Transformations of the Plane Study Guide

Unit Outcomes and Introduction to Transformations of the Plane

• Learning Objectives: By the end of this unit, students will be able to understand the basic concepts of transforming the plane and apply procedures to transform plane figures.
• Unit Contents:
  • 6.1 Introduction
  • 6.2 Translation
  • 6.3 Reflection
  • 6.4 Rotation
  • 6.5 Applications
• Key Vocabulary:
  • Identity transformation: A transformation where the image of every point is itself.
  • Non-rigid motion: A change in the position of a figure that does not preserve its shape or size.
  • Rigid motion: A motion or mapping that preserves distance between points.
  • Initial point: The starting position of a transformation vector.
  • Reflection: A transformation that flips a figure over a line.
  • Rotation: A transformation that turns a figure about a fixed point.
  • Standard position: The position of an angle or vector with its vertex/initial point at the origin.
  • Terminal point: The ending point of a transformation vector.
  • Transformation: A mapping of points in the plane to other points.
  • Vector translation: Every point moves in the same direction and the same distance.

6.1 Introduction to Transformations

• Context: Building upon Unit 5 (Geometric and Algebraic aspects of vector representation), this unit focuses on rigid motions.
• Activity 6.1 Analysis:
  • Condition A: Compressing or stretching a spring changes its size/shape (non-rigid motion).
  • Condition B: The Earth rotating about its axis preserves its shape/size (rigid motion).
  • Condition C: Seeing an image in a plane mirror preserves shape/size (rigid motion).
  • Condition D: Drawing a home’s door involves scaling or mapping, which may or may not be rigid depending on the context of the representation.
• Categorization of Mappings:
  • Some mappings preserve shape, size, or distance between any two points.
  • Others do not preserve these properties.
• Definition 6.1: Rigid Motion: A motion which preserves distance. For any two points $A$ and $B$, the distance between them equals the distance between their images $A'$ and $B'$, such that $AB = A'B'$.
• Properties of Rigid Motion: Rigid motion carries any plane figure to a congruent plane figure (e.g., triangles to congruent triangles).
• Types of Rigid Motion: Translation, Reflection, and Rotation.

6.2 Translation

• Definition 6.2: A translation is a transformation that occurs when every point of a figure is moved from one location to another location along the same direction through the same distance.
• Translation Vector: If point $P$ is translated to $P'$, the vector $\mathbf{PP'}$ is the translation vector.
• Coordinate Formula for Translation:
  • Let $T = (a, b)$ be the translation vector.
  • The origin is translated to $(a, b)$. Formula: $(0, 0) \rightarrow (a, b)$.
  • The image of point $P(x, y)$ under translation vector $T$ is $P'(x + a, y + b)$.

Examples and Solutions for Point Translation

• Example 1.1: Let $T = (2, 3)$. Find the images of $A(2, 1)$, $B(-1, 2)$, and $C(-5, 1)$.
  • For $A(2, 1)$, add 2 to $x$ and 3 to $y$. Result: $A'(4, 4)$.
  • For $B(-1, 2)$, result: $B'(1, 5)$.
  • For $C(-5, 1)$, result: $C'(-3, 4)$.
  • General rule: For any point $R(x, y)$, the image is $R'(x + 2, y + 3)$.
• Example 1.2: If translation takes the origin to $(-2, 2)$, find images of $P(3, 5)$ and $Q(-1, 4)$.
  • Translation vector is $T = (-2, 2)$.
  • $P'(3 + (-2), 5 + 2) = (1, 7)$.
  • $Q'(-1 + (-2), 4 + 2) = (-3, 6)$.
• Example 1.3: The image of $(1, 2)$ is $(2, 4)$. Find the image of $(-2, 3)$.
  • Vector $T = (2 - 1, 4 - 2) = (1, 2)$.
  • Image of $(-2, 3)$ is $(-2 + 1, 3 + 2) = (-1, 5)$.
• Example 2: Vertices of triangle $ABC$ are $A(-4, -4)$, $B(-2, -1)$, and $C(-1, -5)$. Find coordinates under $T = (6, 5)$.
  • $A'(-4 + 6, -4 + 5) = (2, 1)$.
  • $B'(-2 + 6, -1 + 5) = (4, 4)$.
  • $C'(-1 + 6, -5 + 5) = (5, 0)$.

Line and Circle Translation

• Line Translation Property: A translation maps lines onto parallel lines.
• Example 3: Translation takes $(-2, 3)$ to $(1, 2)$. Find images for:
  • $L_1: 3x - y + 4 = 0$:
    • Vector $T = (1 - (-2), 2 - 3) = (3, -1)$.
    • Let $x' = x + 3$ and $y' = y - 1$. Then $x = x' - 3$ and $y = y' + 1$.
    • Substitute into $L_1$: $3(x' - 3) - (y' + 1) + 4 = 0 \rightarrow 3x' - 9 - y' - 1 + 4 = 0 \rightarrow 3x' - y' - 6 = 0$.
    • Resulting line: $3x - y - 6 = 0$.
  • $L_2: 4y + 2x + 1 = 0$:
    • Substitute: $4(y' + 1) + 2(x' - 3) + 1 = 0 \rightarrow 4y' + 4 + 2x' - 6 + 1 = 0 \rightarrow 4y' + 2x' - 1 = 0$.
    • Resulting line: $4y + 2x - 1 = 0$.
• Circle Translation Property: Translation is a rigid motion. To translate a circle, translate the center by vector $T(a, b)$ and keep the radius the same.
• Example 4: Translation takes origin to $(-2, 2)$. Find equation for circle $x^2 + y^2 = 4$.
  • Center $(0, 0)$ translates to $(-2, 2)$. Radius remains $2$.
  • New equation: $(x + 2)^2 + (y - 2)^2 = 4$.

