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# 12.2: Translations and Vectors

Created by: CK-12

## Learning Objectives

• Graph a point, line, or figure and translate it $x$ and $y$ 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 $A(1, -1), B(3, 0), C(5, -5)$ and $D(-3, 0)$?
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, $S$, and her parents live in Paso Robles, $P$. She will be moving to Ukiah, $U$, in a few weeks. All measurements are in miles. Find:

a) The component form of $\stackrel{\rightharpoonup}{PS}, \stackrel{\rightharpoonup}{SU}$ and $\stackrel{\rightharpoonup}{PU}$.

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

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 $A$, then the image would be labeled $A'$, said “a prime.” If there is an image of $A'$, that would be labeled $A''$, 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 $x$ units and $y$ units.

Example 1: Graph square $S(1, 2), Q(4, 1), R(5, 4)$ and $E(2, 5)$. Find the image after the translation $(x, y) \rightarrow (x - 2, y + 3)$. 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.

$(x, y) & \rightarrow (x - 2, y + 3)\\S(1,2) & \rightarrow S'(-1,5)\\Q(4,1) & \rightarrow Q'(2,4)\\R(5,4) & \rightarrow R'(3,7)\\E(2,5) & \rightarrow E'(0,8)$

Example 2: Find the translation rule for $\triangle TRI$ to $\triangle T'R'I'$.

Solution: Look at the movement from $T$ to $T'$. $T$ is (-3, 3) and $T'$ is (3, -1). The change in $x$ is 6 units to the right and the change in $y$ is 4 units down. Therefore, the translation rule is $(x,y) \rightarrow (x + 6, y - 4)$.

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 $\triangle TRI \cong \triangle T'R'I'$ from Example 2.

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

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

## 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 $A$ to $B$, or the distance traveled, is the vector. This vector would be labeled $\stackrel{\rightharpoonup}{AB}$ because $A$ is the initial point and $B$ 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 $\stackrel{\rightharpoonup}{AB}$ combines the horizontal distance traveled and the vertical distance traveled. We write the component form of $\stackrel{\rightharpoonup}{AB}$ as $\left \langle 3, 7 \right \rangle$ because $\stackrel{\rightharpoonup}{AB}$ travels 3 units to the right and 7 units up. Notice the brackets are pointed, $\left \langle 3, 7 \right \rangle$, not curved.

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

a)

b)

Solution:

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

b) The vector is $\stackrel{\rightharpoonup}{EF}$. The component form of $\stackrel{\rightharpoonup}{EF}$ is $\left \langle 4, 1 \right \rangle$.

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

Solution: The graph above is the vector $\stackrel{\rightharpoonup}{ST}$. From the initial point $S$ 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 $\triangle ABC$ has coordinates $A(3, -1), B(7, -5)$ and $C(-2, -2)$. Translate $\triangle ABC$ using the vector $\left \langle -4, 5 \right \rangle$. Determine the coordinates of $\triangle A'B'C'$.

Solution: It would be helpful to graph $\triangle ABC$. To translate $\triangle ABC$, add each component of the vector to each point to find $\triangle A'B'C'$.

$A(3, -1) + \left \langle -4, 5 \right \rangle & = A'(-1, 4)\\B(7, -5) + \left \langle -4, 5 \right \rangle & = B'(3,0)\\C(-2, -2) + \left \langle -4, 5 \right \rangle & = C'(-6, 3)$

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

Solution: To write $\left \langle -4, 5 \right \rangle$ as a translation rule, it would be $(x, y) \rightarrow (x - 4, y + 5)$.

Know What? Revisited

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

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

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

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

## Review Questions

1. What is the difference between a vector and a ray?

Use the translation $(x, y) \rightarrow (x + 5, y - 9)$ for questions 2-8.

1. What is the image of $A(-6, 3)$?
2. What is the image of $B(4, 8)$?
3. What is the preimage of $C'(5, -3)$?
4. What is the image of $A'$?
5. What is the preimage of $D'(12, 7)$?
6. What is the image of $A''$?
7. Plot $A, A', A''$, and $A'''$ from the questions above. What do you notice? Write a conjecture.

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

1. $(x, y) \rightarrow (x - 2, y - 7)$
2. $(x, y) \rightarrow (x + 11, y + 4)$
3. $(x, y) \rightarrow (x, y - 3)$
4. $(x, y) \rightarrow (x - 5, y + 8)$

In questions 13-16, $\triangle A'B'C'$ is the image of $\triangle ABC$. Write the translation rule.

1. Verify that a translation is an isometry using the triangle from #15.
2. If $\triangle A'B'C'$ was the preimage and $\triangle ABC$ 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. $\stackrel{\rightharpoonup}{AB} = \left \langle 4, -3 \right \rangle$
2. $\stackrel{\rightharpoonup}{CD} = \left \langle -6, 8 \right \rangle$
3. $\stackrel{\rightharpoonup}{FE} = \left \langle -2, 0 \right \rangle$
4. The coordinates of $\triangle DEF$ are $D(4, -2), E(7, -4)$ and $F(5, 3)$. Translate $\triangle DEF$ using the vector $\left \langle 5, 11 \right \rangle$ and find the coordinates of $\triangle D'E'F'$.
5. The coordinates of quadrilateral $QUAD$ are $Q(-6, 1), U(-3, 7), A(4, -2)$ and $D(1, -8)$. Translate $QUAD$ using the vector $\left \langle -3, -7 \right \rangle$ and find the coordinates of $Q'U'A'D'$.

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

1. $(x, y) \rightarrow (x - 3, y + 8)$
2. $(x, y) \rightarrow (x + 9, y - 12)$
3. $(x,y) \rightarrow (x, y - 7)$

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

1. $\left \langle-7, 2 \right \rangle$
2. $\left \langle 11, 25 \right \rangle$
3. $\left \langle 15, -9 \right \rangle$

1. $y = -2x+17$
2. Kite
3. $y = -2x+5$
4. $y = \frac{1}{2} x-5$

Feb 23, 2012

Oct 23, 2014