Lesson Video: Reflections | Nagwa Lesson Video: Reflections | Nagwa

# Lesson Video: Reflections Mathematics • First Year of Preparatory School

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In this video, we will learn how to reflect points and shapes in given reflection lines. There are many different ways we can transform an object without altering its shape or size.

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### Video Transcript

In this lesson, we will learn how to reflect points and shapes in given reflection lines. There are many different ways we can transform an object without altering its shape or size. Weβll focus on one of these transformations now. This one is called a reflection.

Reflections can be thought of as mirror images. When we look in a mirror, we see a virtual image of ourselves and our surroundings, as if we were on the other side of the mirror. In geometry, we reflect objects in a mirror line. And we call the result of the reflection an image. A particularly important feature of reflections is that the real object and its image must be the same distance from the mirror line. Weβve experienced this when moving towards or away from a mirror. The reflected image of ourselves moves as well.

Suppose that we wanted to reflect a single point π΄ through a mirror line to give the image, point π΄ prime. If we did so, we would notice that the mirror line is the perpendicular bisector of the line segment between these points. We can use this to define the reflection of any object more formally. A reflection is a transformation that preserves the perpendicular distances of all points from the mirror line.

Letβs now look at an example of how to apply this definition to identify the image of a reflection.

Five points are plotted on the grid in the figure below. When reflecting point π across the dotted line, which of the four other points will it map onto? Is it option (A) point π, option (B) point π, (C) point π, or (D) point π?

Remember, a reflection is a transformation that preserves the perpendicular distances of all points from the mirror line. In particular, if we define the image of π to be the point π prime, this is the reflection through the mirror line where the mirror line is the perpendicular bisector of the line segment that joins these points.

So, we can find the image of π by drawing a line perpendicular to the mirror line. Thatβs the dotted line in this picture. When we do, we see that this perpendicular line passes through point π. In particular, point π is the same perpendicular distance from the dotted line as point π. So, the correct answer is option (D). Point π is the image of π in a reflection through the dotted line.

It is worth noting that we can apply the definition of reflection to more than just points. For instance, a straight line is a geometric object consisting of all points that satisfy a given rule. So, we can reflect these points through a line, which will result in the reflection of the object. In particular, we might reflect a shape by individually reflecting the vertices and then joining these together. Letβs demonstrate these ideas in the next two examples.

Which of the following represents the image of the line segment π΄π΅ after a reflection in the line πΏ?

Remember, we can reflect a line segment through a line by reflecting its endpoints. When we do so, the image of each point will be the same perpendicular distance from the mirror line as the original point. So, letβs do this for each diagram. When we do so, we can see that options (B) and (C) are definitely incorrect. And at first glance, option (D) looks correct. But we can see that the vertices have been reversed. In fact, the correct answer is option (A). This represents the image of the line segment π΄π΅ after a reflection in line πΏ.

Letβs now look at a similar example involving a polygon.

Which of the following represents the image of triangle π΄π΅πΆ after a reflection in the line π?

Remember, when we reflect a single point, we preserve the perpendicular distance of this point from the mirror line. This means we can reflect a polygon by reflecting each of its vertices and then joining these together. Take, for instance, option (A). If we reflect vertex π΅ in the mirror line, we see that its image does not correspond to the image in the second triangle. We have a similar issue with the diagram in option (B).

However, when we reflect vertex π΅ in option (C), it corresponds to point π΅ prime in the second triangle. Reflecting point π΄ through the mirror line corresponds to the point π΄ prime in the second triangle. And when we complete the same process for point πΆ, we do indeed see that this second triangle is a full reflection through the mirror line. Itβs worth noting that for option (B) the triangle did look like it might be correct, but the vertices were in the wrong order. So, we have confirmed that the correct answer is option (C).

Before we consider another example, letβs write down some useful properties of reflections. First, as we observed in our earlier example, reflecting a line segment will give a congruent line segment. In other words, it will give another line segment of the exact same size. We also know that the mirror line is the perpendicular bisector of any line segment between a point and its image. By considering both of these two points, we can now see that reflecting the shape gives a congruent shape. And in particular, the measure of any angles are preserved.