6.3 Reflection

• Definition 6.3: Let $L$ be a fixed line. A reflection $M$ about line $L$ is a transformation that carries point $A$ to point $A'$ such that $L$ is the perpendicular bisector of the segment $\mathbf{AA'}$.
• Basic Properties:
  1. Function: If $A = B$, then $M(A) = M(B)$.
  2. One-to-one: If $A \neq B$, then $M(A) \neq M(B)$.
  3. Onto: For every point $A'$ in the plane, there exists a point $A$ such that $M(A) = A'$.

Case A: Reflection in the line $y = mx$

• Specific Sub-cases (based on angle of inclination $\theta$):
  1. x-axis ($\theta = 0$): Image of $(x, y)$ is $(x, -y)$.
  2. y-axis ($\theta = \frac{\pi}{2}$): Image of $(x, y)$ is $(-x, y)$.
  3. Line $y = x$ ($\theta = \frac{\pi}{4}$): Image of $(x, y)$ is $(y, x)$.
  4. Line $y = -x$ ($\theta = \frac{3\pi}{4}$): Image of $(x, y)$ is $(-y, -x)$.
• General Formula for $y = mx$:
  • Let $\theta$ be the angle such that $m = \tan(\theta)$.
  • The reflection of point $P(x, y)$ is $P'(x', y')$, where:
    • $x' = x \cos(2\theta) + y \sin(2\theta)$
    • $y' = x \sin(2\theta) - y \cos(2\theta)$
• Example 5: Reflect points about $y = \frac{1}{\sqrt{3}}x$.
  • $\theta = \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$. Thus $2\theta = \frac{\pi}{3}$.
  • $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$.
  • For $A(2, 1)$: $x' = 2(\frac{1}{2}) + 1(\frac{\sqrt{3}}{2}) = 1 + \frac{\sqrt{3}}{2}$; $y' = 2(\frac{\sqrt{3}}{2}) - 1(\frac{1}{2}) = \frac{2\sqrt{3} - 1}{2}$. Result: $A'(\frac{2 + \sqrt{3}}{2}, \frac{2\sqrt{3} - 1}{2})$.

Case B: Reflection in the line $y = mx + b$

• Procedure for a point:
  1. Find the slope of line $L$, $m$.
  2. Find the equation of line $s$, which passes through $A(x, y)$ and has slope $-\frac{1}{m}$.
  3. Find intersection point $B$ of $L$ and $s$. $B$ is the midpoint of segment $\mathbf{AA'}$.
  4. Use midpoint formula to find coordinates of $A'$: $B = (\frac{x + x'}{2}, \frac{y + y'}{2})$.
• Example 6: Reflect $(4, 0)$ over $y = 2x + 1$.
  • Slope $m = 2$. Slope of perpendicular line $s = -\frac{1}{2}$.
  • Equation of $s$: $y - 0 = -\frac{1}{2}(x - 4) \rightarrow y = -\frac{1}{2}x + 2$.
  • Intersection $B$: $2x + 1 = -\frac{1}{2}x + 2 \rightarrow \frac{5}{2}x = 1 \rightarrow x = \frac{2}{5}$. Then $y = 2(\frac{2}{5}) + 1 = \frac{9}{5}$.
  • Midpoint formula: $\frac{x' + 4}{2} = \frac{2}{5} \rightarrow x' + 4 = \frac{4}{5} \rightarrow x' = -\frac{16}{5}$. $\frac{y' + 0}{2} = \frac{9}{5} \rightarrow y' = \frac{18}{5}$.
  • Result: $A'(-\frac{16}{5}, \frac{18}{5})$.
• Example 7: Reflection of $(-2, 4)$ over specific lines:
  • Line $y = -2$: Point is 6 units above. Image is 6 units below. Result: $(-2, -8)$.
  • Line $x = 1$: Point is 3 units left. Image is 3 units right. Result: $(4, 4)$.

