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Lesson Explainer: Translations Mathematics • 11th Grade

In this explainer, we will learn how to translate points, line segments, and shapes, given the direction and magnitude of the translation.

There are many different ways we can move objects without changing their shape. We will be focusing on translations, which can be thought of as sliding the object a fixed distance and in a fixed direction. For example, we can translate a square of side length 1 cm to the right by 2 cm as shown in the following diagram.

We call the new position of the square its β€œimage” after the translation of 2 cm to the right. We can note that both squares are of the same size and have the same orientation; this is important since translations never change the size, shape, or orientation. A translation will only ever affect the position of the object.

We can translate any object, provided we are given the direction of the translation and the distance (or magnitude) we need to translate. We can represent each of these parts of the translation in different ways.

For example, we have seen that we can represent the magnitude of the translation with a length such as 2 cm; however, we can also represent the magnitude by using the length of a given line segment.

Similarly, we can represent the direction of the translation using the cardinal directions; however, for more general directions, we can use a ray.

To see these in action, let’s translate a line segment 𝐢𝐷 along a ray 𝐴𝐡 at a magnitude of 𝐴𝐡.

Let’s start by translating the endpoints along 𝐴𝐡; we can do this by constructing rays parallel to 𝐴𝐡 starting at 𝐢 and 𝐷.

We recall that to sketch a line parallel to 𝐴𝐡 starting at 𝐢, we sketch 𝐴𝐢 and we want to duplicate ∠𝐢𝐴𝐡 at 𝐢. We do this by tracing a circle at 𝐴 that intersects the lines as shown. We also sketch a congruent circle centered at 𝐢, labeling the points as shown.

We then set the radius of the compass equal to π‘‹π‘Œ and trace a circle of this radius centered at 𝐸.

οƒͺ𝐢𝐹 is then parallel to 𝐴𝐡. We can sketch a circle of radius 𝐴𝐡 to find the point on οƒͺ𝐢𝐹 that is 𝐴𝐡 units from 𝐢. This will be the image of 𝐢 after the translation. We call this 𝐢′.

We can follow this process for point 𝐷 and every point on the line.

Since every point on the line segment is translated the same amount, we could connect the endpoints of the translated line segment to find its image after the translation.

We can use this method to translate any line segment or polygon by translating each of its vertices. We can notice that as 𝐢𝐢′ and 𝐷𝐷′ are parallel line segments of the same length, 𝐢𝐷𝐷′𝐢′ is a parallelogram. This means that 𝐢𝐷=𝐢′𝐷′ and 𝐢𝐷⫽𝐢′𝐷′. In a similar way, we can see that the directions of rays are kept the same, so 𝐢𝐷 will be translated to 𝐢′𝐷′.

Let’s see how to apply this to translate a given triangle 𝐴𝐡 units in the direction οƒͺ𝐡𝐢.

One way of doing this is to use a compass and straightedge. We can start by noting that translating a triangle will only change its position and not its shape. This means we can translate the triangle by just translating its vertices.

We want to translate each vertex 𝐴𝐡 units in the direction οƒͺ𝐡𝐢. We can set our compass to have a radius of 𝐴𝐡 by putting the point of the compass at 𝐴 and setting the pencil at 𝐡.

To translate point 𝐴 in the direction οƒͺ𝐡𝐢, we first draw a ray parallel to οƒͺ𝐡𝐢 at point 𝐴. We then trace a circle centered at 𝐴 with a radius of 𝐴𝐡; this will make the translation 𝐴𝐡 units. We call the point of intersection between the circle and ray 𝐴′ as shown.

We can see that 𝐴𝐴′=𝐴𝐡 since it is a radius of a circle with radius 𝐴𝐡. We can also note that 𝐴𝐴′⫽οƒͺ𝐡𝐢 and they have the same direction. So, 𝐴′ is the image of 𝐴 after a translation 𝐴𝐡 units in the direction οƒͺ𝐡𝐢. We can apply this process to points 𝐡 and 𝐢.

We extend 𝐡𝐢 to be the ray οƒͺ𝐡𝐢 and we trace a circle of radius 𝐴𝐡 centered at 𝐡 to find 𝐡′ and a congruent circle centered at 𝐢 to find 𝐢′ as shown.

