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CA Geometry: Compass construction

Video transcript
We are on problem 56. Scott is constructing a line perpendicular to line L from point P. Fair enough. Which of the following should be his first step? So he wants to draw a line that looks something like this. He want to draw it going straight up through the point P. And so how do you do that? Obviously, if you just use a ruler, maybe by accident, you draw it at a slight angle or something like that, right? So he wants to have an exact line. Let's see, I don't even know what he's doing in this first one. He's drawing these x's. I don't know how he's determining where those x's go. So that A doesn't look right. Step B, it looks like he kind of picked two points and is drawing and using his compass to pivot off of those to kind of draw arcs there. But that's still not clear how it could help. If he knew that he could do some math and try to figure out that these are equidistant from P. But still, what he wants to do is find a point right around there, right below P, where if he draws a line between P and that point, it'll be perpendicular. So that doesn't seem to be of much help. Here, he picked a point and he's drawing an arc, but that arc doesn't really tell me much information. Let me see, D. This looks interesting. So it looks like he took the pivot of his compass and he drew a circle around with a constant radius. Obviously, that's what makes it a circle. And then what do he could do now is he could take his pivot. He could you could mark each of these points, right? If he marked each of those points and just pivoted around them and drew a circle, he would actually-- so let's say that around that point, he were to pivot and the circles looks something like-- I don't know. And he would want to change the radius a little bit, so let's say it looks something like that. I'm going to try my best to draw it. Let's say it looks something like that. And then around that point, he does the same thing. He pivots around it. I guess the two would have to be big enough to intersect with each other. I know I'm drawing it really horrible. The point at which they do intersect would be equidistant between those two points. Another way to think about it, it would be another point that's equidistant between the two points, because when you do it first with P and you draw that circle, you're saying both of these points are going to be equidistant from P. Just by definition, right? This is a circle and that's a constant radius. And then if you were to take each of those points and draw circles or draw arcs-- so let's say from that on you draw an arc like that and from that one you draw an arc like that-- you'd say, wow, this point is also going to be of a constant distance from both of them. So if I were to draw a line between both of these points using a straight edge, that line will be perpendicular to line L. So if I were to just do that, then it would be perpendicular, so I think D is the first step. All right, problem 57: Which triangle can be constructed using the following steps? OK, this is interesting. A lot of compass work here. Put the tip of the compass on point A. Open the compass so that the pencil tip is on point B. Draw arc above AB. So then they drew this arc right here. That's this thing that I'm trying to color in. Fair enough. Without changing the opening, put the metal tip on points B, so now you put the pivot there, and draw an arc interesting the first point at C. So now they want us to draw that second-- let me do it in another color. They're going to draw that second arc. OK, now draw AC and BC. Now what have we drawn? So when you draw this first arc-- what they drew is really a semicircle-- the radius is constant. So if the radius is constant, you know that this distance right here is going to be equal to this distance. They're just both radiuses of this semicircle or of this arc. They're just radiuses. They're equal to the length of the opening of our compass. So that's going to be equal to that. And then when you put the pivot here and you keep that distance the same, so now the pencil edge goes here, the distance is still there. And now when you do this arc, you now know that this length is equal to this length, because now they're both radiuses of this second arc. So now, you know all three sides are equal, so this is an equilateral triangle. Equilateral, D. OK, the diagram shows triangle ABC. Fair enough. Which statement would prove that ABC is a right triangle? This is clearly not a right angle. This is probably the right angle for the right triangle. And this is something that maybe you learned in algebra class, and if you didn't, you're about to learn it right now. And I'm looking at the choices. They talk a lot about slope. If I just have a line let's say here, and it has slope M, and I want to say what is the slope on a line that is perpendicular to this line right here? Well then, it would look like that. It would be perpendicular. It would have a 90-degree angle, and its slope would be