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# Geometric Translations

## Movement of every point in a figure the same distance in the same direction.

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Geometric Translations

What if Lucy lived in San Francisco, S\begin{align*}S\end{align*}, and her parents lived in Paso Robles, P\begin{align*}P\end{align*}? She will be moving to Ukiah, U\begin{align*}U\end{align*}, in a few weeks. All measurements are in miles. Find:

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

b) Lucy’s parents are considering moving to Fresno, F\begin{align*}F\end{align*}. Find the component form of PF\begin{align*}\stackrel{\rightharpoonup}{PF}\end{align*} and UF\begin{align*}\stackrel{\rightharpoonup}{UF}\end{align*}.

c) Is Ukiah or Paso Robles closer to Fresno?

### Geometric Translations

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 A\begin{align*}A\end{align*}, then the image would be labeled A\begin{align*}A'\end{align*}, said “a prime.” If there is an image of A\begin{align*}A'\end{align*}, that would be labeled A′′\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 x\begin{align*}x\end{align*} units and y\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 A\begin{align*}A\end{align*} to B\begin{align*}B\end{align*}, or the distance traveled, is the vector. This vector would be labeled AB\begin{align*}\stackrel{\rightharpoonup}{AB}\end{align*} because A\begin{align*}A\end{align*} is the initial point and B\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 AB\begin{align*}\stackrel{\rightharpoonup}{AB}\end{align*} combines the horizontal distance traveled and the vertical distance traveled. We write the component form of AB\begin{align*}\stackrel{\rightharpoonup}{AB}\end{align*} as 3,7\begin{align*}\left \langle 3, 7 \right \rangle \end{align*} because AB\begin{align*}\stackrel{\rightharpoonup}{AB}\end{align*} travels 3 units to the right and 7 units up. Notice the brackets are pointed, 3,7\begin{align*}\left \langle 3, 7 \right \rangle\end{align*}, not curved.

#### Graphing and Translating

Graph square S(1,2),Q(4,1),R(5,4)\begin{align*}S(1, 2), Q(4, 1), R(5, 4)\end{align*} and E(2,5)\begin{align*}E(2, 5)\end{align*}. Find the image after the translation (x,y)(x2,y+3)\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.

(x,y)S(1,2)Q(4,1)R(5,4)E(2,5)(x2,y+3)S(1,5)Q(2,4)R(3,7)E(0,8)\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*}

#### Naming and Writing Vectors

1. Name the vector and write its component form.

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

2. Name the vector and write its component form.

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

#### Earlier Problem Revisited

a) PS=84,187,SU=39,108,PU=123,295\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) PF=62,91,UF=185,204\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*}

### Examples

#### Example 1

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

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*}.

#### Example 2

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

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.

#### Example 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*}.

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*}

#### Example 4

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

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*}.

### Review

1. 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.

1. What is the image of \begin{align*}A(-6, 3)\end{align*}?
2. What is the image of \begin{align*}B(4, 8)\end{align*}?
3. What is the preimage of \begin{align*}C'(5, -3)\end{align*}?
4. What is the image of \begin{align*}A'\end{align*}?
5. What is the preimage of \begin{align*}D'(12, 7)\end{align*}?
6. What is the image of \begin{align*}A''\end{align*}?
7. 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.

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

1. 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*}.
2. 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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### Vocabulary Language: English

TermDefinition
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.