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5. Translations of Figures in a Plane
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Chapter 1
5. 

Translations of Figures in a Plane

This lesson focuses on the concept of translation in geometry, which involves moving geometric figures in a plane without altering their shape, size, or orientation. This is particularly useful in fields like engineering, architecture, and design where precise movement of shapes is required. For instance, in architecture, you might need to move a window design from one wall to another while maintaining its dimensions and orientation. The lesson provides exercises and examples to help you understand how to perform these translations effectively.

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Student Learning Objectives:
  • Translate plane figures using vectors
12 Theory slides
10 Exercises - Grade E - A
Each lesson is meant to take 1-2 classroom sessions
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Translations of Figures in a Plane
Slide of 12
Besides rotations, another common type of transformation is translations. As the name suggests, the effect a translation has on an object is a slide across a plane. In this lesson, the formal definition and properties of a translation will be developed.

Catch-Up and Review
Here is some recommended reading before getting started with this lesson.

Explore

Translations Using Vectors

In the following applet, the magnitude and direction of the vector labeled v can be set. Then, △ ABC can be translated along the vector.

Once the translation is performed, is there any relationship between AA', BB', CC', and v?
Discussion

Properties of Translations

From the exploration, the following conclusions can be drawn.

  • The lengths of AA', BB', and CC' are equal to the magnitude of v.
  • The three segments AA', BB', and CC' are parallel to v.

To perform a translation, a vector is required. This implies that the direction of the translation plays an important role. To illustrate this statement, consider the following diagram.

To triangles such that AA', BB', and CC' are all parallel and congruent

It can be seen that the lengths of AA', BB', and CC' are equal to the magnitude of v. Also, AA', BB', and CC' are parallel to v. Therefore, the same two conclusions written above apply to this diagram. AA'=|v| BB'=|v| CC'=|v| and AA' ∥ v BB' ∥ v CC' ∥ v However, △ A'B'C' is not a translation of △ ABC. This is because, although vectors BB' and CC' have the same direction as v, the vector A A' does not have the same direction as v.

Vectors BB' and CC' have the same direction as vector v, while AA' has the opposite direction

Consequently, and referring to what can be inferred from the exploration applet, a third conclusion can be drawn.

The vectors AA', BB', and CC' have the same direction as v.

Discussion

Definition of Translations

Be aware that the three conclusions written before do not depend on the preimage. They hold true even when the preimage is a single point, a segment, a polygon, or any other figure. Then, these properties can be used to define a translation properly.

Concept

Translation of Geometric Objects

A translation is a transformation that moves every point of a figure the same distance in the same direction. More precisely, a translation along a vector v maps every point A in the plane onto its image A' such that the following statements hold true.

These three properties imply that the quadrilateral formed by A, A', the tip of v, and the tail of v is a parallelogram.

Vector v, point A and its image A under a translation along v

Since translations preserve side lengths and angle measures, they are rigid motions.
Explore

Translating a Figure

By definition of translation, there is a relationship between the vector v that defines this transformation and the segment connecting a point with its image. Below it will be explored whether there is a relationship between a figure and its image after a translation. In the applet, the measure of ∠ ABC and the length of its sides can be set. Also, the magnitude and direction of v can be defined. Once done, ∠ ABC can be translated along v.

After performing the translation, consider the following questions.

  • What do the preimage and the image have in common?
  • Apart from the dimensions, is there any other relationship between AB and A'B'?
  • Do BC and B'C' have the same relationship as AB and A'B'?
Example

Identifying Translations

As the previous exploration shows, translations preserve side lengths and angle measures. That confirms translations are rigid motions. Additionally, take note that translations map segments onto parallel segments. Consider the polygon P shown in the middle of the diagram below. The other polygons are images of P after different transformations.

What polygons are the image of P after a translation?

Answer
Only polygons P_4, P_6, and P_(14) are the image of P after a translation.

Hint
Remember that translations map segments onto parallel segments. Since P looks like an arrow pointing up, the images of P after a translation will also look like an arrow pointing up.

Solution
Since translations map segments onto parallel segments, the sides of P are parallel to the sides of its image after any translation. Notice that P looks like an arrow pointing up.

