Lesson Video: Translations | Nagwa Lesson Video: Translations | Nagwa

Lesson Video: Translations Mathematics

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

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Video Transcript

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

A translation can be thought of as sliding an object a fixed distance in a fixed direction. For example, we can translate a square of side length one centimeter to the right by two centimeters, as shown. We call the new position of the square its image after the translation. Both squares are the same size and orientation. The translation only affects the position of the object.

We say that 𝐴 prime is the image of point 𝐴 under the translation π‘₯ units in the direction of the ray 𝐡𝐢 if the length 𝐴𝐴 prime is equal to π‘₯, the rays 𝐴𝐴 prime and 𝐡𝐢 are parallel and have the same direction. We can translate line segments and polygons by translating their endpoints or vertices.

Translations do not affect lengths. So, if the line segment 𝐴𝐡 is translated to the line segment 𝐴 prime 𝐡 prime, then the length of 𝐴𝐡 is equal to the length of 𝐴 prime 𝐡 prime. Translations also do not affect directions. In particular, if the line segment 𝐴𝐡 is translated to the line segment 𝐴 prime 𝐡 prime, then the line segment 𝐴𝐡 is parallel to the line segment 𝐴 prime 𝐡 prime. Similarly, the ray 𝐴𝐡 will be translated to the ray 𝐴 prime 𝐡 prime.

Let’s take a look at an example of determining the correct translation of a point in a given direction.

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

Starting with the point in space 𝑋, we are told that the translation is of magnitude 𝑀𝑁. This means that the image of 𝑋, 𝑋 prime, will lie somewhere on the circle of radius 𝑀𝑁 centered at the original point 𝑋. We also have the ray 𝑀𝑁, which points in the direction of the translation. We can therefore construct a ray starting from the point 𝑋 that is parallel and in the same direction to the ray 𝑀𝑁. And where this new ray touches the circle will be where the image 𝑋 prime lies.

To do this, we start by constructing a ray that passes through 𝑋 in the direction of 𝑀. We seek to duplicate the angle 𝑁𝑀𝑋 at the point 𝑋. We can do this by constructing two arcs of circles of equal radius centered at the points 𝑋 and 𝑀, labeling the points of intersection between the circle centered at 𝑀 and the two rays 𝐴 and 𝐡. We then measure this distance between 𝐴 and 𝐡 and draw another arc of another circle centered on the far point of intersection between the circle arc centered at 𝑋 and the ray passing through 𝑋 and 𝑀. A ray passing through the point 𝑋 and the point of intersection of these two arcs will be parallel to the ray 𝑀𝑁.

We can therefore draw the ray starting at the point 𝑋 and passing through this point of intersection. And the point of intersection between this ray and the circle will be the image of the point 𝑋 prime. Looking at the possible answers, we can see that this matches with the diagram in 𝐴.

Let’s now look at an example where we will determine which of five given diagrams shows the correct translation of a line segment.

The image of the line segment 𝐢𝐷 is the line segment 𝐢 prime 𝐷 prime following a translation of magnitude 𝐴𝐡 in the direction of the ray 𝐴𝐡. Which of the following diagrams represents this?

We can approach this problem with a process of elimination. Recall that translating a line segment is equivalent to translating its endpoints, in this case 𝐢 and 𝐷, to their images 𝐢 prime and 𝐷 prime and connecting the images of the points with a new line segment. In all five diagrams, the ray 𝐴𝐡 is pointing up and to the left. So any translation of the line segment that doesn’t place the image in this direction must be incorrect. This immediately rules out answers (A) and (B).

Next, the translation is of magnitude 𝐴𝐡, meaning the distance between one endpoint of the line segment and its image must be the length 𝐴𝐡. In answer (D), the distance between the points 𝐢 and its image 𝐢 prime is clearly less than the length 𝐴𝐡. So the answer cannot be (D).

And finally translations leave the lengths of line segments unchanged. And in answer (E), we can see that the length of the image line segment 𝐢 prime 𝐷 prime is greater than the length of the original line segment 𝐢𝐷. Therefore, the answer cannot be (E).

Therefore, the only remaining answer is (C). And we can see that the line segment is indeed translated in the correct direction by the correct distance and does not affect the size and orientation of the line.

