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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.