Therefore, the image of P after any translation will also look like an arrow pointing up. In the given diagram, it can be seen that only P_4, P_6, and P_(14) satisfy this condition. Consequently, these polygons are images of P after a translation.

To obtain the other polygons, P must be translated and a rotated.

Example

Translations and Polygon's Vertices

In the previous example, only the shape of polygon P was used to determine which polygons were the image of P after a translation. When the vertices are labeled, keep an eye on them. Consider the following three squares.

Which of the squares S_1 or S_2, if either, is the image of PQRS after a translation?

Hint
Pay close attention to the labels of the vertices. Remember that, for every preimage A and image A', AA' must be parallel to the vector that defines the translation. This means that all the segments connecting a point to its image are parallel to each other.

Solution
At first glance, it seems like both squares S_1 and S_2 are the image of PQRS after a translation. In fact, if the vertices were not labeled, that would be the case.

However, because the vertices are labeled, closer attention needs to be paid. By definition of translation, for every preimage A and its image A', the following relations hold true.

  • The segment whose endpoints are A and A', AA', has the same length as the magnitude of the vector that defines the translation. This means all the segments connecting a point to its image have the same length.
  • The segment whose endpoints are A and A', AA', is parallel to the vector that defines the translation. This means that all the segments connecting a point to its image are parallel to each other.

With the above information in mind, the segments that connect a vertex and its image will be drawn.

Considering only the squares PQRS and S_2, the following observations about the segments that connect the vertices and their corresponding images can be made.

  • These segment have different lengths.
  • These segments are not parallel.

Therefore, S_2 is not the image of PQRS after a translation. Conversely, S_1 satisfies the two conditions previously written. Even more, since every vector that connects a vertex to its image have the same direction, it can be concluded that S_1 is a translation of PQRS.

Discussion

Translations Performed by Hand

Translations can be performed by hand with the help of a straightedge and a compass.

Triangle ABC and vector PQ

To translate △ ABC along v follow the four steps below.

1
Draw the Line l Containing v
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Using a straightedge, draw the line that contains the vector that defines the translation. In this case, this vector is v.

2
Draw Parallel Lines to l Through Each Vertex
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Draw parallel lines to l through each vertex.

3
Copy v Over the Lines Drawn
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Over the three lines drawn in the previous step, using a compass copy vector v in such way that each vertex of △ ABC is the tail of the new vector.

The tip of each vector is the image of each vertex.

4
Draw the Image
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Draw the triangle formed by the image points A', B', and C'.

Extra

 Alternative Method
By definition of translation, vectors AA' and v are parallel and have the same magnitude and direction. This makes PAA'Q a parallelogram. With this in mind, the translation of △ ABC can be performed as follows.

  • Copy the length of v: open the compass to the length of v. Then, with this setting, place the compass tip on each vertex of △ ABC and draw an arc.

  • Copy the length from P to each vertex: place the compass tip on P and open the compass to C. Then, with this setting, place the compass tip on Q and make an arc that intersects the arc centered at C. The point of intersection is C'.
    Here, PCC'Q is a parallelogram. Repeating the same process, the image of the rest of the vertices can be found.
  • Draw the Image: Draw the triangle formed by the image points A', B', and C'.
Discussion

Translations in the Coordinate Plane

In the coordinate plane, the component form of the translation vector v is closely related to the coordinates of the image of a point P(a,b). Investigate this relationship by using the following applet.

Applet to investigate the coordinates of a point after a translation along vector (v1,v2).
As it can be seen in the applet, the image of P(a,b) after a translation along the vector v = ⟨ v_1,v_2 ⟩ is the point P'(a+v_1,b+v_2). In other words, to find the image of P(a,b), add v_1 to the x-coordinate and v_2 to the y-coordinate.

When a figure and its image after a particular translation are given, it is possible to find the translation vector. In fact, the translation vector is any vector connecting a preimage with its image. That is, any vector of the form AA'.
Pop Quiz

Practice Translations

In the following applet, one of the following tasks may be required.

  • Translate △ ABC along the given vector v.
  • Write the component form of the translation vector that maps △ ABC onto △ A'B'C'.

