a = 0$ and $b = 0$ , so the rule becomes $(x, y) \mapsto (kx, ky)$ . Save that shortcut for a centre at the origin.

To describe a sequence, compare a pair of corresponding sides to find $k$ , check whether the figure is flipped or turned, pick rigid motions that fix its direction and position, and write the exact coordinates after every step.

Worked example

A translation, then a dilation about the origin

Triangle $ABC$ has vertices $A(-4, -2)$ , $B(-2, -2)$ and $C(-4, -1)$ . Triangle $DEF$ has vertices $D(3, 3)$ , $E(9, 3)$ and $F(3, 6)$ , with $A \to D$ , $B \to E$ and $C \to F$ . Describe a translation followed by a dilation centred at the origin that maps $ABC$ onto $DEF$ .

Find the scale factor from matching sides, subtracting the smaller coordinate from the larger so each length is positive. DE is 3 times AB, and DF = 6 - 3 = 3 is also 3 times AC = -1 - (-2) = 1. AB and DE are horizontal and run right; AC and DF are vertical and run up. The triangles face the same way, so no flip or turn is needed.
$k = \frac{DE}{AB} = \frac{9 - 3}{-2 - (-4)} = \frac{6}{2} = 3$
Plan the slide by working backwards. The dilation about the origin with k = 3 multiplies both coordinates by 3, so the point that lands on D(3, 3) is (1, 1). Moving A(-4, -2) to (1, 1) takes 5 units right and 3 units up.
$(-4 + 5,\ -2 + 3) = (1,\ 1)$
Translate with the rule (x, y) → (x + 5, y + 3). A translation adds the same amounts to every point, so differences between coordinates do not change: A'B' = 3 - 1 = 2 and A'C' = 2 - 1 = 1, as before, and the right angle stays.
$A' = (1, 1),\ B' = (-2 + 5,\ -2 + 3) = (3, 1),\ C' = (-4 + 5,\ -1 + 3) = (1, 2)$
Dilate about the origin with k = 3. The shortcut (3x, 3y) is valid here only because the centre is (0, 0). The images are exactly D, E and F. Each length is now 3 times as long, DE = 6 and DF = 3, and the sides at D are still horizontal and vertical, so the right angle is kept.
$A'' = (3 \cdot 1,\ 3 \cdot 1) = (3, 3),\ B'' = (3 \cdot 3,\ 3 \cdot 1) = (9, 3),\ C'' = (3 \cdot 1,\ 3 \cdot 2) = (3, 6)$

Answer

Slide triangle $ABC$ 5 units right and 3 units up, then dilate it about the origin with scale factor $3$ . The image is exactly triangle $DEF$ , so the two triangles are similar.

A(-4, -2) \to (1, 1) \to D(3, 3)

,

B(-2, -2) \to (3, 1) \to E(9, 3)

and

C(-4, -1) \to (1, 2) \to F(3, 6)

: slide, then dilate about

(0, 0)

with

k = 3

.

Worked example

A reflection, then a dilation about another centre

Triangle $PQR$ has vertices $P(-3, 3)$ , $Q(-1, 3)$ and $R(-3, 4)$ . Triangle $STU$ has vertices $S(5, 5)$ , $T(1, 5)$ and $U(5, 7)$ , with $P \to S$ , $Q \to T$ and $R \to U$ . Describe a sequence that maps $PQR$ onto $STU$ and ends with a dilation centred at $(1, 1)$ .

Compare matching sides: PQ = -1 - (-3) = 2 and ST = 5 - 1 = 4; PR = 4 - 3 = 1 and SU = 7 - 5 = 2. Both ratios are 2, so k = 2. Q is right of P, but T is left of S, while R and U are both above. The triangle is flipped left to right, so the sequence needs a reflection.
$\frac{ST}{PQ} = \frac{4}{2} = 2,\quad \frac{SU}{PR} = \frac{2}{1} = 2$
Work backwards to choose the mirror. The dilation doubles every offset from the centre (1, 1). S(5, 5) is offset (4, 4) from the centre, so the point that lands on S is offset half as much, (2, 2), which puts it at (3, 3). Reflecting P(-3, 3) across the y-axis gives exactly (3, 3), so no slide is needed.
$(1 + 4 \div 2,\ 1 + 4 \div 2) = (3,\ 3)$
Reflect across the y-axis, a vertical mirror line, with the rule (x, y) → (-x, y). Changing the sign of every x-coordinate keeps each horizontal distance, and the y-coordinates do not change, so lengths and angles stay the same: P'Q' = 3 - 1 = 2 and P'R' = 4 - 3 = 1.
$P'(3, 3),\quad Q'(1, 3),\quad R'(3, 4)$
Dilate about (1, 1) with k = 2. The centre is not the origin, so use the full rule: subtract the centre, double the offsets, then add the centre back.
$(x,\ y) \mapsto \big(1 + 2(x - 1),\ 1 + 2(y - 1)\big)$
Apply the rule. The offsets of P', Q' and R' from (1, 1) are (2, 2), (0, 2) and (2, 3). Doubling gives (4, 4), (0, 4) and (4, 6), and adding (1, 1) gives exactly S, T and U. Every length doubled, ST = 4 and SU = 2, and the sides at S are still horizontal and vertical, so the right angle is kept.
$P'' = (1 + 4,\ 1 + 4) = (5, 5),\ Q'' = (1 + 0,\ 1 + 4) = (1, 5),\ R'' = (1 + 4,\ 1 + 6) = (5, 7)$

Answer

Reflect triangle $PQR$ across the $y$ -axis, then dilate the image about $(1, 1)$ with scale factor $2$ . The image is exactly triangle $STU$ , so the two triangles are similar.

