Geometric transformations

This chapter rigorously examines fundamental geometric transformations, including translation, reflection, and rotation, alongside principles of symmetry-based reasoning. Mastery of these concepts is crucial for solving advanced geometry problems and forms a significant component of the CMI examination.

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Chapter Contents

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|---|-------| | 1 | Translation | | 2 | Reflection | | 3 | Rotation | | 4 | Symmetry-based reasoning |

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We begin with Translation.

Part 1: Translation

Translation

Overview

Translation is the simplest rigid transformation in geometry: every point moves through the same distance in the same direction. Even though the idea is simple, exam problems use translation in coordinate geometry, vector form, locus arguments, symmetry reasoning, and transformation-based simplification. In CMI-style questions, the important skill is to see translation as a rule on all points, not as motion of a single figure. ---

Learning Objectives

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Core Idea

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Coordinate Form

Examples:

• by $(3,-2)$:

$\qquad (x,y)\mapsto (x+3,y-2)$

• by $(-1,5)$:

$\qquad (x,y)\mapsto (x-1,y+5)$ ::: ---

Vector Interpretation

This makes translation especially useful in coordinate geometry and proofs involving parallelograms. ---

What Translation Preserves

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Translation of Common Objects

This shows that translation sends a line to a parallel line. ---

Translation of a Curve

This substitution rule is extremely important in coordinate geometry. ---

Composition of Translations

So translations combine like vectors. ::: ---

Translation and Locus

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Minimal Worked Examples

Example 1 Translate the point $(2,-3)$ by the vector $(4,5)$. Using the formula, $\qquad (2,-3)\mapsto (2+4,-3+5)=(6,2)$ --- Example 2 Find the image of the line $\qquad x+y-1=0$ under translation by $(2,-3)$. Replace $\qquad x\to x-2,\ y\to y+3$ So, $\qquad (x-2)+(y+3)-1=0$ $\qquad x+y=0$ Hence the image line is $\qquad x+y=0$ ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The image of the point $(1,2)$ under translation by the vector $(3,-4)$ is" options=["$(4,-2)$","$(-2,6)$","$(3,-4)$","$(4,6)$"] answer="A" hint="Add the vector componentwise." solution="Under translation by $(3,-4)$, $\qquad (1,2)\mapsto (1+3,2-4)=(4,-2)$ Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="A point $(5,-1)$ is translated by the vector $(-2,7)$. Find the $y$-coordinate of the image." answer="6" hint="Only the second coordinate is needed." solution="The image is $\qquad (5-2,\ -1+7)=(3,6)$ So the required $y$-coordinate is $\boxed{6}$." ::: :::question type="MSQ" question="Which of the following are preserved under a translation?" options=["Distance between two points","Angle between two lines","Parallelism","A line's slope"] answer="A,B,C,D" hint="Translation is a rigid motion." solution="Translation preserves all Euclidean lengths and angles, and it sends every line to a parallel line. Hence slope is preserved as well. Therefore the correct answer is $\boxed{A,B,C,D}$." ::: :::question type="SUB" question="Find the image of the line $2x-y+3=0$ under translation by the vector $(1,2)$." answer="$2x-y+3=0 \mapsto 2x-y-1=0$" hint="Use $x\to x-1$ and $y\to y-2$ in the original equation." solution="To find the image equation under translation by $(1,2)$, replace $\qquad x\to x-1,\quad y\to y-2$ in the original equation: $\qquad 2(x-1)-(y-2)+3=0$ $\qquad 2x-2-y+2+3=0$ $\qquad 2x-y+3=0$ Wait carefully: simplify again: $\qquad 2(x-1)-(y-2)+3 = 2x-2-y+2+3 = 2x-y+3$ So in fact the equation remains the same. This means the line is invariant under this translation. Hence the image is $\boxed{2x-y+3=0}$." ::: ---

Summary

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Part 2: Reflection

Reflection

Overview

Reflection is a rigid transformation that flips a figure across a line, called the mirror line or axis of reflection. In geometry problems, reflection is a very strong tool because it preserves distances and angles while converting difficult path or locus problems into simpler straight-line arguments. In CMI-style questions, reflection is often used in coordinate geometry, symmetry, optimization, and proof problems. ---

Learning Objectives

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Core Idea

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Standard Reflections in Coordinate Geometry

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Reflection in Vertical and Horizontal Lines

These formulas come from the midpoint condition relative to the mirror line. ::: ---

What Reflection Preserves

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Reflection of Equations

If a figure remains unchanged under a reflection, it has symmetry about that line. ::: ---

