## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Related Articles:

### Measuring Segments

- Ruler postulate: The points on any line can be paired with real numbers so that, given any two points P and Q on the line, P corresponds to zero, and Q corresponds to a positive number
- If Q is between P and R, then QP + QR = PR
- If PQ + QR = PR, then Q is between P and R
- Pythagorean Theorem: In a right triangle, the sum of the squares of the measures of the legs equals the square of the measure of the hypotenuse
- Pythagorean Theorem: a
^{2}+ b^{2}= c^{2} - The distance d between any two points with coordinates (x
_{1}, y_{1}) and (x_{2}, y_{2}) is given by the formula:

### Measuring Segments

- AE = | − 8 − 7| = | − 15| = 15
- BD = | − 2 − 3| = | − 5| = 5
- DE = |3 − 7| = | − 4| = 4
- CE = |0 − 7| = | − 7| = 7

1. AB = 3, BC = 8, AC = ?

2. AC = 9, BC = 4, AB = ?

- 1. AC = AB + BC = 3 + 8 = 11
- 2. AB = AC − BC = 9 − 4 = 5

1) AB = 5x + 1, BC = x − 2, AC = 17

2) AB = 2, BC = 2x + 2, AC = 3x + 1

- 1) AB + BC = AC
- 5x + 1 + x − 2 = 17
- x = 3
- BC = x − 2 = 3 − 2 = 1
- 2) AB + BC = AC
- 2 + 2x + 2 = 3x + 1
- x = 3
- BC = 2x + 2 = 2 ×3 + 2 = 8

2. BC = 8

- d = √{(2 − ( − 1))
^{2}+ (4 − ( − 3))^{2}} = √{58}

- A(4, 5), B( − 4, − 3)
- d = √{( − 4 − 4)
^{2}+ ( − 3 − 5)^{2}} = √{8^{2}+ 8^{2}} = √{128} = 8√2

- A( − 3, 2), B (4, − 3)
- AB = √{(4 − ( − 3))
^{2}+ ( − 3 − 2)^{2}} = √{7^{2}+ 5^{2}} = √{74} - BC = AB − AC = √{74} − 3

^{o}, AB = 6, BC = 5, find AC.

- With Pythagorean Theorem, AC = √{AB
^{2}+ BC^{2}} - AC = √{6
^{2}+ 5^{2}} = √{61}

^{o}, AC = 10, AB = 6, find BC.

- With Pythagorean Theorem, BC = √{AC
^{2}− AB^{2}} - BC = √{10
^{2}− 6^{2}} = √{100 − 36} = √{64} = 8

^{o}, ∠ADB = 90

^{o}, AB = 5, BD = 4, BC = 6, D is between points A and C, find CD.

- With Pythagorean Theorem, AC = √{AB
^{2}+ BC^{2}} = √{5^{2}+ 6^{2}} = √{61}

AD = √{AB^{2}− BD^{2}} = √{5^{2}− 4^{2}} = 3 - CD = AC − AD = √{61} − 3

^{o}, ∠ADB = 90

^{o}, AD = 3, CD = 5, BD = 4, D is between points A and C, find AB and BC.

- With Pythagorean Theorem, AB = √{AD
^{2}+ BD^{2}} = √{3^{2}+ 4^{2}} = 5 - BC = √{CD
^{2}+ BD^{2}} = √{5^{2}+ 4^{2}} = √{41}

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Measuring Segments

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Segments
- Ruler Postulate
- Segment Addition Postulate
- Segment Addition Postulate
- Pythagorean Theorem
- Pythagorean Theorem, cont.
- Distance Formula
- Extra Example 1: Find Each Measure
- Extra Example 2: Find the Missing Measure
- Extra Example 3: Find the Distance Between the Two Points
- Extra Example 4: Pythagorean Theorem