the negative inverse. The negative inverse of this first slope. So if the slope from A to B is a negative inverse of the slope for B to C, then we're in business. These are definitely perpendicular, at least line segments, and this would be a 90-degree angle. Let's see. So what did I say? Slope AB, or we could say slope BC, should be equal to the negative inverse of the slope of A to B. See, if you multiply both sides of that times the slope AB, you get slope AB times slope BC is equal to negative 1. I just multiplied both sides times slope of AB. And if we could go here, choice B is exactly what we wrote right there. Next problem, 59: Figure ABC, oh! It's a parallelogram. Fair enough. That's parallel to that. That's parallel to that. What are they asking us? What are the coordinates of the point of intersection of the diagonals? So what are these coordinates? And we've mentioned it before, but the big kind of crux of this problem or what you need to know is that for any parallelogram, the diagonals bisect each other. So that means that the distance from here to here, from O to the intersection point is equal to the intersection point to B. And so this intersection point is the midpoint. It is bisected by line AC. And similarly, you can make the argument that this line segment right here is equal to-- is congruent to that line segment. If this is really the midpoint between O and B, then we just have to find the midpoint of their coordinates. And the way to find the midpoint of two coordinates is actually very intuitive. You just average the coordinates. So if I were to average the x's-- so the x-coordinate here is going to be-- the x-coordinate of point B, which is A plus C, plus the x-coordinate of the origin which is 0 over 2, because I had two points that I'm averaging. So that's going to be the x-coordinate. And then y-coordinate is going to be the y-coordinate coordinate of B plus the y-coordinate of the origin. I'm just averaging them. The y-coordinate of the origin is 0 divided by 2. This will give me the coordinate of the midpoint between origin and B. So that equals A plus C over 2 and then B over 2, so this is A plus C over 2. That makes sense because A plus C is going to be here someplace, and we just took the average of the two between zero and that, and you got the midpoint. And then the y-coordinate is going to be B over 2, which makes sense because that up there is B and we're just halfway between B and zero. So it's A plus C over 2, B over 2. And is that one of the choices? A plus C over 2, I'm assuming that they want us to do choice C and they just forgot to type the B in there. This is supposed to be a B. Because this is definitely not right. That's not right. And we know that this is right, but it's not A plus something over on this side. It's just B over 2 on the y-coordinate. So that's not right. All right, problem 60. I tried to squeeze it in. I don't know if you can see the whole problem. But it says what type of triangle is formed by the points A is 4 comma 2, B is 6 comma negative 1, and C is negative 1 comma 3? So I think the best thing is to just try to graph it and at least start getting an intuition, and then we can see the distances between the points, and hopefully, figure out what type of triangle it is. So let's see, some of the points get a little bit negative. I'll have to draw some of the negative quadrants. So if I were to draw it like that. OK, let's see. So 4 comma 2. That's right there. That's point A. And then I have 6 comma negative 1. That's point B. I don't know if you can see it down there. And C is negative 1 comma 3. So it's out here. Now let me connect the dots. That's one side, that's another side, and that's the other side. So off the bat, I just know this isn't going to be a right triangle. It's not an equilateral triangle. And the only way it's going to be an isosceles triangle is if this length is equal to that length, so let's just try it. Let's just test it out. So what is the distance from A to C? The distance squared from A to C is equal to the differences in their x's. So 4 minus negative 1, so there's a difference of 5, right? So there's a difference in their x's squared plus a difference in their y's. So 2 and 3. You could just say 2 minus 3 or 3 minus is 2. It doesn't matter. We just care about the difference. Plus one squared. So the distance squared is equal to 25 plus 1 is equal to 26. So this distance is the square root of 26. And the distance between A and B, same logic. Let's see, when you go from the distance squared, the difference is in there x's. Between 6 and 4, you have a distance of 2, so it's 2 squared plus the difference in their y's. 2 and negative 1 are 3 apart, right? Plus 3 squared. And so that is equal to 4 plus 9, which is equal to 13, so this is equal to the square root of 13. OK, and this number down here, you can figure it out, but it's going to be bigger than both of them, right? You can just look at it and say that. So this is definitely a scalene triangle. All of the sides are different. Anyway, see you in the next video.