(a) Prove that the midpoint of the line segment from P1(x1, y1, z1) to P2(x2, y2, z2) is ((x1+x2

(a) Prove that the midpoint of the line segment from P1(x1,...

Key Concepts

Distance Formula in 3D

The distance formula in three dimensions is used to calculate the length of a line segment between two points by extending the Pythagorean theorem. It takes into account the differences in the x, y, and z coordinates. This formula is essential for determining lengths of segments like the medians in a triangle, allowing one to measure distances accurately in space.

Medians of a Triangle in 3D

A median in a triangle is a line segment drawn from one vertex to the midpoint of the opposite side. In a three-dimensional setting, finding the midpoints and then using the distance formula enables one to compute the lengths of these medians. This concept is important for understanding the properties of triangles and their centers in a 3D coordinate system.

3D Analytic Geometry

This subject involves studying geometric figures using a three-dimensional coordinate system. It allows points, lines, and shapes to be represented algebraically and solved using equations. In the context of the problem, understanding the coordinates of points and how to manipulate them is fundamental to proving formulas and calculating distances in space.

Midpoint Formula in 3D

The midpoint formula is a method to find the point that lies exactly halfway between two points in three-dimensional space. It is derived by taking the average of the corresponding coordinates of the two points. This concept is crucial when working with geometric constructions and proofs, such as locating the center of a line segment connecting two points.

Transcript

00:04 Let's verify that the midpoint formula actually does give us the point in between points p1 and p2.

00:12 To do that, we're going to find the distance from m to p1 and from m to p2.

00:18 And we're going to verify that it is half, that each one of those is half of the distance from p1 to p2.

00:27 So let's begin by writing an expression for the distance from p1 to m using the distance form.

00:35 We're simply going to subtract x coordinate, subtract y coordinates, and subtract z coordinates, and then do some algebra and see what that simplifies to.

00:45 So x coordinates, it's going to be x1 plus x2 divided by 2 minus x1 squared, plus x1, y1, plus y2 over 2 minus y2 squared, plus y2 squared, plus y2, over 2 minus y2 squared, plus x1 plus z2 divided by 2 minus z 2 squared.

01:30 Each one of these is going to have this expression.

01:33 I'm going to write it in detail for the x's, and then just use the same result for the y and the z.

01:40 I'm going to have x1 plus x2 over 2 minus.

01:48 I'm going to write this as 2 x1.

01:52 Over 2.

01:54 I can do that by looking this as a fraction with one in the denominator and then multiplying by 2 over 2.

02:05 This is 2x1 when i subtract in the numerators i have x1 plus x2 minus 2x1 divided by 2.

02:23 That's going to give me x1 sorry, x2 minus x1 over 2.

02:32 A little sloppy with the notation.

02:34 So check that x1 minus 2x1 is negative x1.

02:39 We're going to want to write it this way as i do this for all three of these.

02:48 But all three of these are going to have the exact same algebra.

02:51 I just didn't want to have to write it out three times and make a line that was incredibly long.

02:57 So the distance from p1 to m we can now see is the square root of x2 minus x1 divided by 2 squared plus y2 minus y1 divided by 2 squared plus z1 z2 minus z 1 over 2 squared i can square each of those, squaring the numerator and denominator.

03:39 I get x2 minus x1 squared over 4 plus y2 minus y1 squared over 4.

03:55 Plus z2 minus z1 squared over 4.

04:04 I can subtract, no, add those numerators together to give me the square root of x squared minus x2 minus x1 squared plus y2 minus y1 squared plus z2 minus y1 squared, plus z2 minus z1 squared, all of that over four.

04:38 And if i take the square root of the numerator and denominator, i get the square root of the numerator.

04:52 Root of x2 minus x1 squared plus y2 minus y1 squared plus y2 minus y1 squared plus z2 minus z1 squared and all of that is over the square root of 4 which is 2 and notice that the numerator is simply the distance formula from p1 to p2 so i have found that the distance from p1 to p2 so i found that the distance from p1 to to m is half the distance from p1 to p2.

05:33 Now that's not enough.

05:35 This doesn't tell us that all three points lie in the same line.

05:39 If i show that the distance from m to p2 is also half of the distance from p1 to p2, then because the distances of the smaller pieces add up to being equal to the longer piece, we can conclude that they're on the same line.

05:56 And we'll also have concluded that the distance from p2 to m is also half of the distance from p1 to p2.

06:11 Let's go ahead and write an expression for the distance from m to p2.

06:19 And again, we're going to use the distance formula, subtracting x coordinates, y coordinates, and z coordinates.

06:29 And then we're going to do a lot of algebra and see how that works out.

06:34 So this is going to be the square root of x2 minus x1 plus x2 over 2 square, plus y2 minus y1 plus y2 over 2 square, plus z2 minus z 1 plus z 1 plus z 2 over 2 over 2, squared.

07:11 I'm going to do the same thing.

07:13 I'm going to show the algebra in detail for how we take care of the x's, and i'm just going to use the same result for the y's and the z's.

07:24 So again, i'm going to observe that i can make this a fraction over one.

07:29 And so i can rewrite this as 2x2 over 1, over 2.

07:37 When i multiply both, you're entering a now when i buy 2 minus x1 plus x2 over 2, which is going to give me 2x2 minus x1 minus x2 as i distribute that negative sign to do the subtraction.

08:08 2x2 minus this x2 is going to give me 1x2 and then minus the x2.

08:16 I'm going to have x2 minus x1 over 2.

08:29 So let me get some separation here, or at least frame some of these things so that they're not running together too much.

08:46 So the xs are going to simplify to x2 minus x1 over 2.

08:55 That same operation, same arithmetic, same algebra is going to convert the ys to y2.

09:01 Minus y1 over 2 and the same thing for the z's so i can say that mp is equal to mp2 is the square root of x2 minus x1 over 2 squared plus y2 minus y 1 over 2 squared plus z 2 minus y 1 over 2 square over to square, which is the exact same expression we had here.

09:50 Exact same expression we had here.

09:52 So we know how that simplifies.

09:55 So i'm not going to write all of those steps again.

09:58 I know that mp2 is going to be equal to the square root of x2 minus x1 squared plus y2 minus y1 squared plus z2 minus z1 squared all divided by two so i get that m p2 is equal to p1 p2 divided by 2 and so p1m plus m p2 is equal to p1 p2 over 2 plus p1, p2 over 2.

10:54 And so this is p1, p2.

10:58 And so the fact that the distance of the two smaller pieces adds up to the distance of the long piece, we know that these three points have to all be on the same line.

11:10 And we've already seen that each one of the smaller pieces is half of the distance from p1 to p2.

11:18 So it truly is in the middle of those two points.

11:23 It is the correct midpoint between p1 and p2.

11:33 Now, let's use the midpoint formula to take a look at this problem.

11:40 Given triangle abc with vertices at a, 123, b, negative 2, 0, 5, and c, 415, what are the lengths of its medians? i'm going to draw a triangle, and i'm not even going to pretend that this triangle has any resemblance to how this might look should we draw this triangle in our coordinate axis system.

12:06 I just want a triangle with points a, b, and c that i can find midpoints of, that i can find medians of, that i can name, and refer to easily.

12:18 So i'm going to call this point a, i want to call this point b, i'm going to call that point c.

12:23 And again, i don't care if that's not where they would look if i try to graph them in our three -dimensional system, because i just want some points for reference.

12:33 The midpoint from b to c will call that d.

12:39 So there is one median...

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