<meta http-equiv="refresh" content="1; url=/nojavascript/"> Translations and Vectors | CK-12 Foundation
Dismiss
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Geometry - Second Edition Go to the latest version.

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

Review Queue Answers

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

Image Attributions

Description

Subjects:

Grades:

Date Created:

Feb 23, 2012

Last Modified:

Oct 23, 2014
Files can only be attached to the latest version of None

Reviews

Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
CK.MAT.ENG.SE.2.Geometry.12.2

Original text