Letβs now look at how to apply these properties to determine the length of the sides and the angle measures in a triangle.

In the following figure, triangle π΄ prime π΅ prime πΆ prime is the image of triangle π΄π΅πΆ by reflection in the line πΏ. (1) Fill in the blanks. The length of π΄ prime πΆ prime equals blank centimeters, and the length of π΄ prime π΅ prime equals blank centimeters. (2) Fill in the blanks. Line segment π΄π΄ prime is blank to line segment π΅π΅ prime, and line segment πΆπΆ prime is blank to line πΏ. (3) Find the measure of angle π΄.

Remember, when we reflect a polygon in a mirror line, we create a second congruent polygon. This means that the two triangles in our diagram are congruent. That in turn means that their line segments and angle measures are equal. This fact helps us to answer part (1). Line segment π΄πΆ is congruent to line segment π΄ prime πΆ prime. They must have the same lengths. Since line segment π΄πΆ is four centimeters, line segment π΄ prime πΆ prime must also be four centimeters. And we put four in the first blank space.

Next, line segment π΄π΅ must be congruent to line segment π΄ prime π΅ prime. And so, π΄ prime π΅ prime must be six centimeters in length. And six goes in our second blank space.

Letβs now consider question (2). First, we add line segments π΄π΄ prime and π΅π΅ prime to the diagram. We know that these line segments must be perpendicular to the mirror line. If theyβre both perpendicular to the mirror line, we can conclude some further information. That is, their alternate angles are equal, and they must in fact be parallel to one another. To find the second blank word in question (2), we add the line segment to the diagram. And of course, we know that πΆπΆ prime is perpendicular to line πΏ.

Finally, we consider question (3). Remember, these two triangles are congruent, which means they share angle measures. In particular, this means that the measure of angle π΄ must be equal to the measure of angle π΄ prime. Angle π΄ prime is 31 degrees, so angle π΄ is also 31 degrees.

And so, we have filled in the blanks. The correct entries were four, six, parallel, perpendicular, and 31 degrees.

It is worth noting that we can reflect any shape through a line by reflecting all of the points. And this includes circles, where we would reflect its center and maintain its radius.

Given a circle with center π that intersects with line πΏ at points π΄ and π΅, draw an image of circle π after a reflection in line πΏ. Which of the following statements is correct? Is it option (A) line segment π΄π is parallel to line segment π΄π prime? Option (B) line segment π΅π is parallel to line segment π΅π prime. Option (C) line segment ππ prime is perpendicular to line segment π΄π΅. Option (D) ππ prime is equal to π΄π΅. Or option (E) π΄ prime π΅ prime is greater than π΄π΅.

To reflect a circle in a mirror line, we must first reflect its center and then preserve its radius. We reflect the center, thatβs point π, by first creating the perpendicular line to πΏ that passes through π. Next, we know that line segment ππ΄ is a radius of our original circle. So, we can trace an arc with center π΄ and with the same radius. The point where this arc intersects our line is the center of our image. So, we have the image of our circle after reflection. We can now use this to identify the correct statement.

Thereβs no way that π΄π and π΄π prime can be parallel to one another. They quite clearly meet and form an acute angle. In fact, π΅π and π΅π prime cannot be parallel either for the same reasons. Of course, we do know that ππ prime must be perpendicular to π΄π΅. And this is because we constructed the perpendicular line bisector of line πΏ at the very start. And line πΏ passes through π΄π΅. The correct answer is option (C). Line segment ππ prime is perpendicular to line segment π΄π΅.

Weβll now recap the key points from this lesson. First, a reflection is a transformation which preserves the perpendicular distances of all points from the mirror line. Next, we learned that when we reflect a point through a mirror line and then join these two points, the mirror line will be the perpendicular line bisector of that newly created line segment.

We saw that we can reflect a line segment by individually reflecting its endpoints. And to reflect a polygon, we reflect its vertices and join them together. We learned that when we reflect a shape, its image preserves its length and angles. So, its image is in fact congruent to the original shape. And finally, we learned that we can reflect a circle by reflecting its center and preserving its radius.

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