P(-3, 3) \to P'(3, 3) \to S(5, 5)

,

Q(-1, 3) \to Q'(1, 3) \to T(1, 5)

and

R(-3, 4) \to R'(3, 4) \to U(5, 7)

: reflect across the

y

-axis, then dilate about

(1, 1)

with

k = 2

.

Common mistake: using $(kx, ky)$ when the centre is not the origin. In the example above, the shortcut $(2x, 2y)$ would send $P'(3, 3)$ to $(6, 6)$ instead of $(5, 5)$ , and the whole triangle to $(6, 6)$ , $(2, 6)$ and $(6, 8)$ . That copy is the right size, but it sits 1 unit right and 1 unit up from $STU$ , so it is the wrong image. The shortcut doubles each point's distance from the origin, while this dilation doubles its distance from $(1, 1)$ . Expanding shows the gap: $a + k(x - a) = kx + (1 - k)a$ and $b + k(y - b) = ky + (1 - k)b$ . For $k \ne 1$ (when $k = 1$ , no point moves), both extra terms are $0$ only if $a = 0$ and $b = 0$ , so $(kx, ky)$ is correct only when the centre is the origin.

To justify whether two figures are similar, match their vertices in order around each figure and test two conditions:

every pair of corresponding angles is equal, and
every pair of corresponding sides gives the same ratio, new length $\div$ original length.

Side lengths are positive, so each ratio is defined. If both conditions hold, the figures are similar. A positive scale factor multiplies every length by the same number, so it keeps shorter sides shorter: compare shortest with shortest and longest with longest. If two of those ratios differ, the figures are not similar, even when all their angles are equal, because any sequence of rigid motions and dilations multiplies every length by one common number. Take every measure from exact coordinates or stated values, never from how a picture looks.

Lower-left corners are labelled:

M

at

(1, 1)

,

N

at

(-8, -6)

and

K

at

(2, -6)

. Upper-right corners:

M

at

(4, 3)

,

N

at

(-2, -2)

and

K

at

(8, -3)

.

Worked example

Justifying with angles and side ratios

Rectangle $M$ has vertices $(1, 1)$ , $(4, 1)$ , $(4, 3)$ and $(1, 3)$ . Rectangle $N$ has vertices $(-8, -6)$ , $(-2, -6)$ , $(-2, -2)$ and $(-8, -2)$ . Rectangle $K$ has vertices $(2, -6)$ , $(8, -6)$ , $(8, -3)$ and $(2, -3)$ . Is $N$ similar to $M$ ? Is $K$ similar to $M$ ? Justify each answer.

Check the angles. In each rectangle, every side is horizontal (its ends share a y-coordinate) or vertical (its ends share an x-coordinate), and each corner joins one of each. A horizontal side and a vertical side meet at exactly 90°, so all corresponding angles are equal. This comes from the coordinates, not from how the corners look.
Find the side lengths in grid units, subtracting the smaller coordinate from the larger so each length is positive. For each rectangle, the longer side is listed first.
$M: 4 - 1 = 3,\ 3 - 1 = 2;\quad N: -2 - (-8) = 6,\ -2 - (-6) = 4;\quad K: 8 - 2 = 6,\ -3 - (-6) = 3$
Compare N with M, longer side with longer side and shorter with shorter. Both ratios are exactly 2, and the angles are equal, so N is similar to M.
$\frac{6}{3} = 2,\quad \frac{4}{2} = 2$
Confirm with a sequence. Dilate M about the origin with k = 2, which doubles both coordinates: (1, 1), (4, 1), (4, 3) and (1, 3) become (2, 2), (8, 2), (8, 6) and (2, 6). Then translate 10 units left and 8 units down to get (-8, -6), (-2, -6), (-2, -2) and (-8, -2), exactly the vertices of N.
$(x,\ y) \mapsto (2x,\ 2y) \mapsto (2x - 10,\ 2y - 8)$
Compare K with M. The longer sides give 2, but the shorter sides give exactly 1.5. The ratios differ, so K is not similar to M, even though every angle is 90°: a sequence would have to multiply the long sides by 2 and the short sides by 1.5, yet every sequence multiplies all lengths by one number.
$\frac{6}{3} = 2 \ne \frac{3}{2} = 1.5$

Answer

Rectangle $N$ is similar to rectangle $M$ : every angle is $90^\circ$ and every side is $2$ times as long. Rectangle $K$ is not similar to $M$ , because its side ratios, $2$ and $1.5$ , are not equal.

Takeaway: Two figures are similar when a sequence of rigid motions and dilations, each dilation with a stated centre and a positive scale factor, maps one exactly onto the other. Check every step with exact coordinates: use $(kx, ky)$ only when the centre is the origin, and $\big(a + k(x - a), b + k(y - b)\big)$ otherwise. To justify similarity, show equal corresponding angles and one common side ratio; two corresponding side ratios that differ show the figures are not similar.