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# 12.2: Translations and Vectors

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Graph a point, line, or figure and translate it x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} units.
• Write a translation rule.
• Use vector notation.

## Review Queue

1. Find the equation of the line that contains (9, -1) and (5, 7).
2. What type of quadrilateral is formed by A(1,1),B(3,0),C(5,5)\begin{align*}A(1, -1), B(3, 0), C(5, -5)\end{align*} and D(3,0)\begin{align*}D(-3, 0)\end{align*}?
3. Find the equation of the line parallel to #1 that passes through (4, -3).
4. Find the equation of the line perpendicular to #1 that passes through (4, -3).

Know What? Lucy currently lives in San Francisco, S\begin{align*}S\end{align*}, and her parents live 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?

## Transformations

Recall from Lesson 7.6, we learned about dilations, which is a type of transformation. Now, we are going to continue learning about other types of transformations. All of the transformations in this chapter are rigid transformations.

Transformation: An operation that moves, flips, or changes a figure to create a new figure.

Rigid Transformation: A transformation that preserves size and shape.

The rigid transformations are: translations, 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.

Also in Lesson 7.6, we learned how to label an image. If the preimage is A\begin{align*}A\end{align*}, 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.”

## Translations

The first of the rigid transformations is a translation.

Translation: 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.

Example 1: 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.

Solution: 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 2: Find the translation rule for \begin{align*}\triangle TRI\end{align*} to \begin{align*}\triangle T'R'I'\end{align*}.

Solution: 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*}.

From both of these examples, we see that a translation preserves congruence. Therefore, a translation is an isometry. We can show that each pair of figures is congruent by using the distance formula.

Example 3: Show \begin{align*}\triangle TRI \cong \triangle T'R'I'\end{align*} from Example 2.

Solution: Use the distance formula to find all the lengths of the sides of the two triangles.

\begin{align*}& \underline{\triangle TRI} && \underline{\triangle T'R'I'}\\ & TR = \sqrt{(-3 - 2)^2 + (3 - 6)^2} = \sqrt{34} && T'R' = \sqrt{(3 - 8)^2 + (-1 - 2)^2} = \sqrt{34}\\ & RI = \sqrt{(2 -(-2))^2 + (6 - 8)^2} = \sqrt{20} && R'I' = \sqrt{(8 - 4)^2 + (2 - 4)^2} = \sqrt{20}\\ & TI = \sqrt{(-3-(-2))^2 + (3 - 8)^2} = \sqrt{26} && T'I' = \sqrt{(3 - 4)^2 + (-1 - 4)^2} = \sqrt{26}\end{align*}

## Vectors

Another way to write a translation rule is to use vectors.

Vector: 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 4: Name the vector and write its component form.

a)

b)

Solution:

a) 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*}.

b) 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*}.

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

Solution: The graph above 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.

The positive and negative components of a vector always correlate with the positive and negative parts of the coordinate plane. We can also use vectors to translate an image.

Example 6: 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*}.

Solution: 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 7: Write the translation rule for the vector translation from Example 6.

Solution: 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*}.

Know What? 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*}

## Review Questions

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.

1. Verify that a translation is an isometry using the triangle from #15.
2. If \begin{align*}\triangle A'B'C'\end{align*} was the preimage and \begin{align*}\triangle ABC\end{align*} was the image, write the translation rule for #16.

For questions 19-21, name each vector and find its component form.

For questions 22-24, plot and correctly label each vector.

1. \begin{align*}\stackrel{\rightharpoonup}{AB} = \left \langle 4, -3 \right \rangle\end{align*}
2. \begin{align*}\stackrel{\rightharpoonup}{CD} = \left \langle -6, 8 \right \rangle\end{align*}
3. \begin{align*}\stackrel{\rightharpoonup}{FE} = \left \langle -2, 0 \right \rangle\end{align*}
4. 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*}.
5. 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*}.

For problems 27-29, write the translation rule as a translation vector.

1. \begin{align*}(x, y) \rightarrow (x - 3, y + 8)\end{align*}
2. \begin{align*}(x, y) \rightarrow (x + 9, y - 12)\end{align*}
3. \begin{align*}(x,y) \rightarrow (x, y - 7)\end{align*}

For problems 30-32, write the translation vector as a translation rule.

1. \begin{align*}\left \langle-7, 2 \right \rangle\end{align*}
2. \begin{align*}\left \langle 11, 25 \right \rangle\end{align*}
3. \begin{align*}\left \langle 15, -9 \right \rangle\end{align*}

1. \begin{align*}y = -2x+17\end{align*}
2. Kite
3. \begin{align*}y = -2x+5\end{align*}
4. \begin{align*}y = \frac{1}{2} x-5\end{align*}

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