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### Course: Geometry (all content)>Unit 10

Lesson 8: Symmetry

# 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?
• 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.
• 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?
• 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.
• 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?
• I would suggest looking up "equations of a line" and "slope-intercept form" on the KA search bar.
• 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?

Monir
• 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.
• Is a trapezoid essentially the same as a trapezium?
• Yes, they are the same thing, trapezium in British, trapezoid in American.
• Anyone else got the working wrong but the answer right?
• 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.
• I believe it is exclusive to Khan Academy, though I may be wrong.
• It says a quadrilateral so why is it a triangle?
• 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.
• 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?
• 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.
• Is there a better way to reflect points on a graph other than continuing along the line like what Sal did at around to plot the points?