- Intro 0:00
- Segments 0:06
- Examples of Segments
- Ruler Postulate 1:30
- Ruler Postulate
- Segment Addition Postulate 5:02
- Example and Definition of Segment Addition Postulate
- Segment Addition Postulate 8:01
- Example 1: Segment Addition Postulate
- Example 2: Segment Addition Postulate
- Pythagorean Theorem 12:36
- Definition of Pythagorean Theorem
- Pythagorean Theorem, cont. 15:49
- Example: Pythagorean Theorem
- Distance Formula 16:48
- Example and Definition of Distance Formula
- Extra Example 1: Find Each Measure 20:32
- Extra Example 2: Find the Missing Measure 22:11
- Extra Example 3: Find the Distance Between the Two Points 25:36
- Extra Example 4: Pythagorean Theorem 29:33

### Geometry Online Course

### Transcription: Measuring Segments

*Welcome back to Educator.com.*0000

*This lesson is on measuring segments; let's begin.*0002

*Segments: if we have a segment AB, here, this looks like a line; we know that this is a line, because there are arrows at the end of it.*0009

*But if we are just talking about this part from point A to point B, that is a segment,*0019

*where we are not talking about all of this--just between point A and point B.*0027

*That would be a segment, and we call that segment AB.*0035

*And it is written like this: instead of having...if it was just a line, the whole thing, that we were talking about, then we would write it like this.*0038

*But for the segments, we are just writing a line like that--a segment above it.*0048

*A segment is like a line with two endpoints.*0057

*And if we are talking about the measure of AB, the measure of AB is like the distance between A and B.*0062

*And when you are talking about the measure, you don't write the bar over it--you just leave it as AB.*0074

*So, when you are just talking about the segment itself, then you would put the bar over it;*0080

*if not, then you are just leaving it as AB; OK.*0084

*Ruler Postulate: this has to do with the distance: The points on any line can be paired with real numbers*0093

*so that, given any two points, P and Q, on the line, P corresponds to 0, and Q corresponds to a positive number,*0102

*just like when you want to measure something--you use a ruler, and you put the 0 at the first point;*0111

*and then you see how long whatever you are trying to measure is.*0122

*In that same way, if I have two points on a number line--let's say I have a point at 2 and another point at 8--*0128

*then if I were to use a ruler to find the distance between 2 and 8, I would place my 0 here, on the 2.*0138

*That is what I am saying: the first number corresponds to 0.*0146

*It is as if this becomes a 0, and then we go 1, 2, 3, 4, 5, 6; so whatever this number becomes, when this is 0--that would be the distance.*0149

*And when you use the ruler postulate, you can also find the distance of two points on the number line, using absolute value.*0166

*I can just subtract these two numbers, 2 minus 8; but I am going to use absolute value.*0176

*So then, 2 - 8 is -6; the absolute value of it is going to make it 6.*0186

*Remember: absolute value is the distance from 0; so if it is -6, how far away is -6 from 0? 6, right?*0192

*So, absolute value just makes everything positive.*0200

*You can also...if I want to find the distance from 8 to 2, it is the same thing: the absolute value of 8 minus 2.*0205

*OK, if I measure the distance from here to here, it is the same thing as if I find the distance from this to this.*0215

*That is also 6; so either way, your answer is going to be 6.*0223

*Another example: Find the distance between this point, -4, and...let's see...5.*0230

*I am going to find the distance from -4 to +5.*0240

*I can do the same thing: the absolute value of -4 - 5; this is the absolute value of -9, which is 9.*0244

*From -4 to 5...they are 9 units apart from each other.*0260

*And you can also do the other way: the distance from 5 to -4...minus -4...a minus negative becomes a positive, so this is absolute value of 9, which is 9.*0266

*You don't have to do it twice; I am just trying to show you that you will have the same distance,*0284

*whether you start from this number and go to the other number, or you start from the other one and you go the other way.*0292

*That is the Ruler Postulate.*0299

*The next one, the Segment Addition Postulate: you will use this postulate many, many, many times throughout the course.*0304

