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12.2: Translations

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

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

Review Queue

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

Know What? The distances between San Francisco, \begin{align*}S\end{align*}S, Paso Robles, \begin{align*}P\end{align*}P, and Ukiah, \begin{align*}U\end{align*}U, are given in miles the graph. Find:

a) The translation rule for \begin{align*}P\end{align*}P to \begin{align*}S\end{align*}S.

b) The translation rule for \begin{align*}S\end{align*}S to \begin{align*}U\end{align*}U.

c) The translation rule for \begin{align*}P\end{align*}P to \begin{align*}U\end{align*}U.

d) The translation rule for \begin{align*}U\end{align*}U to \begin{align*}S\end{align*}S. It is not the same as part b.


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

Rigid Transformation: A transformation that does not change the size or shape of a figure.

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 or congruence transformations.

In Lesson 7.6, we learned how to label an image. If the preimage is \begin{align*}A\end{align*}A, then the image would be \begin{align*}A’\end{align*}A, said “a prime.” If there is an image of \begin{align*}A’\end{align*}A, that would be labeled \begin{align*}A”\end{align*}A, said “a double prime.”


Translation: A transformation that moves every point in a figure the same distance in the same direction.

This transformation moves the parallelogram to the right 5 units and up 3 units. It is written \begin{align*}(x,y) \rightarrow (x+5,y+3)\end{align*}(x,y)(x+5,y+3).

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

Solution: 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*}(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)

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

Solution: Look at the movement from \begin{align*}T\end{align*}T to \begin{align*}T’\end{align*}T. The translation rule is \begin{align*}(x,y) \rightarrow (x+6,y-4)\end{align*}(x,y)(x+6,y4).

Example 3: Show \begin{align*}\triangle TRI \cong \triangle T’R’I’\end{align*}TRITRI 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*}TRITR=(32)2+(36)2=34RI=(2(2))2+(68)2=20TI=(3(2))2+(38)2=26TRITR=(38)2+(12)2=34RI=(84)2+(24)2=20TI=(34)2+(14)2=26

This verifies our statement at the beginning of the section that a translation is an isometry or congruence translation.

Example 4: 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*} to the left 4 units and up 5 units. Determine the coordinates of \begin{align*}\triangle A’B’C’\end{align*}.

Solution: Graph \begin{align*}\triangle ABC\end{align*}. To translate \begin{align*}\triangle ABC\end{align*}, subtract 4 from each \begin{align*}x\end{align*} value and add 5 to each \begin{align*}y\end{align*} value.

\begin{align*}& A(3,-1) \rightarrow (3-4,-1+5)=A’(-1,4)\\ & B(7,-5) \rightarrow (7-4,-5+5)=B’(3,0)\\ & C(-2,-2) \rightarrow (-2-4,-2+5)=C’(-6,3)\end{align*}

The rule would be \begin{align*}(x,y) \rightarrow (x-4,y+5)\end{align*}.

Know What? Revisited

a) \begin{align*}(x,y) \rightarrow (x-84,y+187)\end{align*}

b) \begin{align*}(x,y) \rightarrow (x-39,y+108)\end{align*}

c) \begin{align*}(x,y) \rightarrow (x-123,y+295)\end{align*}

d) \begin{align*}(x,y) \rightarrow (x+39,y-108)\end{align*}

Review Questions

  • Questions 1-13 are similar to Example 1.
  • Questions 14-17 are similar to Example 2.
  • Questions 18-20 are similar to Example 3.
  • Questions 21-23 are similar to Example 1.
  • Questions 24 and 25 are similar to Example 4.

Use the translation \begin{align*}(x,y) \rightarrow (x+5,y-9)\end{align*} for questions 1-7.

  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 image 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?

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*}
  5. \begin{align*}(x,y) \rightarrow (x+1,y)\end{align*}
  6. \begin{align*}(x,y) \rightarrow (x+3,y+10)\end{align*}

In questions 14-17, \begin{align*}\triangle A’B’C’\end{align*} is the image of \begin{align*}\triangle ABC\end{align*}. Write the translation rule.

Use the triangles from #17 to answer questions 18-20.

  1. Find the lengths of all the sides of \begin{align*}\triangle ABC\end{align*}.
  2. Find the lengths of all the sides of \begin{align*}\triangle A’B’C’\end{align*}.
  3. What can you say about \begin{align*}\triangle ABC\end{align*} and \begin{align*}\triangle A’B’C’\end{align*}? Can you say this for any translation?
  4. 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 #14.
  5. 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 #15.
  6. Find the translation rule that would move \begin{align*}A\end{align*} to \begin{align*}A’(0, 0)\end{align*}, for #16.
  7. 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*} to the right 5 units and up 11 units. Write the translation rule.
  8. 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*} to the left 3 units and down 7 units. Write the translation rule.

Review Queue Answers

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

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Date Created:
Feb 22, 2012
Last Modified:
Feb 03, 2016
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