Learn about CA Geometry: Compass Construction

Khan Academy Presents: CA Geometry: 56-60, compass construction, coordinates

We are on problem 56. Scot 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 wants to draw it, you know, going straight up through the point P. And so, how do you do that? Obviously, if you know, you just use a ruler, maybe by accident you draw it a slight angle or something like that, right? So, he wants to have an exact line. So, let’s see. I don’t even know what he’s doing in this first, when he’s drawing these x’s. I don’t know how he’s determining where those x’s go. So, that’s step A. 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 – It’s still not clear how it could help. I mean, if he knew that he you know, he could do some Math and try to figure out that these are equidistant from P and – but still, I mean, what he wants to do is find the point right around there – right below P, where if he draws a line between P and that point, it will 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, he took – it looks like he took the pivot of his compass and he drew a circle around with a concept radius of it. You see, that’s what makes it a circle. And then, what he could do now is he could take his pivot He 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 would have pivoted and the circle 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 this. I would try my best to draw it. So, let’s say it looks something like that. And 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, right? But where the point that they do intersect – and I’m doing it really horrible – The point at which they do intersect would be equidistant between those two points, right? And another way of thinking 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, “Oh, both of these points are going to be equidistant from P”, right? Just by definition, right? This is a circle and that is a constant radius. And then if you would take each of those points and draw circles or draw arcs – So, let say from that one you would 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, that would be perpendicular. So, I think D is the first step. All right. Problem 57. Which triangle can be constructed using the following steps? Okay, this is interesting. A lot of compass work here. Put the tip of the compass on point A. So, we go down there. So, they put the tip of the compass on point A. Open the compass so that the pencil tip is on point B. Okay, pencil tip is here. Draw arc above AB so then, they drew this arc right here. Draw arc above AB. That’s this thing that I’m trying to color in. Fair enough. Without changing the opening, put the metal tip on point B. So now, you put the pivot there and draw an arc of intersecting – and draw an arc intersecting at point C. So now, they want us to draw that second – Let me do that another color. They’re going to draw that second arc, right? Okay. Now, draw AC and BC. All right. Now, what have we drawn? So when you draw this first arc, I mean an arc it’s – we know we draw is like a semicircle, right? The radius is constant. So if the radius is constant, you know that this distance right here – that distance is going to be equal to this distance, right? They’re just both radiuses of this semicircle or of this arc. They’re just radiuses, right? They’re equal to the length of the opening of our compass. So, that is going to be equal to that. And then when you put the pivot here, and you keep the distance the same – So now, the pencil edge goes here, right? 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 – They’re both radiuses of this second arc. So now you know that three sides are equal. So, we have to be dealing this. This is an equilateral triangle. Equilateral – D. Okay. The diagram shows triangle ABC. Fair enough. Which statement would prove that ABC is a right triangle – is a right triangle? And this time, well I mean, you know, this is clearly not a right angle. This is probably with a 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. But if I wanted to find, and I’m looking at the choices, they talk a lot about slope. So if I just want to find – If I just have a line let’s say here and has slope M, and I want to say, “What is the slope of a line that is perpendicular to this line right here. Well then, – well, it would look like that, right? It will be perpendicular. We would have a 90-degree angle. And its slope would be the negative inverse – the negative inverse of this first slope, right. So, if the slope from A to B is a negative inverse of the slope from B to C, then we’re in business. Then, D is definitely perpendicular, at least line segments. This would be a 90-degree angle. So, let’s see. Slope A –So, what did I say? Slope AB, or we could say slope BC – slope of segment BC should be equal to the negative inverse of the slope A to B. So if we 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 of the equation times slope of AB. And if we 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, and 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 with a big kind of, I guess, cracks of this problem, or what you need to know is that for any parallelogram, the diagonals, bisect each other. They bisect each other. So that means that the distance from here to here, from O to the midpoint, or from I shouldn’t say the midpoint, from O to the intersection point is equal to the intersection 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. So, if – Now, let’s just do D. If we want to find – 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, right? So, if I were to average the x’s between – So, let’s see. 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 of that two points that I’m averaging. So, that’s going to be x coordinate. This is going to be – and then the y coordinate is going to be the y coordinate of B. So, that’s 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. And so that equals what? A plus C over 2 and then B over 2, so this is – this is right here, A plus C over 2, right? And that makes sense because A plus C is going to be here some place. And we just took the average of the two between 0 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 half way between B and 0, right. So this is A plus C over 2 B over 2 and this is not 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 the sides. It’s just B over 2 on the y coordinate. So, that’s just not right. All right Problem 60. I tried to squeeze it in, I don’t know if you can see the problem, but it says what type of triangle is formed by the – I think that gets cut off for you – form by the point A is 4,2. B is 6, negative 1. And C is negative 1,3. So, I think the best thing is just try to graph and at least start getting an intuition and we could see the distances between the points and hopefully, figure out what type of triangle that it is. So, let’s see. Some of the points get a little bit negative. We have a negative – so let me also draw some of the negative quadrants. So, if I were to draw it like that – Okay, and let see. So, 4,2. So, if I were to say 1 – 2 – 3 – 4,1- 2. That’s right there. That’s point A. Then if 6, negative 1. So, 4 – 5 – 6,negative 1. Negative 1 is right there. That’s point B. I don’t know if you can see it right there. And C is negative 1,3. So, negative 1,3. So, it’s out here – negative 1,3. Okay, 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 mean I just know, this isn’t going to be a right triangle. It’s not equilateral triangle. And if – 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, right? Well, the distance is equal to or we can say the distance squared but let’s say the distance squared from A to C is equal to the differences in their x’s. So, 4 minus negative 1, that’s the difference of 5, right? So, the difference of their x’s squared plus a difference in their y’s. So, 2 and 3. You can either say 2 minus 3 or 3 minus 2. It doesn’t matter. We just care about the difference. 1 squared – that’s 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 this is between A and B – same logic. Let’s see. When you go from the distance squared. The differences in their x’s. 6- 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 three squared, and so that is equal to 4 plus 9, which is equal to 13. So, it’s equal to the square root of 13. Okay, and this number – this number down here, you can figure out. That’s going to be bigger than all of the – both of them, right? You can just look at it and say that. So this is definitely a scale in triangle. All of the sides are different. Anyway, see you in the next video.