*A postulate, to review, is a math statement that is assumed to be true.*0314

*Unlike theorems...theorems are also math statements, but theorems have to be proved*0321

*in order for us to use them, to accept it as true; but postulates we can just assume to be true.*0330

*So, any time there is a postulate, then we don't have to question its value or its truth.*0334

*We can just assume that it is true, and then just go ahead and use it.*0344

*Segment Addition Postulate: if Q is between P and R, then QP + PR = PR.*0348

*If Q...I think this is written incorrectly...is between P and R...this is supposed to be QR; so let me fix that really quickly.*0365

*QP + QR = PR: so Q is between P and R--let me just write that out here.*0382

*If this is P, and this is R, and Q is between P and R; then they are saying that QP or PQ plus QR, this one, is going to equal the whole thing.*0390

*It is...if I have a part of something, and I have another part of something, it makes up the whole thing.*0408

*And if PQ + QR equals PR, then Q is between P and R; so you can use it both ways.*0414

*And it is just saying that this whole thing is...let's say that this is 5, and this is 7; well, then the whole thing together is 12.*0425

*Or if I give you that this is 10, and then the whole thing is 15, then this is going to be 5, right?*0442

*That is all that it is saying: the whole thing can be broken up into two parts, or the two parts can be broken up into two things...*0456

*it just means that, if it is, then Q is between P and R; or if they give you that Q is between P and R--*0463

*if that is given, that a point is between two other points on the segment, then you can see that these two parts equal the whole thing.*0471

*That is the Segment Addition Postulate.*0480

*Find BC if B is between A and C and AB is 2x - 4; BC is 3x - 1; and AC equals 14.*0484

*Find BC if B is between A and C...let's draw that out: here is A and C, and B is between them.*0500

*It doesn't have to be in the middle, just anywhere in between those two points.*0510

*If I have B right here, then we know that AB, this segment, plus this segment, equals the whole segment, AC.*0514

*So, AB is 2x - 4; and BC is 3x - 1; the whole thing, AC, is 14.*0525

*I need to be able to find BC; well, I know that, if I add these two segments, then I get the whole segment, right?*0541

*So, I am going to do 2x - 4, that segment, plus 3x - 1, equals 14.*0548

*So, here, to solve this, 2x + 3x is 5x; and then, -4 - 1 is -5; that equals 14.*0563

*If I add 5 to that, 5x = 19, and x = 19/5; OK.*0577

*And they want you to find BC; now, you found x, but always look to see what they are asking for.*0586

*BC = 3(19/5) - 1; and then, this is going to be 57/5, minus 1; so I could change this 1 to a 5/5,*0602

*because if I am going to subtract these two fractions, then I need a common denominator.*0625

*Minusing 1 is the same thing as minusing 5/5; and that is only so that they will have a common denominator, so that you can subtract them.*0631

*And then, this will be 52/5; you could just leave it as a fraction.*0639

*And notice one thing: how these BC's, these segments, don't have the bars over them.*0649

*And that is because you are dealing with measure: whenever you have a segment equaling its value,*0656

*equaling some number, some distance, some value, then you are not going to have the bar over it, because you are talking about measure.*0661

*The next one: Write a mathematical sentence given segments ED and EF.*0676

*This is using the Segment Addition Postulate; that is the kind of mathematical sentence it wants you to give.*0682

*ED and EF--that is all you are given, segments ED and EF.*0689

*Well, here is E; there is an E in both; that means that E has to be in the middle.*0696

*E has to be here, because since it is in both segments, that is the only way I can have E in both.*0707

*So then, here, this will be D, and this can be F.*0719

*Now, this can be F, and this can be D; it doesn't really matter,*0722

*as long as you have E in the middle, somewhere in between, and then D and F as the endpoints of the whole segment.*0726

