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### Course: High school geometry > Unit 1

Lesson 5: Reflections# Reflecting shapes: diagonal line of reflection

Sal is given two line segments on the coordinate plane and the definition of a reflection, and he draws the image of the segments under that reflection.

## Want to join the conversation?

- How can I do this without a scratch pad? The actual activity does not allow drawing and even Sal had to go to another page to draw. I am struggling on questions that you can't just eyeball like -3x and stuff. Please help, thanks.(30 votes)
- Actually, I was able to eyeball the stuff on the exercises. For diagonal lines, just count diagonally how many points it takes to get to the line of reflection, and count that many points on the other side.(20 votes)

- not going to lie this stuff not making sense let me mind u again a senior in high school(22 votes)
- That's school for ya LOL.(7 votes)

- In the exercise for this, it just gave me a line segment with no way to add vertices. How do I do the excercise?(11 votes)
- In the following exercise, you need to click the graph to add vertices. For the exercise after that, click and drag the points to their proper positions depicting the line of reflection.(10 votes)

- How would you reflect a shape if it was on both sides of the line of reflection?(10 votes)
- Just invert the coordinate values of every point; every positive changes to a negative and vice versa.

For example, -X becomes X, and -Y becomes Y.(5 votes)

- My moms 36 and its breaking HER head, much less mine.(7 votes)
- This doesn't help me because I'm trying to reflect a triangle over a diagonal line, not a line segment.(5 votes)
- Can't I just count the diagonal squares to find the correct point on the reflection? I guess it would depend on what the slope is though.(4 votes)
- So you are correct in that it depends on the slope. the line that you would draw would be perpendicular to the line of reflection, so as long as the line of reflection has either a positive or negative 1 slope, counting diagonals works well.(3 votes)

- So that little triangle thing means change? Like, change in the number or variable? Is there a video on that?(3 votes)
- The little triangle essentially stands for "change in". In this scenario, it is part of "change in x" which just means that the x value changes that much in that direction -- (7, y) reflects to (1.5, y).(4 votes)

- What happened to this exercise? There isn't one. Can you please add an exercise regarding this topic? That would be greatly appreciated.

Thanks in advance!(4 votes) - Forgive me if I sound silly but couldn't you just count the squares diagonally instead of finding the slope?(3 votes)
- You might be able to now, but it's good to learn the process because it might become a lot harder and more complex later and your process might no longer work or become very tiresome. That really killed me in Algebra.(2 votes)

