High school geometry
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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- Why would it not be an isosceles trapezoid?(28 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.(5 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?(16 votes)
- Well the math is a little more complicated when the slope of the line isn't
-1, but yes, you can use the perpendicular line (which has a slope of
-1/mcompared to the original line's slope of
m) to calculate the reflected point mathematically.(12 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?(7 votes)
- I would suggest looking up "equations of a line" and "slope-intercept form" on the KA search bar.(10 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
- 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.(5 votes)
- Is a trapezoid essentially the same as a trapezium?(8 votes)
- Yes, they are the same thing, trapezium in British, trapezoid in American.(4 votes)
- It says a quadrilateral so why is it a triangle?(2 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.(4 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.(3 votes)
- Anyone else got the working wrong but the answer right?(3 votes)
- I realize this is a simple reflection line with a slope of -1, however if the coefficient was different so that perhaps it was a steeper slope, would we know what is perpendicular by taking the opposite reciprocal of the slope and then just shifting up or down and right or left so many units based on that calculation? For example, let's say the line we are reflecting over is y = -2/3x-2. A line perpendicular to this one would have a slope of +3/2. We would then use up 3 units and to the right 2 units as a measurement to determine a point's reflection across the original line (or work backwards if the point was above the line), counting how many units we have moved to get to the line of reflection? I'm fairly bad at visualizing reflections, etc. and I have to take an important math test where I cannot write anything down so I am attempting to see if this is a correct method that I can use in my head when I am challenged with this no-writing test. Thank you!(2 votes)
- Yes but you have to calculate the distance from the point to the line, and then double it to reflect across to the other side.(2 votes)
- Why are two of the lines parallel?(2 votes)
- If you're asking how we know that the lines are parallel, it's because we can see that their slopes are equal.
For example, the lines y=-7x+12 and y=-7x-8 are parallel, because their slopes (the slope is always a coefficient of x) are the same. The b-value, AKA the y-intercept, is irrelevant in the context of parallel lines, because regardless of a line's vertical translation, it cannot EVER intersect a line that has its same slope (parallel). If you play around with a graph and look at different lines, it might help you to visualize this better.
Also, to identify perpendicular lines (though I know this isn't what you asked), just remember that two lines are perpendicular if their slopes are the negative reciprocals of one another. For example, y=6x+4 is perpendicular to the line y=-1/6x+2 because the first line's slope of 6 can be turned into 1/6, and then made negative. This gives you -1/6, which - if we check again, is the exact slope of the other line, so they must be perpendicular.(2 votes)
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.