*And to use the Segment Addition Postulate, I can say that DE or ED, plus EF, equals DF; we just write it like that.*0737

*OK, the Pythagorean Theorem: In a right triangle, the sum of the squares of the measures of the legs equals the square of the measure of the hypotenuse.*0758

*You probably remember this from algebra: if you have a right triangle...*0770

*now, you have to keep that in mind; the Pythagorean Theorem can only be used on right triangles;*0776

*a right triangle, and you use it to find a missing side.*0785

*You have to be given two out of the three sides--any two of the three sides--to find the missing side.*0789

*That is what you use the Pythagorean Theorem for--only for right triangles, though.*0796

*So, a ^{2} + b^{2} = c^{2}: that is the formula.*0799

*You have to make the hypotenuse c; this has to be c.*0809

*Now, just to go over, briefly, the Pythagorean Theorem, we have a right triangle, again.*0815

*Now, let's say that this is 3; then, if a ^{2} + b^{2} = c^{2}, then a and b are my two sides, my two legs, a and b.*0829

*The hypotenuse will always be c; it doesn't have to be c, but from the formula, whatever you make this equal to--*0844

*the square of the sum of the two sides--has to be...*0852

*I'm sorry: you have to square each side, and then you take the sum of that; it equals the hypotenuse squared.*0858

*OK, and let me just go over this part right here.*0864

*If we have this side as 3, then you square it, and it becomes 9.*0870

*Now, you can also think of it as having a square right there; so if this is 3, then this has to be 3; this whole thing is 9.*0877

*If this is 4, if I make a square here, then this has to be 4; this whole thing is 16--the area of the square.*0890

*And then, it just means that, when you add up the two, it is going to be the area of this square right here.*0907

*Then, the area of this square is going to be 25, because you add these up, and then that is going to be the same.*0917

*And then, that just makes this side 5.*0925

*a ^{2}...back to this formula...+ b^{2} = c^{2}: you just have to square the side,*0935

*square the other side, add them up, and then you get the hypotenuse squared.*0942

*Let's do a problem: Find the missing side.*0950

*I have the legs, the measure of the two legs, and I need to find the hypotenuse.*0956

*So, that is 4 squared, plus 3 squared, equals the hypotenuse squared; so I can just call that c squared.*0961

*So, 4 squared is 16, plus 9, equals c squared; 25 = c ^{2}.*0972

*And then, from here, I need to square root both; so this is going to become 5.*0983

*Now, normally, when you square root something, you are going to have a plus/minus that number;*0991

*but since we are dealing with distance, the measure of the side, it has to be positive; so this right here is 5.*0997

*OK, the distance formula: The distance between any two points with coordinates (x _{1},y_{1})*1010

*and (x _{2},y_{2}), is given by the formula d = the square root*1018

*of the difference of the x's, squared, plus the difference of the y's, squared.*1025

*Here, this distance formula is used to find the distance between two points.*1035

*And we know that a point is (x,y); and the reason why it is labeled like this...*1044

*you have to be careful; I have seen students use these numbers as exponents.*1051

*Instead of writing it like that, they would say (x ^{2},y^{2}); that is not true.*1056

*This is just saying that it is the first x and the first y; so this is from the first point.*1061

*They are saying, "OK, well, this is (x,y) of the first point; and this is the second point."*1065

*And that is all that these little numbers are saying; they are saying the first x and first y,*1076

*from the first point, and the second x and the second y from the second point.*1082

*x _{2} just means the second x, the x in the second point.*1090

*And it doesn't matter which one you make the first point, and which one you make the second point;*1094

*just whichever point you decide to make first and second, then you just keep that as x _{2}, x_{1}, y_{2}, and y_{1}.*1100

*Find DE for this point and point D and E; so then, I can make this (x _{1},y_{1}), my first point;*1108

*and then this would be (x _{2},y_{2})--not (x^{2},y^{2}); it is (x_{2},y_{2}), the second point.*1119

