Transformation Sequences - Math 1

1. 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 $A'B'C'$?
   Answer: Rotation $180°$ about the origin followed by reflection across the line $y = x$ 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).
2. 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?
   Answer: Lines $m$ and $n$ must be the same line, and point $P$ must lie on this line 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.
3. 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?
   Answer: $225°$ 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 $1$).
4. 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?
   Answer: Rotation $180°$ about the center followed by rotation $60°$ clockwise about the center 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.
5. 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 $k$?
   Answer: $k = 1$ or $k = -1$ 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.
6. 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?
   Answer: The sequence must contain an odd number of reflections and any number of other isometries 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.
7. 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 $T_2$?
   Answer: Rotation of $288°$ counterclockwise about the center only 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.
8. 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?
   Answer: Reflection across a vertical line, then reflection across a horizontal line 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.

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