# 12.3: Geometric Translations

**At Grade**Created by: CK-12

**Practice**Geometric Translations

What if Lucy lived in San Francisco, \begin{align*}S\end{align*}

a) The component form of \begin{align*}\stackrel{\rightharpoonup}{PS}, \stackrel{\rightharpoonup}{SU}\end{align*}

b) Lucy’s parents are considering moving to Fresno, \begin{align*}F\end{align*}

c) Is Ukiah or Paso Robles closer to Fresno?

After completing this Concept, you'll be able to answer these questions.

### Watch This

CK-12 Foundation: Chapter12TranslationsA

### Guidance

*A* *transformation**is an operation that moves, flips, or changes a figure to create a new figure. A* *rigid transformation**is a transformation that preserves size and shape. The rigid transformations are: translations (discussed here), reflections, and rotations. The new figure created by a transformation is called the* *image**. The original figure is called the* *preimage**. Another word for a rigid transformation is an* *isometry**. Rigid transformations are also called* *congruence transformations**. If the preimage is \begin{align*}A\end{align*} A, then the image would be labeled \begin{align*}A'\end{align*}, said “a prime.” If there is an image of \begin{align*}A'\end{align*}, that would be labeled \begin{align*}A''\end{align*}, said “a double prime.”*

A **translation** is a transformation that moves every point in a figure the same distance in the same direction. In the coordinate plane, we say that a translation moves a figure \begin{align*}x\end{align*} units and \begin{align*}y\end{align*} units. Another way to write a translation rule is to use vectors. A **vector** is a quantity that has direction and size.

In the graph below, the line from \begin{align*}A\end{align*} to \begin{align*}B\end{align*}, or the distance traveled, is the vector. This vector would be labeled \begin{align*}\stackrel{\rightharpoonup}{AB}\end{align*} because \begin{align*}A\end{align*} is the **initial point** and \begin{align*}B\end{align*} is the **terminal point**. The terminal point always has the arrow pointing towards it and has the half-arrow over it in the label.

The **component form** of \begin{align*}\stackrel{\rightharpoonup}{AB}\end{align*} combines the horizontal distance traveled and the vertical distance traveled. We write the component form of \begin{align*}\stackrel{\rightharpoonup}{AB}\end{align*} as \begin{align*}\left \langle 3, 7 \right \rangle \end{align*} because \begin{align*}\stackrel{\rightharpoonup}{AB}\end{align*} travels 3 units to the right and 7 units up. Notice the brackets are pointed, \begin{align*}\left \langle 3, 7 \right \rangle\end{align*}, not curved.

#### Example A

Graph square \begin{align*}S(1, 2), Q(4, 1), R(5, 4)\end{align*} and \begin{align*}E(2, 5)\end{align*}. Find the image after the translation \begin{align*}(x, y) \rightarrow (x - 2, y + 3)\end{align*}. Then, graph and label the image.

The translation notation tells us that we are going to move the square to the left 2 and up 3.

\begin{align*}(x, y) & \rightarrow (x - 2, y + 3)\\ S(1,2) & \rightarrow S'(-1,5)\\ Q(4,1) & \rightarrow Q'(2,4)\\ R(5,4) & \rightarrow R'(3,7)\\ E(2,5) & \rightarrow E'(0,8)\end{align*}

#### Example B

Name the vector and write its component form.

The vector is \begin{align*}\stackrel{\rightharpoonup}{DC}\end{align*}. From the initial point \begin{align*}D\end{align*} to terminal point \begin{align*}C\end{align*}, you would move 6 units to the left and 4 units up. The component form of \begin{align*}\stackrel{\rightharpoonup}{DC}\end{align*} is \begin{align*}\left \langle -6, 4 \right \rangle\end{align*}.

#### Example C

Name the vector and write its component form.

The vector is \begin{align*}\stackrel{\rightharpoonup}{EF}\end{align*}. The component form of \begin{align*}\stackrel{\rightharpoonup}{EF}\end{align*} is \begin{align*}\left \langle 4, 1 \right \rangle\end{align*}.

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter12TranslationsB

#### Concept Problem Revisited

a) \begin{align*}\stackrel{\rightharpoonup}{PS}= \left \langle -84, 187 \right \rangle, \stackrel{\rightharpoonup}{SU} = \left \langle -39, 108 \right \rangle, \stackrel{\rightharpoonup}{PU} = \left \langle -123, 295 \right \rangle\end{align*}

b) \begin{align*}\stackrel{\rightharpoonup}{PF} = \left \langle 62, 91 \right \rangle,\stackrel{\rightharpoonup}{UF} = \left \langle 185, -204 \right \rangle\end{align*}

c) You can plug the vector components into the Pythagorean Theorem to find the distances. Paso Robles is closer to Fresno than Ukiah.

\begin{align*}UF = \sqrt{185^2 + (-204)^2} \cong 275.4 \ miles, PF = \sqrt{62^2 + 91^2} \cong 110.1 \ miles\end{align*}

### Vocabulary

A ** transformation** is an operation that moves, flips, or otherwise changes a figure to create a new figure. A

**(also known as an**

*rigid transformation***or**

*isometry***) is a transformation that does not change the size or shape of a figure. The new figure created by a transformation is called the**

*congruence transformation***. The original figure is called the**

*image***. A**

*preimage***is a transformation that moves every point in a figure the same distance in the same direction. A**

*translation***is a quantity that has direction and size. The**

*vector***of a vector combines the horizontal distance traveled and the vertical distance traveled.**

*component form*### Guided Practice

1. Find the translation rule for \begin{align*}\triangle TRI\end{align*} to \begin{align*}\triangle T'R'I'\end{align*}.

2. Draw the vector \begin{align*}\stackrel{\rightharpoonup}{ST}\end{align*} with component form \begin{align*}\left \langle 2, -5 \right \rangle\end{align*}.

