### Video Transcript

In this video, weβll learn how to
find the images of points, lines, and shapes after their reflection in the π₯- or
π¦-axis on the coordinate plane. When we reflect a shape in a line,
we flip it. And there are a number of ways we
can do this. We could reflect the shape in the
π₯-axis. Thatβs this line. We could reflect it in the
π¦-axis. Thatβs this line. Or we might even reflect it in a
different vertical line. Those are of the form π₯ equals π,
where π is a constant.

The vertical line π₯ equals
negative one, for example, is this one. It passes through the π₯-axis at
negative one. We could use a different horizontal
line π¦ equals π, where π is a real constant. For example, the line π¦ equals two
passes through the π¦-axis at two as shown. Finally, we might even look at
reflecting in a diagonal line like π¦ equals π₯. Remember, all the π₯- and
π¦-coordinates on this line are equal, or the line π¦ equals negative π₯ as
shown.

We say that the line in which we
reflect the shape is called the mirror line. And the reflection of the shape is
called its image. And we use the prime symbol to
represent the image of an object. For example, the reflection of
point π΄ would be π΄ prime. And the image, the reflection, of
shape π΄π΅πΆ would be π΄ prime π΅ prime πΆ prime. Finally, itβs important that we
realize that an object and its image have the same size and theyβll be located the
same distance from the mirror line on opposite sides. Letβs see what this might look
like.

Which pair of triangles represents
a reflection in the π₯-axis?

The π₯-axis is this horizontal
line. We could call that the line with
equation π¦ equal zero. Now, weβre looking for the pair of
triangles which represent a reflection in this line. Now, when we reflect a shape, we
flip it. The two triangles will have the
same size, and theyβll be located the same distance from the π₯-axis but on opposite
sides. So, letβs look at some of these
pairs.

Weβll begin by looking at shape π΄
and π΅. For shape π΄ and π΅, this first
vertex is two units away from our mirror line. For both shapes, this second vertex
is five units away from the mirror line on opposite sides. And our third vertices are both
three units away from the mirror line on opposite sides. We see that each of the points are
the same distance from the π₯-axis on opposite sides. And the shape is flipped but
otherwise unchanged. That is indeed a reflection in the
π₯-axis. So, thatβs a good indication of us
that the pair of triangles that represent the relevant reflection are π΄ and π΅. But letβs check and see whatβs
happened between the other pairs.

Letβs look at shape π΄ and πΆ. Once again, comparing the relevant
vertices of our shapes, we see that they are the same distance from the π¦-axis on
opposite sides. And the shape is flipped over the
π¦-axis but otherwise unchanged. In this case then, shapes π΄ and πΆ
represent a reflection in the π¦-axis.

Now, what about shapes π΅ and
πΆ? Well, there are two ways to
describe this. We add in a diagonal line with the
equation π¦ equals negative π₯. Now, we compare the vertices of π΅
and πΆ. This time, our vertices are the
same distance from the line π¦ equals negative π₯ but on opposite sides. Otherwise, shapes π΅ and πΆ, apart
from being flipped β remember, thatβs a reflection β remain the same shape and
size. So, we could say that shapes π΅ and
πΆ represent a reflection in the line π¦ equals negative π₯.

If we look carefully though, we can
even say the shape has been rotated 180 degrees about the origin. Thatβs the point zero, zero. In this case though, the pair of
triangles we were looking for were π΄ and π΅.

In our next example, weβll look at
how to reflect a shape in the π¦-axis.

Find the images of π΄π΅πΆπ· after
reflection in the π¦-axis.

We begin by identifying the
location of our mirror line. Weβre told to reflect the shape in
the π¦-axis. Thatβs this vertical line here. We might say this line has the
equation π₯ equals zero. The images of π΄π΅πΆπ· are the
vertices of our rectangle after it has been reflected. And we use prime notation, π΄ prime
π΅ prime πΆ prime π· prime. Now, when we reflect π΄π΅πΆπ· in
the line π₯ equals zero or the ~~π₯-axis~~ [π¦-axis], we know it ends up
being flipped. Otherwise it should end up the
exact same size and the exact same distance from our mirror line on opposite
sides.

Now, a really sensible way to
perform the reflection is to do it vertex by vertex. Letβs start with vertex πΆ. We measure the distance between
vertex πΆ and the π¦-axis. And of course, thatβs the
perpendicular distance. Thatβs one, two units. This means vertex πΆ prime must be
two units away from the π¦-axis in the opposite direction. That has Cartesian coordinates
negative two, one.

Letβs repeat this process with
π·. Once again, itβs two units away
from the π¦-axis. And so, π· prime will also be two
units away from the π¦-axis in the opposite direction. It has Cartesian coordinates
negative two, six. Weβll now look at vertex π΄. This time thatβs one, two, three,
four, five, six, seven, eight units away from our mirror line. π΄ prime is, therefore, eight units
away from the mirror line in the opposite direction. So, itβs at the point negative
eight, six.

Now, notice that since the size of
the shape remains unchanged, we could have simply measured the distance between π΄
and π· and then repeated that on the other side. Letβs see what that might look like
for the side πΆπ΅. The length of this side is one,
two, three, four, five, six units. And so, the distance between
vertices π΅ prime and πΆ prime must also be six units.

And so, we have the image of
π΄π΅πΆπ·. π΄ prime has coordinates negative
eight, six. Remember, we go along the corridor
and up the stairs. π΅ prime is negative eight,
one. πΆ prime is negative two, one. And π· prime is negative two,
six.