Reflection as a Problem-Solving Tool

A broken path often becomes a straight segment after reflecting one part of the diagram. ---

Minimal Worked Examples

Example 1 Reflect the point $(3,-2)$ in the $x$-axis. Using $\qquad (x,y)\mapsto (x,-y)$, we get $\qquad (3,-2)\mapsto (3,2)$ --- Example 2 Reflect the point $(4,1)$ in the line $x=2$. Using $\qquad (x,y)\mapsto (2a-x,y)$ with $a=2$, $\qquad (4,1)\mapsto (4-4,1)=(0,1)$ More carefully, $\qquad 2a-x = 4-4 = 0$ So the image is $(0,1)$. ---

Symmetry Tests

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The reflection of the point $(2,-5)$ in the $y$-axis is" options=["$(-2,-5)$","$(2,5)$","$(-2,5)$","$(5,-2)$"] answer="A" hint="Only the $x$-coordinate changes sign." solution="Reflection in the $y$-axis sends $\qquad (x,y)\mapsto (-x,y)$ So $\qquad (2,-5)\mapsto (-2,-5)$ Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="A point $(7,3)$ is reflected in the line $x=4$. Find the $x$-coordinate of the image." answer="1" hint="Use $x\mapsto 2a-x$ for reflection in $x=a$." solution="For reflection in the line $x=4$, $\qquad x\mapsto 2\cdot 4 - x = 8-x$ So the image of $(7,3)$ has $x$-coordinate $\qquad 8-7=1$ Therefore the answer is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following are preserved under reflection?" options=["Distance","Angle measure","Area","Orientation"] answer="A,B,C" hint="Reflection is a rigid motion but not orientation-preserving." solution="Reflection preserves distances, angles, and area. But it reverses orientation. Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Find the image of the point $(3,5)$ under reflection in the line $y=x$." answer="$(5,3)$" hint="Swap the coordinates." solution="Reflection in the line $y=x$ sends $\qquad (x,y)\mapsto (y,x)$ Therefore $\qquad (3,5)\mapsto (5,3)$ Hence the required image is $\boxed{(5,3)}$." ::: ---

Summary

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Part 3: Rotation

Rotation

Overview

Rotation is one of the central rigid motions in geometry. It preserves distances, angles, collinearity, and shape, but changes direction and position relative to the axes. In CMI-style problems, rotation is often tested together with coordinates, polar form, matrices, composition of transformations, and especially the interaction between rotations and reflections. ---

Learning Objectives

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Core Idea

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Rotation in Cartesian Coordinates

This follows directly from angle addition formulas. ---

Rotation in Polar Form

This is often the cleanest form in theoretical problems. ---

Basic Examples

Example 1 $\qquad R_{90^\circ}(1,0) = (0,1)$ because the point $(1,0)$ is turned by $90^\circ$ counterclockwise. --- Example 2 $\qquad R_{180^\circ}(x,y)=(-x,-y)$ because a half-turn reverses the direction of the vector from the origin. ---

Rotation Matrix

This is especially useful in composition questions. ---

Reflection Across a Line Through the Origin

This is one of the most important formulas in this topic. ---

Special Reflection Cases

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Composition of Rotations

In particular, $\qquad (R_\alpha)^n = R_{n\alpha}$ ---

Composition of Reflections

This is a high-value result and appears directly in advanced exam problems. ---

Why This Formula Is True

Using the angle form of reflection:

• first reflection sends $\theta$ to $2\beta-\theta$

• second reflection sends that to

$\qquad 2\alpha-(2\beta-\theta)=\theta+2(\alpha-\beta)$ So the net effect is exactly a rotation by angle $2(\alpha-\beta)$. Hence, $\qquad F_\alpha \circ F_\beta = R_{2(\alpha-\beta)}$ ---

PYQ-Relevant Consequences

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Worked PYQ-Type Example

Example Let $\qquad G = (F_{20^\circ}\circ F_{25^\circ})^9$ Then $\qquad F_{20^\circ}\circ F_{25^\circ} = R_{2(20^\circ-25^\circ)} = R_{-10^\circ}$ So $\qquad G = (R_{-10^\circ})^9 = R_{-90^\circ}$ Thus $G$ is rotation by $90^\circ$ clockwise. For a point $(x,y)$, $\qquad R_{-90^\circ}(x,y)=(y,-x)$ Hence for $P=(20,25)$, $\qquad G(P)=(25,-20)$ This is exactly the kind of reduction exam questions expect. ---