Reflection of Lines and Circles

• Line Reflection:
  1. Choose any point $A$ on the line $s$.
  2. Find the image $A' = M(A)$.
  3. Find intersection point $C$ of line $s$ and reflection axis $L$.
  4. The image line $s'$ passes through $A'$ and $C$.
• Example 8: Reflect $-3x + y = 2$ over $y = x + 4$.
  • Intersection point $C(1, 5)$. Point $A(0, 2)$ maps to $A'(-2, 4)$.
  • New line equation: $s': x - 3y + 14 = 0$.
• Circle Reflection:
  1. If the center is on the axis, the circle is its own image.
  2. Otherwise, find the image of center $O$ as $O'$. Radius remains identical.
• Example 9: Reflect $x^2 + y^2 - 4x - 2y + 4 = 0$ in $y = x - 1$.
  • Center is $(2, 1)$. Checking center: $1 = 2 - 1$. Center is on the line.
  • Image circle is the same as pre-image.

6.4 Rotation

• Definition 6.4: A rotation $R$ about point $O$ through angle $\theta$ is a transformation where $OA = OA'$ and the measure of angle $\angle AOA' = \theta$.
• Convention: $\theta > 0$ is counter-clockwise; $\theta < 0$ is clockwise.

Case A: Rotation about the Origin

• Theorem 6.1: Let $R_{\theta}(x, y) = (x', y')$. Then:
  • $x' = x \cos(\theta) - y \sin(\theta)$
  • $y' = x \sin(\theta) + y \cos(\theta)$
• Example 2: Rotate points about origin.
  • $(1, 3)$ through $30^\circ$:
    • $x' = 1 \cos(30^\circ) - 3 \sin(30^\circ) = \frac{\sqrt{3} - 3}{2}$.
    • $y' = 1 \sin(30^\circ) + 3 \cos(30^\circ) = \frac{1 + 3\sqrt{3}}{2}$.
  • $(3, -4)$ through $810^\circ$:
    • $810^\circ = 2 \times 360^\circ + 90^\circ$, which is equivalent to $90^\circ$.
    • $x' = 3 \cos(90^\circ) - (-4) \sin(90^\circ) = 3(0) + 4(1) = 4$.
    • $y' = 3 \sin(90^\circ) + (-4) \cos(90^\circ) = 3(1) - 4(0) = 3$.
• Standard Rotations:
  • $90^\circ$ ($\frac{\pi}{2}$): $(x, y) \rightarrow (-y, x)$.
  • $180^\circ$ ($\pi$): $(x, y) \rightarrow (-x, -y)$.
  • $270^\circ$ ($\frac{3\pi}{2}$): $(x, y) \rightarrow (y, -x)$.
  • $360^\circ$ ($2n\pi$): Identity transformation ($(x, y) \rightarrow (x, y)$).

Case B: Rotation about arbitrary point $(a, b)$

• Corollary 6.1: This involves three steps:
  1. Translate by $T = (-a, -b)$ to bring center of rotation to origin.
  2. Rotate by angle $\theta$ about the origin.
  3. Translate back by $T = (a, b)$.
• General Formula:
  • $x' = a + (x - a) \cos(\theta) - (y - b) \sin(\theta)$
  • $y' = b + (x - a) \sin(\theta) + (y - b) \cos(\theta)$
• Example 6: Rotate $A(3, 4)$ through $\theta = \pi$ about center $(6, 5)$.
  • $x' = 6 + (3 - 6) \cos(\pi) - (4 - 5) \sin(\pi) = 6 + (-3)(-1) - 0 = 9$.
  • $y' = 5 + (3 - 6) \sin(\pi) + (4 - 5) \cos(\pi) = 5 + 0 + (-1)(-1) = 6$.
  • Result: $(9, 6)$.

6.5 Applications and Problem Solving

• Invariant Points: Points that do not move under a transformation. In reflections, points on the axis are invariant.
• Preservation of Distance:
  • If $FG$ is reflected to $GH$, then $Length(FG) = Length(GH)$.
  • Example: In a symmetric shape where $L$ is the axis, if $FH = 16\,cm$, then $EH$ (the perpendicular segment to the base) is half the span if $E$ is on the axis and symmetry holds, or determined by the geometry (Solution provided: $EH = \frac{1}{2} \times 16 = 8\,cm$).
• Composite Problems:
  • Problem 3: Triangle $ABC$ vertices $A(2, 3)$, $B(5, 7)$, $C(5, 3)$. Reflect over x-axis then rotate $180^\circ$ about origin.
  1. Reflect over x-axis: $A'(2, -3)$, $B'(5, -7)$, $C'(5, -3)$.
  2. Rotate $180^\circ$ about origin: $(x, y) \rightarrow (-x, -y)$.
  3. Final images: $A''(-2, 3)$, $B''(-5, 7)$, $C''(-5, 3)$.

Questions & Discussion

• Activity 6.2 Conclusion: When sliding a triangle (translation), the pre-image and image have the same size, shape, and orientation.
• Activity 6.3 Conclusion: When folding/flipping a figure (reflection), shape and size are the same, but the figure faces the opposite direction.
• Exercise 6.13 Highlight:
  • Image of point $I$ under horizontal reflection: point $A$.
  • Image of line segment $CD$: line segment $EF$.
  • Invariant points: points situated on the horizontal axis of reflection (e.g., $J, G$).
  • Length calculations: If $HI = 3\,cm$, and reflection occurs, its image $HA$ is also $3\,cm$. If $FH = 3\,cm$, then $EF = FH = 3\,cm$.