Math 1 Quiz: Transformation Sequences

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Transformation SequencesQuestion 1 of 8

A regular octagon is inscribed in a circle centered at the origin. One vertex is located at $(8, 0)$ . After applying a transformation sequence, this vertex maps to $(-4\sqrt{2}, -4\sqrt{2})$ . If the sequence consists of a rotation about the origin followed by a dilation centered at the origin, what is the angle of rotation?

135°

180°

225°

270°

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Math 1 Quiz

Math 1 Quiz: Transformation Sequences

Practice Transformation Sequences in Math 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Transformation Sequences, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$135°$
$180°$
$225°$ (correct answer)
$270°$

Explanation: The original vertex is at

(8, 0)

, which has angle

0°

from the positive

x

-axis. The final vertex is at

(-4\sqrt{2}, -4\sqrt{2})

. To find its angle:

\tan(\theta) = \frac{-4\sqrt{2}}{-4\sqrt{2}} = 1

, and since both coordinates are negative, the point is in the third quadrant. Therefore,

\theta = 180° + 45° = 225°

. The rotation angle is

225° - 0° = 225°

. We can verify: the distance from origin to the final point is

\sqrt{(-4\sqrt{2})^2 + (-4\sqrt{2})^2} = \sqrt{32 + 32} = 8

. After rotation by

225°

, we would have

(8\cos(225°), 8\sin(225°)) = (8 \cdot (-\frac{\sqrt{2}}{2}), 8 \cdot (-\frac{\sqrt{2}}{2})) = (-4\sqrt{2}, -4\sqrt{2})

. Since this matches the final position exactly, no dilation is needed (scale factor

1

Question 2

A transformation sequence consists of three steps applied to triangle $RST$ : (1) dilation by scale factor $k$ centered at the origin, (2) rotation of $\alpha$ degrees about the origin, (3) translation by vector $\vec{v}$ . If this sequence maps triangle $RST$ onto triangle $R'S'T'$ , and both triangles have the same area, what must be true about $k$ ?

$k = 1$ only
$k = -1$ only
$k = 1$ or $k = -1$ (correct answer)
$k$ can be any nonzero real number

Explanation: If the original triangle and final triangle have the same area, and the only transformation that can change area is dilation, then the dilation must preserve area. The area changes by a factor of

k^2

under dilation with scale factor

k

. For the areas to be equal:

k^2 \cdot \text{(original area)} = \text{(original area)}

, so

k^2 = 1

, which gives

k = \pm 1

. Note that

k = -1

corresponds to a dilation that includes a

180°

rotation, but this is still a valid dilation. Rotations and translations preserve area regardless of their parameters.

Question 3

Rhombus $WXYZ$ has vertices $W(-2, 3)$ , $X(1, 5)$ , $Y(4, 3)$ , and $Z(1, 1)$ . A transformation sequence maps this rhombus onto rhombus $W'X'Y'Z'$ where all vertices have integer coordinates and the rhombus has the same area but opposite orientation. Which property must be true about any such transformation sequence?

The sequence must contain exactly one reflection and an even number of rotations
The sequence must contain an odd number of reflections and any number of other isometries (correct answer)
The sequence must contain exactly two reflections and may include translations or rotations
The sequence must contain at least one dilation with scale factor $-1$ and one translation

Explanation: For the orientation to be opposite (reversed), the transformation sequence must include an odd number of orientation-reversing transformations. Reflections reverse orientation, while rotations, translations, and positive dilations preserve orientation. Since the area is preserved, we cannot have dilations with scale factors other than

\pm 1

. A dilation with scale factor

-1

is equivalent to a

180°

rotation (preserves orientation). Therefore, to reverse orientation, we need an odd number of reflections. We can include any number of orientation-preserving transformations (rotations, translations) without changing this requirement. Options A and C specify exact numbers incorrectly, and option D incorrectly suggests dilation is required.

Question 4

A regular pentagon is transformed by a sequence that first applies transformation $T_1$ , then transformation $T_2$ . If $T_1$ is a rotation of $72°$ counterclockwise about the center, and the overall sequence maps the pentagon onto itself, which of the following could be $T_2$ ?

Rotation of $288°$ counterclockwise about the center, followed by reflection across any diagonal
Rotation of $72°$ clockwise about the center, followed by rotation of $216°$ counterclockwise
Reflection across a line through the center and any vertex, followed by rotation of $144°$
Rotation of $288°$ counterclockwise about the center only (correct answer)

Explanation: Since

T_1

is a

72°

counterclockwise rotation and the overall sequence maps the pentagon onto itself, we need

T_2

to be the inverse of

T_1

. The inverse of a

72°

counterclockwise rotation is a

72°

clockwise rotation, which equals a

288°

counterclockwise rotation. Option A includes an extra reflection that would change the final result. Option B:

72°

clockwise

+ 216°

counterclockwise

= 144°

counterclockwise, which combined with

T_1

gives

72° + 144° = 216°

, not the identity. Option C involves reflection which changes orientation. Option D gives exactly

72° + 288° = 360° = 0°

, which is the identity transformation.