We can check if this is correct by constructing the translation. We’ve cleared the incorrect answers to give some space. Taking the line segment 𝐢𝐷, we construct two rays in the same direction as the ray 𝐴𝐡, one starting at 𝐢 and the other starting at 𝐷. Next, we set the radius of a compass equal to the length 𝐴𝐡 and trace circles centered at 𝐢 and 𝐷. We label the points of intersection between the circles and the rays 𝐢 prime and 𝐷 prime. 𝐢 prime and 𝐷 prime are a distance 𝐴𝐡 from 𝐢 and 𝐷, respectively, and in the direction of the ray 𝐴𝐡. So they are the images of 𝐢 and 𝐷 under the translation. Therefore, the line segment joining the points 𝐢 prime and 𝐷 prime is the image of the line segment 𝐢𝐷 under the translation. And this matches visually with option (C).

Let’s now look at an example of how to determine the correct translation of a line segment in a square grid.

Fill in the blank. In the figure, 𝐴𝐡𝐢𝐷 is a square, where all interior squares are congruent and 𝐴𝑀 is equal to one centimeter. Then, the image of the line segment π»π‘Œ by a translation of magnitude two centimeters in the direction of the ray 𝐿𝐡 is what.

To start with, the line segment π»π‘Œ is this one here and the ray 𝐿𝐡 is this ray here. Translating a line segment is equivalent to translating its endpoints, in this case 𝐻 and π‘Œ, and then connecting the images of these points with a new line segment. This means we can answer this question by finding the images of the endpoints of this line segment under the translation.

All of the squares are congruent and have a side length of one centimeter. The sides of the squares in the left-to-right direction are also parallel. Therefore, the translation of magnitude two centimeters in the direction of the ray 𝐿𝐡 is equivalent to the ray 𝐻𝐸. Therefore, the image of 𝐻 under the translation 𝐻 prime is the point 𝐸. Similarly, the image of the point π‘Œ under the translation π‘Œ prime is the point 𝑁. Therefore, the image of the line segment π»π‘Œ by a translation of magnitude two centimeters in the direction of the ray 𝐿𝐡 is the line segment 𝐸𝑁.

So 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 point 𝐴 and its image under a translation, 𝐴 prime. The ray 𝐴𝐴 prime is in the direction of the translation, since this is the direction between 𝐴 and its image 𝐴 prime. Likewise, the magnitude of this translation must be the length 𝐴𝐴 prime, since this is the distance 𝐴 is translated. This leads to the following property. If the point 𝐴 is translated on 𝐴 prime by a translation, then the translation has magnitude equal to the length 𝐴𝐴 prime and in the direction of the ray 𝐴𝐴 prime.

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

Fill in the blank. In the figure, the triangles 𝑋𝐿𝑀, πΏπ‘Œπ‘, 𝑁𝑀𝐿, and 𝑀𝑁𝑍 are congruent. The triangle πΏπ‘Œπ‘ is the image of the triangle 𝑀𝑁𝑍 by a translation of magnitude what in the direction of what.

The magnitude of a translation is equivalent to the distance between a point and its image under the translation. And the direction of a translation can be described by the ray from a point passing through its image. The whole triangle 𝑀𝑁𝑍 can be translated by translating its vertices. Therefore, we can determine the magnitude and direction of the translation of the triangle by finding the image under the translation of just one of its vertices.

So we can pick a vertex, for example, 𝑀, and determine its image under the translation. The two triangles 𝑀𝑁𝑍 and πΏπ‘Œπ‘ are highlighted here as the orange and purple triangles, respectively. Translations do not affect the scale or orientation of a shape. Therefore, the relative positions of the vertices in the triangle will remain the same. The top vertex will remain at the top, and so on.

Therefore, the top vertex of the image triangle 𝐿 is the image of the top vertex of the original triangle 𝑀. The magnitude of the translation is therefore given by the distance between 𝑀 and 𝐿, 𝑀𝐿. And the direction is given by the direction of the ray starting at the vertex 𝑀 and passing through 𝐿, 𝑀𝐿. Therefore, the answer is (A).

Let’s finish this video by going over some 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 prime is the image of point 𝐴 under the translation 𝑋 units in the direction of the ray 𝐡𝐢 if 𝐴𝐴 prime is equal to 𝑋 and the rays 𝐴𝐴 prime and 𝐡𝐢 have the same direction.

Translations do not affect the length of line segments or their direction. In particular, if the line segment 𝐴𝐡 is translated onto the line segment 𝐴 prime 𝐡 prime, then the length 𝐴𝐡 is equal to the length 𝐴 prime 𝐡 prime and the line segments 𝐴𝐡 and 𝐴 prime 𝐡 prime are parallel. And finally, 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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