To translate △ ABC, place points A', B', and C' where they should be after the translation is applied.

Performing random translations to random triangles
Example

Composition of Translations

When learning about rotations, it was said that the composition of two rotations could be a translation. Now, the composition of two translations will be examined.

Consider the following pair of quadrilaterals P_1 and P_2. Also, consider a pair of different translations. One translation along vector u = ⟨ 5,1⟩ and the other along v = ⟨ -2,-4⟩.

Pair of quadrilaterals in the coordinate plane
a

In which order must the translations be applied to P_1 so that it is mapped onto P_2? Only one option is correct.

b

Is there a single translation that maps P_1 onto P_2?

If so, write the component form of the translation vector.

Hint
a

Translating along u is the same as translating 5 units to the right and 1 unit up. Translating along v is the same as translating 2 units to the left and 4 unit down. Perform both compositions and compare the resulting images.

b

Consider the vectors that connect the vertices and their images. Are they all parallel and with the same magnitude? Do all of them have the same direction?

Solution
a

Translating P_1 along u = ⟨ 5,1⟩ is equivalent to translating it 5 units to the right and 1 unit up. Similarly, translating P_1 along v = ⟨ -2,-4⟩ is equivalent to translating it 2 units to the left and 1 unit down.

Translating Along Is Equivalent To
u=⟨ 5,1⟩ Translating 5 units to the right and 1 unit up.
v=⟨ -2,-4⟩ Translating 2 units to the left and 4 unit down.

To determine the correct order, both compositions should be tried. First, perform the translation along u followed by the translation along v.

Translation along u first followed by the translation along v

As it can be seen, the above composition maps P_1 onto P_2. Next, perform the translation along v followed by the translation along u.

Translation along v first followed by the translation along u

The last composition also mapped P_1 onto P_2. Consequently, the order in which the translations are applied is insignificant. This implies that the composition of translations is commutative.

b

The fact that the corresponding sides of both polygons are parallel suggests that P_2 can be the image of P_1 under a single translation. To confirm this, draw the vectors connecting corresponding vertices.

Vectors connecting each vertex with its image

All the vectors drawn seem to be parallel and with the same magnitude. Even more, they all have the same direction. This could be checked by finding the component form of each vector.

Vertex Image Vector Component Form
( -4, 0) ( -1, -3) ⟨ -1-( -4), -3- 0 ⟩ ⟨ 3, -3 ⟩
( -1, 1) ( 2, -2) ⟨ 2-( -1), -2- 1 ⟩ ⟨ 3, -3 ⟩
( -1, 2) ( 2, -1) ⟨ 2-( -1), -1- 2 ⟩ ⟨ 3, -3 ⟩
( -3, 3) ( 0, 0) ⟨ 0-( -3), 0- 3 ⟩ ⟨ 3, -3 ⟩

As the table shows, all the vectors connecting a preimage with its image have the same component form. This confirms that the vectors are parallel and have the same magnitude and direction. Consequently, a translation along ⟨ 3, -3 ⟩ maps P_1 onto P_2.
Closure

Translations in the Real World

The two conclusions obtained in the previous example are not a coincidence. In fact, these are general results when performing a composition of translations.

The composition of translation is commutative and its translation vector is the sum of the corresponding translation vectors

Be aware that a composition of transformations might involve translations and rotations. This combination can produce interesting images.

In interior design, it is pretty common to see designs consisting of a single preimage and its images under different transformations such as translations and rotations. Below, two different kitchen tile designs are made using just four right triangles.

Showing two different kitchen tile designs by translating four right triangles to different positions of a square
Although four triangles were used, notice that three of them can be seen as the image of the fourth triangle after three different rotations. To finish this lesson, try creating your own design by performing translations on the given triangles.

Applet to create its own kitchen tile design



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Level 3
Translations of Figures in a Plane
Exercise 2.1

The vector PQ = ⟨ 4,1 ⟩ describes the translation of A(a-1,b) to B(2b+1,4+a). Determine the coordinates of A.