*Then, the distance between these two points...I take my second x (that is 1) minus the other x, so minus -6, squared,*1127

*plus the second y, 5, minus the other y (minus 2), squared.*1143

*1 - -6: minus negative is the same thing as plus the whole thing, so this will be 7 squared plus 3 squared.*1156

*7 squared is 49; plus 3 squared is 9; this is going to be 58.*1170

*Now, 58--from here, you would have to simplify it.*1182

*To see if you can simplify it, the easiest way to simplify square roots--you can just do the factor tree.*1187

*I just want to do this quickly, just to show you.*1194

*A factor of 58 is going to be 2...2 and 29.*1199

*Now, 2 is a prime number, so I am going to circle that.*1207

*And then, 29: do we have any factors of 29? No, we don't.*1211

*So, this will be the answer; we know that we can't simplify it.*1217

*The distance between these two points is going to be the square root of 58.*1224

*Let's do a few examples that have to do with the whole lesson.*1234

*Find each measure: AC: here is A, and here is C.*1238

*You can use the Ruler Postulate, and you can make this point correspond to 0.*1246

*And then, you see what C will become, what number C will correspond to.*1251

*Or, you can just use the absolute value; so for AC, this right here...AC is the absolute value of -6 minus...C is 2;*1259

*so that is going to be the absolute value of -8, which becomes 8.*1275

*BE: absolute value...where is B? -1, minus E (is 9)...so this is the absolute value of -10, which is 10.*1284

*And then, DC: the absolute value of...D is 5, minus 2.*1301

*Now, see how I went backwards, because that was DC.*1311

*It doesn't matter: you can do CD or DC; with segments, you can go either way.*1313

*So, DC is 5 - 2 or 2 - 5; it is going to be the absolute value of 3, which is 3.*1319

*The next example: Given that U is between T and V, find the missing measure.*1332

*Here, let's see: there is T; there is V; and then, U is just anywhere in between.*1342

*TU, this right here, is 4; TV, the whole thing, is 11.*1356

*So, if the whole thing is 11, and this is 4, well, I know that this plus this is the whole thing, right?*1362

*So, you can do this two ways: you can make UV become x; I can make TU;*1370

*or plus...UV is x...equals the whole thing, which is 11.*1377

*You can solve it that way, or you can just do the whole thing, minus this segment.*1384

*If you have the whole thing, and you subtract this, then you will get UV.*1391

*You can do it that way, too; if you subtract the 4, you get 7, so UV is 7.*1395

*The next one: UT, this right here, is 3.5; VU, this right here, is 6.2; and they are asking for the whole thing.*1407

*So, I know that 3.5 + 6.2 is going to give me TV.*1417

*If I add this up, I get 9.7 = TV.*1428

*And the last one: VT (is the whole thing) is 5x; UV is 4x - 1; and TU is 2x - 1; so they want to define TU.*1440

*I know that VU, this one, plus TU--these are the parts, and this is the whole thing.*1459

*The whole thing, 5x, equals the sum of its parts, 4x - 1 plus (that is the first part; the second part is) 2x - 1.*1471

*I am just going to add them up; so this will be 6x - 1 - 1...that is -2.*1487

*And then, if I subtract this over, this is going to be -x = -2, which makes x 2.*1499

*Now, look what they are asking for, though: you are not done here.*1505

*They are asking for TU, so then you have to take that x-value that you found and plug it back into this value right here, so you can find TU.*1508

*TU is going to be 2 times 2 minus 1, which is 4 minus 1, which is 3; so TU is 3.*1520

*The next example: You are finding the distance between the two points.*1537

*The first one has A at (6,-1) and B at (-8,0); so again, label this as (x _{1},y_{1});*1542

*this has to be x; this has to be y; you are just labeling as the first x, first y;*1554

*this is also (x,y), but you are labeling it as x _{2}, the second x, and then the second y.*1559