## Video transcript

- Line segments IN, this
is segment IN over here, and TO, this is TO here, are reflected over the line Y is equal to negative X minus two. So this is the line that
they're reflected about this dashed, purple line. And it is indeed Y equals
negative X minus two. This right over here is
in slope intercept form. The slope should be negative one, and we see that the
slope of this purple line is indeed negative one. If X changes by a certain amount, Y changes by the negative of that. If X changes by one, Y
changes by negative one to get back to that line. If X changes by positive two, Y changes by negative two to get back to another point on that line. And the Y intercept? We see when X is equal to zero, Y should be negative two. When X is equal to zero, Y is indeed negative two. So we've validated that. Then they say, "Draw the
image of this reflection, using the interactive graph." Alright, so we can move
these lines around, and we want to reflect these. And I could try to eyeball it, you know, maybe it's something like this. I don't know, it doesn't exactly right. That looks close to the reflection of IN, and for TO I'd want to
move this down here. TO looks like it would be, I don't know, I'm eyeballing it. This is close, but it can't be close. I want to get exact. So let's, I've copied and
pasted the original problem on my scratch pad so we
can find the exact points and so I don't just have to estimate this. So let's go to the scratch pad. So exactly what we just saw. And the main realization is, is that if we want to
reflect a given point, if we want to reflect a given point, say point I right over here, what we want to do is we want to drop a perpendicular. We want to find a line
that's perpendicular, or a line that has the point I on it, and it's perpendicular to
this line right over here. Remember this is the line, let me do this is that purple color. This is the line, Y is equal
to negative X minus two. Its slope is equal to negative one. So I want a line that
goes through I, point I, that is perpendicular to this line, and I want to drop it, I want to drop it to, I want to drop it to the line
that I'm going to reflect on, and then I'm going to go the same distance onto the other side to find to find the corresponding
point in the image. So how do I do that? Well if this line, if this purple line has a slope of negative one, a line that is perpendicular to it a line that is perpendicular to it so this thing that I'm drawing
in purple right over here, its slope is going to be the
negative reciprocal of this. So the reciprocal of negative one is still just negative one. One over negative one
is still negative one. But we want the negative of that. So the slope here needs to be one. And luckily, that's how I drew it. The slope here needs to be equal to one which is however much I
change in the X direction I change in the Y direction. We see that. To go from this point to
this point right over here, we decrease Y by four and we decrease X by four. Now, if we want to stay on this line to find the reflection,
we just do the same thing. We could decrease X by four, so we'll go from negative
two to negative six, and decrease Y by four, and we end up at this
point right over here. So we end up at the point, this is X equals negative six,
Y is equal to negative four. So this is, this point corresponds to this point right over there. Now, let's do the same thing. Let's do the same thing for point N. For point N, we already know, if we drop a perpendicular, and this is perpendicular, it's
going to have a slope of one because this purple line
has a slope of negative one. The negative reciprocal of
negative one is positive one. And let's see, to go from this point to this point of intersection, we have to go down one and a half. We're going down one and a half, and we're going to the
left one and a half. So we want to do that on the other side. We want to stay on this
perpendicular line. So we want to go left one and a half, and down one and a half. And we get to this point right
over here, which is the point X equals three, Y is
equal to negative eight. So we are now equidistant. We're on this perpendicular line, still, but we're equidistant on the other side. So the image of IN is going to go through negative six, negative four,
and three, negative eight. So let me draw that. Let me see if I can remember, negative six, negative four,
and three, negative eight. So I have a bad memory. So negative six, negative four,
and three, negative eight. And I was close when I estimated, but I wasn't exactly right. So that's looking pretty good. And actually, we can
do the exact same thing with points T and point O. Let me do that. So point T, to point T, To get from point T to the
line in the shortest distance, once again, we drop a perpendicular. This line is going to have a slope of one, because it's perpendicular to the line that has a slope of negative one. And so to get there, we
have to decrease our X by we have to decrease our X. We're going from X
equals five to X equals, it looks like, half. So X went down by four and
a half in the X direction and Y also needs to go
down by four and a half. So if we want to stay on that line, let's decrease our X by four and a half. So that's half, one, two, three, four. And Y needs to go down by four and a half. So that's half, one, two, three, four, and we get to this point right over here, which is the point X equals negative four, Y is equal to negative seven. And it's negative four, negative seven. Let me, so this should be at, X equals negative four,
Y equals negative seven. And there's a couple of
things you could do here. You could just say, hey, this is two long, or two units long, not like,
too long in length, somehow. This is two units long, so
maybe this is two units long. So this is feeling pretty good. But let's just go through the
exercise for point O as well. So point O, once again,
this is going to have, if we drop a perpendicular, it's going to have a slope of one. So whatever our change
in X between this point and this point, we're going to have the same change in Y. And our change in X,
to go from seven to... Seven to one and a half, our change in X is five and a half. So then we do it this way. So the change in X here, so the change in X is equal
to negative five and a half, five point five. If you subtract five
point five from seven, you get to one point five. And our change in Y. Our change in Y is also
negative five point five. Change is Y is negative, you're not going to see that so I'm going to do it differently, negative five point five. And so we need to stay on this line. So we need to change by those same amounts onto the other side of that line. So if we decrease our
X by five and a half. So half, one, two, three,
four, five, we get there, and Y by five and a half, half, one, two, three, four, five. We get to the point, X
equals negative four, Y is equal to negative nine. So we get to the point
X equals negative four, Y is equal to negative
nine, and we're done! We have figured out the
reflection of these two segments.