The translation vector tells us that we have to add 4 to the x-coordinate and 1 to the y-coordinate of A in order to move from A to B. With this information, we can write two equations. Translation:& PQ=⟨ 4, 1⟩ A ( x_A,y_A):& ( a-1,b) B ( x_B,y_B):& ( 2b+1,4+a) [-1em] Equation (I):& a-1 + 4= 2b+1 Equation (II):& b + 1= 4+a If we combine these equations, we get a system of equations which can be solved by using the Substitution Method.

Now we can determine the coordinates of A by substituting a= -4, and b= -1 into the expressions for the x- and y-coordinate of A. A( -4-1, -1) → A(-5,-1)

E
Hint & Answer
S
Solution

A linear function with a y-intercept of b and a slope of m is translated according to the vector ⟨ a,b ⟩. What is the y-intercept of the translated line?

It is given that a line is translated using the vector ⟨ a,b⟩. Let's recall that if we translate a linear function along a vector ⟨ a, b⟩, then we should subtract a from x in the equation and add b to the entire function rule. y=m(x- a)+b+ b Let's simplify the right-hand side.

Now that we have found the equation of the translated line, we can evaluate the y-intercept by substituting 0 for x. Then solve for y.

The y-intercept of the translated line is 2b-ma.

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Hint & Answer
S
Solution

Heichi is visiting colleges to gather information before deciding which one to apply to. He comes up with an example itinerary. He would leave his home and drive 120 miles east and 60 miles south to visit Harvard. From there, he would drive 80 miles west and 40 miles north to visit Princeton. Finally, he would a fly 150 miles northeast to check out Yale. What is the distance from his house to Yale, if he were to fly directly. Round the answer to the nearest mile.

Let's place Heichi's home at the origin of a coordinate plane and let 1 unit represent 1 mile. To go to Harvard, Heichi has to drive 120 miles east and 60 miles south. H_x:& 0 + 120 =120 H_y:& 0 - 60 =-60 This places Harvard at point H(120,- 60) in the coordinate plane.

Next up is Princeton. By using Harvard as a reference point, Heichi has to go 80 miles west and 40 miles north. Therefore, we can find the coordinates of Princeton P, by subtracting 80 from the x-coordinate and adding 40 to the y-coordinate of Harvard. P_x:& 120 - 80 =40 P_y:& -60 + 40 =-20 Now we know that the location of Princeton in our coordinate system is at P(40,- 20).

Finally, Heichi would fly 150 miles northeast from Princeton to Yale. In the diagram, this journey can be described as a right isosceles triangle.

To determine the position of Yale J, we have to find the lengths of this triangle's legs. Since this is an isosceles triangle, its legs are congruent. Furthermore, because the triangle is right, we can find the length of the legs by using the Pythagorean Theorem.

Each leg of the isosceles triangle that describes the journey from Princeton to Yale has the length 75sqrt(2).

To obtain the x-coordinate of J, we have to add 75sqrt(2) to 40. Similarly, to obtain the y-coordinate of J, we have to add 75sqrt(2) to - 20. J_x:& 40 + 75sqrt(2) J_y:& - 20 + 75sqrt(2) Now that we know the coordinates of J, we can make out another right triangle where the distance between Heichi's house and Yale is the hypotenuse.

With this information, we can solve for the distance between Heichi's home and Yale by using the Pythagorean Theorem.

The distance from Heichi's home to Yale if he were to fly directly is about 170 miles.

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Hint & Answer
S
Solution

Triangle ABC has vertices at A(-5,-2), B(-3,4), and C(1,-2). What translation was performed on this triangle to create triangle A'B'C' if the midpoint of A'B' is D'(4,-2).

Let's first plot △ ABC and the midpoint D'.

Since a translation is a rigid motion, every point on the perimeter of △ ABC will be moved the same way. The midpoint on AB, is the point on △ ABC corresponding to D'. Let's find this midpoint D using the Midpoint Formula.

The midpoint of AB is at D(-4,1). Now we can determine the translation by measuring the horizontal and vertical distance between D and D'.

Let's formalize the translation. (x,y)→ (x + 8,y - 3)

E
Hint & Answer
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Solution

Translations of Figures in a Plane

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