*The distance formula is the square root of x _{2}, the second x, minus the other x, squared, plus the difference of the y's, squared.*1565

*x _{2} is -8, minus 6, squared, plus...and then y_{2} is 0, minus -1, squared; this is -14 squared, plus 1 squared.*1586

*-14 squared is 196, plus 1...and then this is just going to be the square root of 197.*1615

*And then, the next one: I have two points here: I have A at this point, and I have B at this point.*1644

*Now, even though it is not like this problem, where they give you the coordinates, they are showing you the coordinates.*1653

*They graphed it for you; so then, you have to find the coordinates of the points first.*1660

*This one is at...this is 0; this is 1; this is 2; then this is 2; A is going to be at (2,1), and then B is at (-1,-2)...and -2.*1664

*So then, to find the distance between those...let's do it right here:*1686

*it is going to be...you can label this, again: this is (x _{1},y_{1}), (x_{2},y_{2}).*1691

*So, it is -2, the second x, minus the first x, minus 2, squared, plus the second y, -2, minus 1, squared.*1704

*And then, let me continue it right here, so that I have more room to go across:*1722

*the square root of...-2 - 2 is -4, squared; and then plus...this is -3, squared;*1728

*-4 squared is 16, plus...this is 9; and that is the square root of 25, which is a perfect square, so it is going to be 5.*1747

*The distance between these two points, A and B, is 5.*1764

*And you could just say 5 units.*1768

*The last example: we are going to use the Pythagorean Theorem to find the missing links.*1774

*It has a typo...find...*1781

*And these are both right triangles; I'll just show that...OK.*1784

*The first one: the Pythagorean Theorem is a ^{2} + b^{2} = c^{2}.*1789

*Here, I am missing this side; I am going to just call that, let's say, b.*1801

*So, a...it doesn't matter if you label this a or this a; just make sure that a and b are the two sides.*1807

*a ^{2} is 5^{2}; plus b^{2}, equals 13^{2}.*1816

*5 squared is 25, plus b squared, equals 169; and then, subtract the 25; so b ^{2} = 144.*1825

*Then, b equals 12, because you square root that; so this is 12.*1840

*The second one: I am going to label this c, because it is the hypotenuse.*1851

*Then: a ^{2} + b^{2} = c^{2}, so 6^{2} + 8^{2} = c^{2}.*1857

*6 ^{2} is 36, plus 64, equals c^{2}; these together make 100, so c is 10.*1868

*OK, well, that is it for this lesson; thank you for watching Educator.com.*1883

2 answers

Last reply by: Kevin Zhang

Sun Aug 21, 2016 3:14 PM

Post by Kevin Zhang on August 21, 2016

do you make your own slides?

0 answers

Post by Nancy Reyes on September 22, 2014

How do I know where to put the segments in addition postulate? Does the order matter?

0 answers

Post by Henriana Tommy on March 1, 2014

How do you know when to write it as 58 squared and not just 58? Does it matter?

0 answers

Post by Alexis Rodriguez-Gilbert on October 22, 2013

section segment addition postulate

0 answers

Post by Alexis Rodriguez-Gilbert on October 22, 2013

im not understanding how you are working the problem out and getting the ultimate answer....are we adding multiplying???

0 answers

Post by julius mogyorossy on September 15, 2013

I spoke too soon, she corrected that mistake, could not E be to the left of D and F.

1 answer

Last reply by: Professor Pyo

Thu Jan 2, 2014 3:40 PM

Post by julius mogyorossy on September 15, 2013

QP + PR does not = PR, that is a mistake.

0 answers

Post by Shahram Ahmadi N. Emran on July 12, 2013

Just a quick heads up, your distance formula in your quick notes is incorrect and you might want to include the correlation between the PyT formula and the distance formula, so that it is clear why you are supposed to add the two components instead of the subtraction that you highlight. It super confused at me at first until I looked it up.

