# Lesson Explainer: Translations Mathematics

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 “image” 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 along a ray at a magnitude of .

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

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

We then set the radius of the compass equal to and trace a circle of this radius centered at .

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

We can follow this process for point 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 and are parallel line segments of the same length, is a parallelogram. This means that and . In a similar way, we can see that the directions of rays are kept the same, so will be translated to .

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

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 units in the direction . We can set our compass to have a radius of by putting the point of the compass at and setting the pencil at .

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

We can see that since it is a radius of a circle with radius . We can also note that and they have the same direction. So, is the image of after a translation units in the direction . We can apply this process to points and .

We extend to be the ray and we trace a circle of radius centered at to find and a congruent circle centered at to find as shown.

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

We say that is the image of under this translation.

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

### Definition: Translation of a Point

We say that is the image of point under the translation units in the direction if , , and and 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 is translated to , then .

Translations do not affect directions. In particular, if is translated to , then . Similarly, will be translated to .

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 is following a translation of magnitude in the direction of . Which of the following diagrams represents this?

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

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 .

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

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

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

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

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

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

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

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 is following a translation of magnitude in the direction of . Which of the following diagrams represents this?

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 and .

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

However, the translation should be in the direction of , 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 with an endpoint at as shown.

We see that 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 and 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 , one starting at and the other starting at .

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

We note that and are now a distance of from and , respectively, and both are in the direction of . So, they are the images of and 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 , where , , and , then draw the image of by a translation of a 15-centimetre magnitude in the direction of . Choose which one of the following figures matches your drawing.

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

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

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

Our triangle now matches the size and orientation of given in each of the options. We now want to translate the triangle 15 cm in the direction . We can do this by noting that is a horizontal line and traveling from to 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 , , and respectively.

These are the vertices of the triangle after a translation 15 cm in the direction of , 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, is a square, where all interior squares are congruent and . Then, the image of by a translation of a magnitude of 2 cm in the direction of is .

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 is the image of point under the translation 2 cm in the direction if the following two conditions hold:

1. .
2. and have the same direction.

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

This is the direction we need to translate , so we want to draw a ray in the same direction as from . 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  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 .

Thus, is the image of under a translation 2 cm in the direction . If we follow the same process for , we get the following.

Therefore, is the image of under a translation 2 cm in the direction . 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 by a translation of a magnitude of 2 cm in the direction of is .

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 and its image under a translation.

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

This gives us the following result.

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

If is translated on by a translation, then the translation has magnitude and direction .

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 , , , and are congruent. is the image of by a translation of magnitude in the direction of .

1. ,
2. ,
3. ,
4. ,
5. ,

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 under the translation. We are told that is translated onto ; we can highlight these triangles as shown.

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

Hence, the translation is of magnitude and in the direction , 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 is the image of point under the translation units in the direction if the following two conditions hold:
1. .
2. and have the same direction.
• Translations do not affect the length of line segments or their direction. In particular, if is translated to , then and .
• We can translate under the translation units in the direction by constructing a ray parallel to and in the same direction as starting at and then tracing a circle of radius centered at . The point of intersection between the ray and circle will be units away from in the direction .
• 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.