Now, in fact, we can generalize
something from this question. Letβs look at the original
coordinates. They are eight, six; eight, one;
two, one; and two, six; respectively. Notice that the π¦-coordinate in
each case remains completely unchanged. The π₯-coordinate is almost the
same, but we notice that itβs the negative version of it. In fact, we can say that the
reflection of a point π₯, π¦ across the π¦-axis is the point negative π₯, π¦.

In a similar way, when we reflect a
point across the π₯-axis, the π₯-coordinate remains unchanged, and we get the
negative version of the π¦-coordinate. So, the reflection of π₯, π¦ across
the π₯-axis is π₯, negative π¦.

In our next example, weβll look at
how to reflect a shape in a vertical line of the form π₯ equals π for real
constants π.

Reflect the rhombus π΄π΅πΆπ· in the
line π₯ equals two.

Letβs begin by identifying our
mirror line. We know that lines of the form π₯
equals π are vertical lines; they pass through the point π, zero. In other words, they pass through
the π₯-axis at π. Now, the line π₯ equals two then
must also be a vertical line, but it must pass through the π₯-axis at two as
shown. So, weβre going to reflect our
rhombus in this line.

Letβs do that vertex by vertex. Weβll begin with vertex π΄. We measure the perpendicular
distance of this vertex from the mirror line. Itβs one, two, three, four
units. This means the image of π΄,
remember we call that π΄ prime, must also be four units away from the line π₯ equals
two but in the opposite direction. And so, π΄ prime is here. Itβs got coordinates negative two,
six.

Letβs now measure the perpendicular
distance of vertex π΅ from our mirror line. This time thatβs one, two, three,
four, five, six units. This means that the image of π΅
will be six units away from the mirror line in the opposite direction. Thatβs here. We repeat this process for the
remaining vertices. πΆ is one, two, three, four units
away from the mirror line. And so, its image, πΆ prime, will
also be four units away.

Finally, π· is two units away from
the line π₯ equals two, which means that π· prime will be here. Itβs two units away from our line
in the opposite direction. And so, weβve reflected the rhombus
π΄π΅πΆπ· in the line π₯ equals two. Weβve given its vertices π΄ prime,
π΅ prime, πΆ prime, and π· prime.

In our next example, weβll consider
how we can find the position of a point after reflecting it in a given diagonal
line.

What is the image of the point
nine, eight under reflection in the straight line π¦ equals π₯?

These sorts of questions can be
quite difficult to visualize. So, letβs plot our point and the
line π¦ equals π₯ on a Cartesian plane. Here is our point nine, eight. Its π₯-coordinate is nine, and its
π¦-coordinate is eight. The line π¦ equals π₯ is a diagonal
line. Every point on this line has equal
π₯- and π¦-coordinates. For example, it will pass through
the point one, one; three, three; five, five; negative two, negative two; and so
on. In fact, it looks a little
something like this.

Weβre going to reflect our point in
this line. So, we look at the perpendicular
distance from our point to the line. We can see thatβs half of the
diagonal of one square. The image of our point will be the
exact same distance away from the line in the opposite direction. So, what are its coordinates? Well, theyβre eight, nine. It now has an π₯-coordinate of
eight and a π¦-coordinates of nine. And so, weβve worked out the image
of the point nine, eight under reflection in the line π¦ equals π₯. Itβs eight, nine.

But we can, in fact, generalize
this. We take a point π₯, π¦. When we reflect it in the line π¦
equals π₯, it becomes π¦, π₯. In other words, the values of the
π₯- and π¦-coordinates interchange. In general, we say that the
reflection of π₯, π¦ across the line π¦ equals π₯ is the point π¦, π₯. And the reflection of π₯, π¦ across
the line π¦ equals negative π₯ gives us the point negative π₯, negative π¦.

In our final example, weβll
consider how we can reflect a full shape in the diagonal line π¦ equals π₯.

Reflect triangle π΄π΅πΆ in the line
π¦ equals π₯.

Remember, the line π¦ equals π₯ is
a straight diagonal line that passes through the origin. Thatβs the point zero, zero. Every point on this line has equal
π₯- and π¦-coordinates. Itβs this line. Notice it passes through the point
two, two; 10, 10; negative six, negative six; and so on. And so, weβre looking to reflect
triangle π΄π΅πΆ in this line. And so, we recall that when we
reflect a shape, itβs flipped, that each vertex of an original figure and its image,
thatβs the reflection, are the same distance away from the line of reflection but
out the other side.

So, we could count the
perpendicular distance from each vertex to our line. Alternatively, we can quote this
fact. The reflection of a point π₯, π¦
across the line π¦ equals π₯ gives us the point π¦, π₯. Now, point π΄ has coordinates
negative four, 12; point π΅ has coordinates negative four, six; and point πΆ has
coordinates two, six. Notice that the values of our π₯-
and π¦-coordinates are interchanged. So, the image of π΄ will have
coordinates 12, negative four; the image of π΅ will have coordinates six, negative
four; and the image of πΆ, πΆ prime, has coordinates six, two. π΄ prime is therefore here, π΅
prime is here, and πΆ prime is here.

And so, weβve reflected triangle
π΄π΅πΆ in the line π¦ equals π₯. Itβs the image π΄ prime π΅ prime πΆ
prime as shown.

In this video, weβve learnt that a
reflection in a mirror line flips the original shape. The image, which is the reflection
of the original and is given using prime notation, will have the same shape and size
as the original object. Each vertex will be an equal
distance away from the mirror line but on the opposite side.

We also saw that we can generalize
the image of certain points after being reflected in various lines. The reflection of the point π₯, π¦
across the π₯-axis is the point π₯, negative π¦. Across the π¦-axis, and we get the
point negative π₯, π¦. Across the line π¦ equals π₯, we
get π¦, π₯. And across the line π¦ equals
negative π₯, we get negative π¦, negative π₯.