0 answers

Post by Shahram Ahmadi N. Emran on July 12, 2013

For the question: "Write a mathematical sentence given segments ED and EF." E does not have to be in the middle:

in the case:

E-------D------F

the mathematical sentence would be EF - ED = DF

In the case:

E----F-------------D

the mathematical sentence would be ED - EF = DF

Also, in the question, ED and EF should have a bar over them because you are talking about segments and not measures

1 answer

Last reply by: Professor Pyo

Fri Aug 2, 2013 2:40 AM

Post by julius mogyorossy on June 3, 2013

Ms. Pyo, I am probably making a fool out of myself again, ask Dr. Carleen, but what would the absolute value of 2 + -5, be, it seems to me it would be 3, but it seems that mathematicians think it should be 7. I was very interested to learn that quadratic equations lie when they are factored, they lie and tell the truth at the same time. Luckily I don't have to factor or use a test point to see where the solution set is, I can see it when I see the equation(s). I found Algebra 2 very interesting, I can't wait to see what Calculus is all about, I have no clue what Trigonometry is about. I have been blessed with incredible instincts, super human, life over death, they are the reason I am still alive. I think I am going to make an incredible discovery having to do with Geometry, I have always sensed incredible potential in Geometry, I have read that incredible things have been allegedly proven having to do with Geometry, I wonder if it is true. Some day I shall do these experiments myself. The knowledge Dr. Carleen gave to me forced scientists to surrender to me, unofficially, some day, officially. I never thought I would see that day, I thought they would do me like Pasteur. Some day you shall know what I am referring to.

1 answer

Last reply by: Professor Pyo

Fri Aug 2, 2013 2:33 AM

Post by Manfred Berger on May 27, 2013

So what the Segment Addition Postulate is saying is basically that colinearity is transitve, isn't it?

1 answer

Last reply by: Manfred Berger

Mon May 27, 2013 11:47 AM

Post by bo young lee on February 19, 2013

at the educator.com where can i find more about the

pythagorean theorem???

0 answers

Post by Kenneth Montfort on February 18, 2013

Just a quick heads up, your distance formula in your quick notes is incorrect and you might want to include the correlation between the PyT formula and the distance formula, so that it is clear why you are supposed to add the two components instead of the subtraction that you highlight. It super confused at me at first until I looked it up.

0 answers

Post by Edward Hook on February 18, 2013

Everybody loves your teaching style Mary and I have to say that I do too!

1 answer

Last reply by: Habibo Ali

Wed Feb 5, 2014 10:19 AM

Post by Mohammed Abdullah on December 13, 2012

In the video it shows d=square root x1-x2)squared+(x1+x2) squared, in the quick notes they subtract the square roots.

1 answer

Last reply by: Professor Pyo

Sat Mar 2, 2013 1:53 AM

Post by chun yung on November 27, 2012

I have a question on the distance formula, why do they have to make it (x2-x1)+(y2-y1) if u put (x1-x2)+(y1-y2)u could get the same answer.

0 answers

Post by Catherine Henderson on July 31, 2012

Wow,

That was great!

0 answers

Post by Joseph Reich on June 20, 2012

For the question: "Write a mathematical sentence given segments ED and EF." E does not have to be in the middle:

in the case:

E-------D------F

the mathematical sentence would be EF - ED = DF

In the case:

E----F-------------D

the mathematical sentence would be ED - EF = DF

Also, in the question, ED and EF should have a bar over them because you are talking about segments and not measures

0 answers

Post by Giri Iyer on November 10, 2011

Very well done and explained.. I am teaching my son geometry and these lessons are so clear conceptually that even I can recall these concepts now :)

0 answers

Post by Pangayar Selvi Shanmugasundaram on August 9, 2011

realy iam telling, its wonderful vedio...

0 answers

Post by Ahmed Shiran on June 4, 2011

Great work by Mary...

0 answers

Post by Suneet Dash on May 31, 2011

i must say, great video