Invariants Under Rotation

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The image of $(1,0)$ under counterclockwise rotation by $90^\circ$ about the origin is" options=["$(0,1)$","$(0,-1)$","$(1,0)$","$(-1,0)$"] answer="A" hint="A $90^\circ$ counterclockwise turn sends the positive $X$-axis to the positive $Y$-axis." solution="A counterclockwise rotation by $90^\circ$ sends $(1,0)$ to $(0,1)$. Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="Find $R_{180^\circ}(3,-5)$." answer="(-3,5)" hint="A half-turn changes $(x,y)$ to $(-x,-y)$." solution="Rotation by $180^\circ$ sends $\qquad (x,y)\mapsto(-x,-y)$. So $\qquad R_{180^\circ}(3,-5)=(-3,5)$. Hence the answer is $\boxed{(-3,5)}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$R_\alpha(r\cos\theta,r\sin\theta)=(r\cos(\theta+\alpha),r\sin(\theta+\alpha))$","$F_\alpha(r\cos\theta,r\sin\theta)=(r\cos(2\alpha-\theta),r\sin(2\alpha-\theta))$","$R_\alpha\circ R_\beta = R_{\alpha+\beta}$","$F_\alpha\circ F_\beta = R_{\alpha-\beta}$"] answer="A,B,C" hint="The composition of two reflections doubles the angle difference." solution="1. True. Rotation adds $\alpha$ to the polar angle.

True. Reflection in a line at angle $\alpha$ sends $\theta$ to $2\alpha-\theta$.

True. Rotations about the same center add their angles.

False. The correct formula is

$\qquad F_\alpha\circ F_\beta = R_{2(\alpha-\beta)}$. Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Geometrically describe $F_\alpha\circ F_\beta$ for arbitrary angles $\alpha$ and $\beta$." answer="Rotation by $2(\alpha-\beta)$ about the origin" hint="Track what happens to a point of polar angle $\theta$." solution="Let a point have polar form $\qquad (r\cos\theta,r\sin\theta)$. Reflection in the line at angle $\beta$ sends $\theta$ to $\qquad 2\beta-\theta$. Applying reflection in the line at angle $\alpha$ next sends this to $\qquad 2\alpha-(2\beta-\theta)=\theta+2(\alpha-\beta)$. Thus the overall effect is to add the angle $2(\alpha-\beta)$ while preserving the distance from the origin. Therefore $\qquad F_\alpha\circ F_\beta$ is rotation by angle $\qquad 2(\alpha-\beta)$ about the origin. Hence the answer is $\boxed{\text{Rotation by }2(\alpha-\beta)\text{ about the origin}}$." ::: ---

Summary

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Part 4: Symmetry-based reasoning

Symmetry-Based Reasoning

Overview

Symmetry-based reasoning is the art of solving geometry problems by identifying mirror structure, equal halves, repeated angles, or invariant distances under transformations. Instead of long coordinate or algebraic work, symmetry lets us prove equal lengths, equal angles, midpoint facts, locus facts, and shortest-path properties very quickly. In exam-level geometry, this topic often appears in reflections, isosceles configurations, regular polygons, and locus problems. ---

Learning Objectives

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Core Idea

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Reflection Symmetry

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Axis of Symmetry in Common Figures

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Symmetry and Equal Distances

These are among the most important symmetry-based loci in geometry. ---

Isosceles Triangle as a Symmetric Figure

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Rotational Symmetry

This helps in angle and congruence arguments. ---

Mirror Method

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Minimal Worked Examples

Example 1 A point $P$ is reflected in a line $\ell$ to $P'$. If $PP'=14$, find the distance from $P$ to $\ell$. Since the line of reflection is the perpendicular bisector of $PP'$, the distance from $P$ to $\ell$ is half of $PP'$. So the distance is $\qquad \dfrac{14}{2}=7$ Hence the answer is $\boxed{7}$. --- Example 2 Why is the perpendicular bisector of $AB$ the locus of points equidistant from $A$ and $B$? If a point lies on the perpendicular bisector, the two triangles formed with the endpoints are mirror images, so the distances to $A$ and $B$ are equal. Conversely, if a point is equidistant from $A$ and $B$, symmetry forces it to lie on the perpendicular bisector. ---