We can now note that each of 𝐴′, 𝐡′, and 𝐢′ is translated 𝐴𝐡 units in the direction 𝐴𝐡, so if we connect these points with line segments, then we have translated triangle 𝐴𝐡𝐢 for 𝐴𝐡 units in the direction 𝐴𝐡.

We say that △𝐴′𝐡′𝐢′ is the image of △𝐴𝐡𝐢 under this translation.

We can define a translation formally and the notation we use as follows.

Definition: Translation of a Point

We say that 𝐴′ is the image of point 𝐴 under the translation π‘₯ units in the direction οƒͺ𝐡𝐢 if 𝐴𝐴′=π‘₯, 𝐴𝐴′⫽οƒͺ𝐡𝐢, and 𝐴𝐴′ and οƒͺ𝐡𝐢 have the same direction.

We can extend this to translating any shape by translating all of the points on the shape.

We can also describe the properties of translations.

Property: Properties of Translations

We can translate line segments and polygons by translating their endpoints or vertices.

Translations do not affect lengths, so if 𝐴𝐡 is translated to 𝐴′𝐡′, then 𝐴𝐡=𝐴′𝐡′.

Translations do not affect directions. In particular, if 𝐴𝐡 is translated to 𝐴′𝐡′, then 𝐴𝐡⫽𝐴′𝐡′. Similarly, 𝐴𝐡 will be translated to 𝐴′𝐡′.

Let’s now see an example of determining the correct translation of a point in a given direction.

Example 1: Translating a Point given the Direction and Magnitude of the Translation

The image of point 𝑋 is 𝑋′ following a translation of magnitude 𝑀𝑁 in the direction of 𝑀𝑁. Which of the following diagrams represents this?

Answer

We first recall that translating a point 𝑀𝑁 units in the direction 𝑀𝑁 means that we move its position on the plane so that this new point 𝑋′ is 𝑀𝑁 units away from 𝑋, so 𝑋𝑋′=𝑀𝑁 and 𝑋𝑋′ has the same direction as 𝑀𝑁.

This gives us two ways of answering the question: either we check the five given options to see if they satisfy both conditions or we use these two conditions to find point 𝑋′.

First, we can trace a circle of radius 𝑀𝑁 centered at 𝑋. We set the radius of the compass equal to 𝑀𝑁 by placing the point of the compass at 𝑀 and the pencil at 𝑁. This circle will contain all points of distance 𝑀𝑁 from 𝑋. So, 𝑋′ must lie on this circle.

We can then construct a ray parallel to 𝑀𝑁 and in the same direction as 𝑀𝑁 starting at 𝑋.

We do this by sketching ⃖⃗𝑋𝑀 and then tracing circles centered at 𝑀 and 𝑋, labeling the points of intersection as shown.

We then set the radius of our compass equal to 𝐴𝐡 and sketch the arc of a circle of this radius centered at 𝐢. The ray between 𝑋 and the point of intersection of these circles is parallel to 𝑀𝑁.

The point of intersection between the ray and the circle of radius 𝑀𝑁 will be 𝑋′.

We see that 𝑋′ is a distance of 𝑀𝑁 from 𝑋 and 𝑋𝑋′ has the same direction as 𝑀𝑁. Thus, 𝑋′ is the image of 𝑋 under the translation 𝑀𝑁 units in the direction 𝑀𝑁.

We can see that this is the same as the diagram given in option 𝐴.

In our next example, we will determine which of five given diagrams shows the correct translation of a line segment.

Example 2: Translating a Line Segment given the Direction and Magnitude of the Translation

The image of 𝐢𝐷 is 𝐢′𝐷′ following a translation of magnitude 𝐴𝐡 in the direction of 𝐴𝐡. Which of the following diagrams represents this?

Answer

We first recall that translating a line segment is the same as translating its endpoints and then connecting the images of these points with a line segment.

This means we can answer this question by finding the images of the endpoints of this line segment under the translation. However, this would require us to translate each of the line segments in the given examples.

Instead, we can recall that translations leave the lengths and directions of line segments unchanged. In particular, we should have that 𝐢𝐷=𝐢′𝐷′ and 𝐢𝐷⫽𝐢′𝐷′.

This allows us to eliminate options. First, in options A and B, we see that the image of 𝐢𝐷 is downward and to the right.

However, the translation should be in the direction of 𝐴𝐡, which is upward and to the left; so this cannot be correct.