3. Triangle \begin{align*}\triangle ABC\end{align*} has coordinates \begin{align*}A(3, -1), B(7, -5)\end{align*} and \begin{align*}C(-2, -2)\end{align*}. Translate \begin{align*}\triangle ABC\end{align*} using the vector \begin{align*}\left \langle -4, 5 \right \rangle\end{align*}. Determine the coordinates of \begin{align*}\triangle A'B'C'\end{align*}.

4. Write the translation rule for the vector translation from #3.

**Answers:**

1. Look at the movement from \begin{align*}T\end{align*} to \begin{align*}T'\end{align*}. \begin{align*}T\end{align*} is (-3, 3) and \begin{align*}T'\end{align*} is (3, -1). The change in \begin{align*}x\end{align*} is 6 units to the right and the change in \begin{align*}y\end{align*} is 4 units down. Therefore, the translation rule is \begin{align*}(x,y) \rightarrow (x + 6, y - 4)\end{align*}.

2. The graph is the vector \begin{align*}\stackrel{\rightharpoonup}{ST}\end{align*}. From the initial point \begin{align*}S\end{align*} it moves down 5 units and to the right 2 units.

3. It would be helpful to graph \begin{align*}\triangle ABC\end{align*}. To translate \begin{align*}\triangle ABC\end{align*}, add each component of the vector to each point to find \begin{align*}\triangle A'B'C'\end{align*}.

\begin{align*}A(3, -1) + \left \langle -4, 5 \right \rangle & = A'(-1, 4)\\ B(7, -5) + \left \langle -4, 5 \right \rangle & = B'(3,0)\\ C(-2, -2) + \left \langle -4, 5 \right \rangle & = C'(-6, 3)\end{align*}

4. To write \begin{align*}\left \langle -4, 5 \right \rangle\end{align*} as a translation rule, it would be \begin{align*}(x, y) \rightarrow (x - 4, y + 5)\end{align*}.

### Practice

- What is the difference between a vector and a ray?

Use the translation \begin{align*}(x, y) \rightarrow (x + 5, y - 9)\end{align*} for questions 2-8.

- What is the image of \begin{align*}A(-6, 3)\end{align*}?
- What is the image of \begin{align*}B(4, 8)\end{align*}?
- What is the preimage of \begin{align*}C'(5, -3)\end{align*}?
- What is the image of \begin{align*}A'\end{align*}?
- What is the preimage of \begin{align*}D'(12, 7)\end{align*}?
- What is the image of \begin{align*}A''\end{align*}?
- Plot \begin{align*}A, A', A''\end{align*}, and \begin{align*}A'''\end{align*} from the questions above. What do you notice? Write a conjecture.

The vertices of \begin{align*}\triangle ABC\end{align*} are \begin{align*}A(-6, -7), B(-3, -10)\end{align*} and \begin{align*}C(-5, 2)\end{align*}. Find the vertices of \begin{align*}\triangle A'B'C'\end{align*}, given the translation rules below.

- \begin{align*}(x, y) \rightarrow (x - 2, y - 7)\end{align*}
- \begin{align*}(x, y) \rightarrow (x + 11, y + 4)\end{align*}
- \begin{align*}(x, y) \rightarrow (x, y - 3)\end{align*}
- \begin{align*}(x, y) \rightarrow (x - 5, y + 8)\end{align*}

In questions 13-16, \begin{align*}\triangle A'B'C'\end{align*} is the image of \begin{align*}\triangle ABC\end{align*}. Write the translation rule.

For questions 17-19, name each vector and find its component form.

- The coordinates of \begin{align*}\triangle DEF\end{align*} are \begin{align*}D(4, -2), E(7, -4)\end{align*} and \begin{align*}F(5, 3)\end{align*}. Translate \begin{align*}\triangle DEF\end{align*} using the vector \begin{align*}\left \langle 5, 11 \right \rangle\end{align*} and find the coordinates of \begin{align*}\triangle D'E'F'\end{align*}.
- The coordinates of quadrilateral \begin{align*}QUAD\end{align*} are \begin{align*}Q(-6, 1), U(-3, 7), A(4, -2)\end{align*} and \begin{align*}D(1, -8)\end{align*}. Translate \begin{align*}QUAD\end{align*} using the vector \begin{align*}\left \langle -3, -7 \right \rangle\end{align*} and find the coordinates of \begin{align*}Q'U'A'D'\end{align*}.

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Term | Definition |
---|---|

Center of Rotation |
In a rotation, the center of rotation is the point that does not move. The rest of the plane rotates around this fixed point. |

Image |
The image is the final appearance of a figure after a transformation operation. |

Preimage |
The pre-image is the original appearance of a figure in a transformation operation. |

Transformation |
A transformation moves a figure in some way on the coordinate plane. |

Translation |
A translation is a transformation that slides a figure on the coordinate plane without changing its shape, size, or orientation. |

### Image Attributions

Here you'll learn what a translation is and how to perform translation rules.