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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.

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  • piceratops ultimate style avatar for user Jasmine Lou
    Why would it not be an isosceles trapezoid?
    (30 votes)
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  • old spice man green style avatar for user Jason
    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?
    (17 votes)
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  • aqualine ultimate style avatar for user Luke Jones
    At Sal 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?
    (8 votes)
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  • leaf green style avatar for user Monir Lashgai
    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
    (7 votes)
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  • leaf grey style avatar for user chenrui.zhang
    Is a trapezoid essentially the same as a trapezium?
    (9 votes)
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  • blobby green style avatar for user Daniel
    Anyone else got the working wrong but the answer right?
    (7 votes)
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  • aqualine ultimate style avatar for user Paladan
    It says a quadrilateral so why is it a triangle?
    (3 votes)
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  • leaf blue style avatar for user William Carrier
    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.
    (6 votes)
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  • blobby green style avatar for user Ahmed Nasret
    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?
    (5 votes)
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    • blobby green style avatar for user kwametoo
      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.
      (1 vote)
  • male robot donald style avatar for user ehsonmaleki2003
    At is there any mathematical way you can find out the vertices of a quadrilateral for a reflection?
    (3 votes)
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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.