Question 5

Rectangle $EFGH$ undergoes a transformation sequence that maps it onto rectangle $E'F'G'H'$ . The sequence preserves all side lengths but reverses the orientation. Which of the following sequences is NOT possible?

Reflection across a vertical line, then reflection across a horizontal line (correct answer)
Rotation of $90°$ counterclockwise, then reflection across the line $y = x$
Reflection across the line $y = -x$ , then rotation of $180°$ about the origin
Translation by $(3, -2)$ , then reflection across the $x$ -axis

Explanation: For orientation to be reversed, the sequence must include an odd number of reflections. Option A: Two reflections preserve orientation (equivalent to a rotation), so orientation is preserved, not reversed. Option B: Rotation preserves orientation, reflection reverses it, so net effect reverses orientation. Option C: Reflection reverses orientation, rotation preserves it, so net effect reverses orientation. Option D: Translation preserves orientation, reflection reverses it, so net effect reverses orientation. Therefore, option A cannot reverse orientation.

Question 6

Triangle $ABC$ has vertices at $A(2, 4)$ , $B(6, 2)$ , and $C(4, 8)$ . After applying a sequence of transformations, the triangle maps onto triangle $A'B'C'$ with vertices at $A'(-4, -2)$ , $B'(-2, -6)$ , and $C'(-8, -4)$ . Which sequence of transformations could map triangle $ABC$ onto triangle $A'B'C'$ ?

Reflection across the $x$ -axis followed by rotation $90°$ counterclockwise about the origin
Rotation $180°$ about the origin followed by reflection across the line $y = x$ (correct answer)
Reflection across the $y$ -axis followed by rotation $90°$ clockwise about the origin
Rotation $90°$ clockwise about the origin followed by reflection across the $x$ -axis

Explanation: To find the correct sequence, we need to trace the transformation of the vertices. First, applying a 180° rotation about the origin to point A(2,4) gives A''(-2,-4). Then reflecting across y = x swaps coordinates to get A'(-4,-2). This matches the given final position. We can verify with the other vertices: B(6,2) → B''(-6,-2) → B'(-2,-6) ✓, and C(4,8) → C''(-4,-8) → C'(-8,-4) ✓. Choice A would give A'(-4,2), Choice C would give A'(-4,2), and Choice D would give A'(4,-2).

Question 7

A regular hexagon is transformed by a sequence of transformations. The sequence maps vertex $A$ to position $A'''$ , which coincides with the position of vertex $D$ in the original hexagon (where vertices are labeled consecutively $A, B, C, D, E, F$ ). Which of the following transformation sequences could produce this result?

Rotation $120°$ counterclockwise about the center followed by reflection across a line through the center
Rotation $60°$ clockwise about the center followed by rotation $180°$ about the center
Reflection across a diagonal followed by rotation $60°$ counterclockwise about the center
Rotation $180°$ about the center followed by rotation $60°$ clockwise about the center (correct answer)

Explanation: In a regular hexagon with vertices labeled consecutively A, B, C, D, E, F, vertex D is 3 positions away from vertex A, which corresponds to a 180° rotation about the center (since each vertex is separated by 60°, and 3 × 60° = 180°). Let's check option D: First, rotate A by 180° about the center, which moves A to position E (the vertex directly opposite). Then rotate by 60° clockwise, which moves the point from position E to position D. This gives the desired result. Option A would move A by 120° to position C, then reflect it, which could map to various positions depending on the reflection line. Option B: 60° clockwise moves A to F, then 180° moves F to C. Option C depends on which diagonal is chosen for reflection.

Question 8

An isosceles triangle undergoes a transformation sequence that maps it onto itself. The sequence consists of a reflection across line $m$ , followed by a rotation about point $P$ , followed by another reflection across line $n$ . If the triangle has exactly one line of symmetry, what can be concluded about the transformation sequence?

Line $m$ must be the line of symmetry, and the rotation angle must be $180°$
Lines $m$ and $n$ must be the same line, and point $P$ must lie on this line (correct answer)
The rotation angle must be $360°$ divided by the number of vertices in the triangle
Point $P$ must be the centroid of the triangle, and lines $m$ and $n$ must be perpendicular

Explanation: Since the triangle maps onto itself and has exactly one line of symmetry, the transformation sequence must be equivalent to either the identity transformation or a reflection across the line of symmetry. A sequence of reflection-rotation-reflection has a specific structure. If lines m and n are the same line, and point P lies on this line, then the sequence becomes: reflection across the line, rotation about a point on the line, reflection across the same line. The two reflections across the same line cancel each other out, leaving only the rotation about point P. For the triangle to map onto itself with only one line of symmetry, this rotation must be 180° about a point on the line of symmetry, or the rotation must be 0° (identity). Choice A is incorrect because m doesn't have to be the line of symmetry initially. Choice C is incorrect because the triangle is not a regular polygon. Choice D is incorrect because the lines don't need to be perpendicular, and P doesn't need to be the centroid.