Common Patterns in Questions

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The locus of points equidistant from two fixed points $A$ and $B$ is" options=["the line $AB$","the perpendicular bisector of $AB$","the angle bisector at $A$","a circle with center $A$"] answer="B" hint="Use mirror symmetry of the endpoints." solution="A point is equidistant from $A$ and $B$ exactly when it lies on the perpendicular bisector of $AB$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="A point $P$ is reflected in a line $\ell$ to $P'$. If $PP'=18$, find the distance from $P$ to $\ell$." answer="9" hint="The mirror line bisects $PP'$ perpendicularly." solution="The reflecting line $\ell$ is the perpendicular bisector of $PP'$. So the distance from $P$ to $\ell$ is half of $PP'$: $\qquad \dfrac{18}{2}=9$ Therefore the answer is $\boxed{9}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["Reflection preserves distances","Every point on the axis of reflection remains fixed","If $P$ reflects to $P'$ in a line $\ell$, then $\ell$ is the perpendicular bisector of $PP'$","Every triangle has a line of symmetry"] answer="A,B,C" hint="Think about what reflection does and does not guarantee." solution="1. True. Reflection is an isometry.

True. Points on the axis of reflection do not move.

True. This is the defining geometric property of reflection.

False. Only some triangles, such as isosceles and equilateral triangles, have a reflection symmetry line.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="In an isosceles triangle $ABC$ with $AB=AC$, prove that the median from $A$ to $BC$ is perpendicular to $BC$." answer="The median from $A$ to $BC$ is also the altitude." hint="Use symmetry of the isosceles triangle." solution="Let $M$ be the midpoint of $BC$. Since $AB=AC$, the triangle is symmetric about the line $AM$. Under this symmetry:

• point $B$ maps to $C$

• the line $AM$ remains fixed

• the midpoint $M$ of $BC$ lies on the symmetry axis

Therefore the segment $BC$ is mapped onto itself with $B$ and $C$ interchanged, so the axis of symmetry $AM$ must be the perpendicular bisector of $BC$. Hence $\qquad AM\perp BC$ So the median from $A$ to $BC$ is also the altitude. This proves the result." ::: ---

Summary

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="A transformation maps a point $(x, y)$ to $(6-x, y)$. Which of the following best describes this transformation?" options=["Reflection about the x-axis","Translation by vector $\begin{pmatrix} 6 \\ 0 \end{pmatrix}$","Reflection about the line $x=3$","Rotation by $180^\circ$ about the point $(3,0)$"] answer="Reflection about the line $x=3$" hint="Consider the midpoint between $(x,y)$ and $(6-x,y)$." solution="The midpoint of $(x,y)$ and $(6-x,y)$ is $\left(\frac{x+6-x}{2}, \frac{y+y}{2}\right) = (3,y)$. Since the midpoint always lies on the line $x=3$, this transformation is a reflection about the line $x=3$."
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:::question type="NAT" question="A point $P(2, 5)$ is translated by the vector $\begin{pmatrix} -3 \\ 1 \end{pmatrix}$ to point $P'$. Then, $P'$ is rotated $90^\circ$ counter-clockwise about the origin to point $P''$. What is the y-coordinate of $P''$?" answer=" -1 " hint="First find $P'$ by applying the translation. Then apply the rotation rule $(x,y) \to (-y,x)$ for $90^\circ$ counter-clockwise rotation about the origin." solution="1. Translation: $P(2,5)$ translated by $\begin{pmatrix} -3 \\ 1 \end{pmatrix}$ gives $P'(2-3, 5+1) = P'(-1, 6)$.

Rotation: $P'(-1, 6)$ rotated $90^\circ$ counter-clockwise about the origin $(0,0)$ maps $(x,y)$ to $(-y,x)$. So $P''(-6, -1)$.

The y-coordinate of $P''$ is $-1$."
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:::question type="MCQ" question="A figure has exactly 5 lines of reflectional symmetry. Which of the following statements must be true about this figure?" options=["It is a square.","It has rotational symmetry of order 5.","It is a regular pentagon.","It has no rotational symmetry."] answer="It has rotational symmetry of order 5." hint="For a regular n-gon, the number of lines of reflectional symmetry is n, and the order of rotational symmetry is also n." solution="A figure with exactly 5 lines of reflectional symmetry must be a regular pentagon (or a star pentagon). A regular n-gon has rotational symmetry of order n. Therefore, this figure has rotational symmetry of order 5."
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:::question type="NAT" question="What is the minimum positive angle of rotation, in degrees, such that a regular 12-sided polygon (dodecagon) maps onto itself?" answer=" 30 " hint="For a regular n-sided polygon, the order of rotational symmetry is n, and the smallest angle of rotation is $\frac{360^\circ}{n}$." solution="A regular 12-sided polygon has rotational symmetry of order 12. The minimum positive angle of rotation for it to map onto itself is $\frac{360^\circ}{12} = 30^\circ$."
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