In option D, we can note that the translation is not the correct distance by adding a congruent line segment parallel to 𝐴𝐡 with an endpoint at 𝐷 as shown.

We see that 𝐷′ must be the other endpoint of this line segment to be the correct translation, so this is not the correct option.

Finally, we can note in option E that the line segments 𝐢𝐷 and 𝐢′𝐷′ have different lengths. So, this cannot be a translation.

This only leaves option C. We can check if this is correct by constructing the translation.

We will translate both endpoints at the same time. First, we will construct two rays in the same direction as 𝐴𝐡, one starting at 𝐢 and the other starting at 𝐷.

Second, we set the radius of our compass equal to 𝐴𝐡 and trace circles centered at 𝐢 and 𝐷. We label the points of intersection between the circles and rays 𝐢′ and 𝐷′ as shown.

We note that 𝐢′ and 𝐷′ are now a distance of 𝐴𝐡 from 𝐢 and 𝐷, respectively, and both are in the direction of 𝐴𝐡. So, they are the images of 𝐢 and 𝐷 under this translation.

Hence, option C is the correct translation.

In our next example, we will construct a given triangle and then translate the triangle a given distance in the direction of a ray between two of its vertices.

Example 3: Translating a Triangle by a Given Amount

Draw △𝐴𝐡𝐢, where 𝐴𝐢=3cm, 𝐴𝐡=4cm, and 𝐡𝐢=5cm, then draw the image of △𝐴𝐡𝐢 by a translation of a 15-centimetre magnitude in the direction of οƒͺ𝐢𝐡. Choose which one of the following figures matches your drawing.

Answer

We first need to sketch △𝐴𝐡𝐢, where 𝐴𝐢=3cm, 𝐴𝐡=4cm, and 𝐡𝐢=5cm. There are two ways of doing this. First, we can note that 3, 4, 5 is a Pythagorean triple, which tells us that △𝐴𝐡𝐢 is a right triangle with a hypotenuse of length 5 cm.

This method to sketch △𝐴𝐡𝐢 only works for Pythagorean triples. So instead, we will use a more general method to sketch the triangle. In the options, we see that 𝐡𝐢 is horizontal. So, we start by sketching a horizontal line of length 5 cm and we mark the endpoints 𝐢 and 𝐡.

Next, we trace a circle of radius 3 cm centered at 𝐢 and a circle of radius 4 cm centered at 𝐡. We can label either point of intersection between the circles 𝐴, since these points are a distance of 3 cm from 𝐢 and 4 cm from 𝐡. To match the diagrams in the options, we choose the point above 𝐡𝐢 as follows.

Our triangle now matches the size and orientation of △𝐴𝐡𝐢 given in each of the options. We now want to translate the triangle 15 cm in the direction οƒͺ𝐢𝐡. We can do this by noting that οƒͺ𝐢𝐡 is a horizontal line and traveling from 𝐢 to 𝐡 moves to the right. So, we need to translate the triangle 15 cm to the right.

We do this by translating each vertex 15 cm to the right and labeling these vertices 𝐴′, 𝐡′, and 𝐢′ respectively.

These are the vertices of the triangle after a translation 15 cm in the direction of οƒͺ𝐢𝐡, and we can see that this is the diagram given in option D.

In our next example, we will determine the correct translation of a line segment in a square grid.

Example 4: Identifying the Image of a Shape after a Translation

Fill in the blank: In the figure, 𝐴𝐡𝐢𝐷 is a square, where all interior squares are congruent and 𝐴𝑀=1cm. Then, the image of π»π‘Œ by a translation of a magnitude of 2 cm in the direction of οƒͺ𝐿𝐡 is .

Answer

We begin by recalling that translating a line segment is the same as translating its endpoints and then connecting the images of these points with a line segment.

This means we can answer this question by finding the images of the endpoints of this line segment under the translation. To do this, we recall that 𝐻′ is the image of point 𝐻 under the translation 2 cm in the direction οƒͺ𝐿𝐡 if the following two conditions hold:

  1. 𝐻𝐻′=2cm.
  2. 𝐻𝐻′ and οƒͺ𝐿𝐡 have the same direction.

Let’s start by translating 𝐻 Β 2 cm in the direction of οƒͺ𝐿𝐡. We do this by first sketching οƒͺ𝐿𝐡 on the grid.

