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### Course: 6th grade (WNCP) > Unit 3

Lesson 6: Transformations- Intro to reflective symmetry
- Rigid transformations intro
- Performing translations
- Translate points
- Rotating points
- Rotate points
- Performing reflections
- Reflect points
- Finding a quadrilateral from its symmetries
- Finding a quadrilateral from its symmetries (example 2)
- Reflective symmetry of 2D shapes

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# Finding a quadrilateral from its symmetries

Two of the points that define a certain quadrilateral are (0,9) and (3,4). The quadrilateral has reflective symmetry over the line y=3-x. Draw and classify the quadrilateral. Created by Sal Khan.

## Want to join the conversation?

- Why would it not be an isosceles trapezoid?(34 votes)
- I think he was just identifying it as a trapezoid in a general sense. I did the same thing when I worked it out on my own.(8 votes)

- Sometimes it's difficult to see the perpendicular to the line of reflection. Therefore, I've been using the following technique: plot the "transform" (I don't know the correct terminology) of the point [e.g. if the point is (0,9), then plot (-9,0) OR if the point is (3,4), then plot (-4,-3)] then move the point to the final, correct reflection in both the x & y directions using the x-intercept & y-intercept of the line of reflection as offsets. In the same example, (-9,0) will move +3 in the x-direction since the x-intercept of the line of reflection is +3 and also move +3 in the y-direction since the y-intercept is also +3 to the final reflection point of (-6,3) and for the point at (3,4), the final reflection point is (-4+3, -3+3) or (-1,0). Is this true in all cases? Even if the reflection isn't over a straight line but perhaps some other 2-dimensional shape such as a circle?(18 votes)
- Well the math is a little more complicated when the slope of the line isn't
`1`

or`-1`

, but yes, you can use the perpendicular line (which has a slope of`-1/m`

compared to the original line's slope of`m`

) to calculate the reflected point mathematically.(14 votes)

- At0:58Sal talks about "when x is 0, y is 3 - that's our y intercept" and then talks about how the slope goes down from there. I've been following everything I can on geometry but I seemed to have missed exactly how these slopes work. Is there another unit I can look at that describes how the whole y = 3 - x thing works?(9 votes)
- I would suggest looking up "equations of a line" and "slope-intercept form" on the KA search bar.(14 votes)

- I do not know how to solve Y=3-X. I did not find any explanation about it in previous videos in this section. Could you explain it to me please?

Thanks for your invaluable services

Monir(8 votes)- You don't really solve "y=3-x." He just put the line on the graph. You can rewrite it as "y=-x+3." the "x" is the slope which is almost like the distance between two points. 3 is the y-intercept which you plot on the y-axis.(7 votes)

- Is a trapezoid essentially the same as a trapezium?(9 votes)
- Yes, they are the same thing, trapezium in British, trapezoid in American.(6 votes)

- Anyone else got the working wrong but the answer right?(8 votes)
- Where is the widget to make polygons? I tried finding it on the site in vain. I need either (1) a URL or (2) search keywords that will yield very few results that include what I'm looking for.(7 votes)
- I believe it is exclusive to Khan Academy, though I may be wrong.(3 votes)

- It says a quadrilateral so why is it a triangle?(4 votes)
- At what point do you see it as a triangle? He even stated at the end that it ends up as a trapezoid which is a quadrilateral.(7 votes)

- why did you connect both points and connect each point to its reflection to derive the quadrilateral? who said that the line y-3-x is the symmetry line of the shape? the question is about a quadrilateral which when reflected (supposed to be the whole shape with its 4 sides) over the line y=3-x the shape does not change. I supposed the following: to connect the two given points and suppose it is the LHS of the quad. then reflect the line using the reflection of the points, then we propose that the reflected line is the RHS of the image and hence conclude that it should be corresponding to the RHS of the original shape, that way we could conclude the RHS of the original shape which is required to be defined. and the same way, we suppose that the original line (the line connecting (0,9) and (3,4) should correspond (parallel) to the LHS of the image.

this way we conclude a quad shape reflected and unchanged .. the RHS of the original and the image correspond to each others, also LHS correspond to each other.

but you answered as if the question is "use these points to find the quad which y=3-x is its symmetry line.

is that true?(6 votes)- Excellent insight! However, if I understand what you are saying with (0,9) and (3,4) being the LHS of the quad, the RHS will be further away from the line. If you reflect that quad across the line, you will have a new quad sitting completely below the line with new coordinates. In KA's solution you end up with the exact same quad - i.e. it has exactly the same coordinates after the reflection.(0 votes)

- Is there a better way to reflect points on a graph other than continuing along the line like what Sal did at around2:07to plot the points?(4 votes)

## Video transcript

Two of the points that define
a certain quadrilateral are 0 comma 9 and 3 comma 4. The quadrilateral is left
unchanged by a reflection over the line y is
equal to 3 minus x. Draw and classify
the quadrilateral. Now, I encourage you
to pause this video and try to draw and
classify it on your own before I'm about to explain it. So let's at least plot the
information they give us. So the point 0
comma 9, that's one of the vertices of
the quadrilateral. So 0 comma 9. That's that point
right over there. And another one of the
vertices is 3 comma 4. That's that right over there. And then they tell us
that the quadrilateral is left unchanged by reflection
over the line y is equal to 3 minus x. So when x is 0, y is 3--
that's our y-intercept-- and it has a slope
of negative 1. You could view
this as 3 minus 1x. So it has a slope of negative 1. So the line looks like this. So every time we increase our
x by 1, we decrease our y by 1. So the line looks something like
this. y is equal to 3 minus x. Try to draw it relatively,
pretty carefully. So that's what it looks like. y is equal to 3 minus x. So that's my best attempt
at drawing it. y is equal to 3 minus x. So the quadrilateral is
left unchanged by reflection over this. So that means if I were to
reflect each of these vertices, I would, essentially,
end up with one of the other vertices
on it, and if those get reflected you're going to
end up with one of these so the thing is not
going to be different. So let's think about
where these other two vertices of this
quadrilateral need to be. So this point, let's just
reflect it over this line, over y is equal to 3 minus x. So if we were to try to drop
a perpendicular to this line-- notice, we have gone
diagonally across one, two, three of
these squares so we need to go diagonally
across three of them on the left-hand side. So one, two, three gets
us right over there. This is the reflection of
this point across that line. Now, let's do the same
thing for this blue point. To drop a perpendicular
to this line, we have to go diagonally
across two of these squares. So let's go diagonally across
two more of these squares just like that to get to that
point right over there. And now we've defined
our quadrilateral. Our quadrilateral
looks like this. Both of these lines
are perpendicular to that original
line, so they're going to have the same slope. So that line is parallel
to that line over there. And then we have this line
and then we have this line. So what type of
quadrilateral is this? Well, I have one pair
of parallel sides, so this is a trapezoid.