This is the direction we need to translate 𝐻, so we want to draw a ray in the same direction as οƒͺ𝐿𝐡 from 𝐻. We can do this by noting that this is a grid of squares, so all of the horizontal lines will have the same direction. This gives us the following direction for the translation.

We want to translate 𝐻  2 cm in this direction, and we can do this by noting that all of the squares have lengths of 1 cm. This means that 𝐸𝐻=2cm.

Thus, 𝐸 is the image of 𝐻 under a translation 2 cm in the direction οƒͺ𝐿𝐡. If we follow the same process for π‘Œ, we get the following.

Therefore, 𝑁 is the image of π‘Œ under a translation 2 cm in the direction οƒͺ𝐿𝐡. It is worth noting that this is a translation of 2 cm to the left and we can apply this translation directly to the line segment.

In either case, we get that the image of π»π‘Œ by a translation of a magnitude of 2 cm in the direction of οƒͺ𝐿𝐡 is 𝐸𝑁.

Thus far, we have applied translations to given geometric objects. However, it is also possible to determine the magnitude and direction of a translation by using a point and its image under the translation. For example, imagine we are given 𝐴 and its image 𝐴′ under a translation.

We can note that 𝐴𝐴′ is the direction of the translation, since this is the direction between 𝐴 and its image. Similarly, we can see that the magnitude of this transition must be 𝐴𝐴′, since the distance 𝐴 is translated.

This gives us the following result.

Property: Magnitude and Direction of a Translation from a Point and Its Image

If 𝐴 is translated on 𝐴′ by a translation, then the translation has magnitude 𝐴𝐴′ and direction 𝐴𝐴′.

In our final example, we will use this property to determine the magnitude and direction of the translation of a given triangle onto its image.

Example 5: Identifying the Direction and Magnitude of a Given Translation

Fill in the blank: In the figure, the triangles 𝑋𝐿𝑀, πΏπ‘Œπ‘, 𝑁𝑀𝐿, and 𝑀𝑁𝑍 are congruent. β–³πΏπ‘Œπ‘ is the image of △𝑀𝑁𝑍 by a translation of magnitude in the direction of .

  1. 𝑀𝐿, 𝑀𝐿
  2. 𝑀𝐿, 𝐿𝑀
  3. 𝑀𝑋, 𝑀𝑋
  4. 𝑀𝑋, 𝑋𝑀
  5. 𝑀𝑁, 𝑀𝑁

Answer

We start by recalling that the magnitude of a translation means the distance between a point and its image under the translation, while the direction of a translation can be described by the ray from a point to its image. We can also recall that we translate triangles by translating their vertices. Therefore, we can determine the magnitude and direction of the translation by finding the image of one of the vertices of the triangle under the translation.

Let’s determine the image of 𝑀 under the translation. We are told that △𝑀𝑁𝑍 is translated onto β–³πΏπ‘Œπ‘; we can highlight these triangles as shown.

We see that the triangle is translated to the left. In particular, 𝑀 is translated onto 𝐿. The distance between these points is the magnitude of the translation and the direction of the ray from 𝑀 to 𝐿 is its direction.

Hence, the translation is of magnitude 𝑀𝐿 and in the direction 𝑀𝐿, which is option A.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • A translation of an object can be thought of as sliding the object in space without changing its shape, size, or orientation.
  • We say that 𝐴′ is the image of point 𝐴 under the translation π‘₯ units in the direction οƒͺ𝐡𝐢 if the following two conditions hold:
    1. 𝐴𝐴′=π‘₯.
    2. 𝐴𝐴′ and οƒͺ𝐡𝐢 have the same direction.
  • Translations do not affect the length of line segments or their direction. In particular, if 𝐴𝐡 is translated to 𝐴′𝐡′, then 𝐴𝐡=𝐴′𝐡′ and 𝐴𝐡⫽𝐴′𝐡′.
  • We can translate 𝐴 under the translation π‘₯ units in the direction οƒͺ𝐡𝐢 by constructing a ray parallel to and in the same direction as οƒͺ𝐡𝐢 starting at 𝐴 and then tracing a circle of radius π‘₯ centered at 𝐴. The point of intersection between the ray and circle will be π‘₯ units away from 𝐴 in the direction οƒͺ𝐡𝐢.
  • We can find the magnitude and direction of a translation by finding a point and its image under the translation and determining the distance between these points and the direction of the ray from the point to its image.

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