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Beg Algebra: Find distance & midpoint betwn points


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About this Lesson

  • Type: Video Tutorial
  • Length: 10:57
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 118 MB
  • Posted: 12/02/2008

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
Trigonometry: Full Course (152 lessons, $148.50)
College Algebra: Relations and Functions (57 lessons, $74.25)
Trigonometry: Algebra Prerequisites (60 lessons, $69.30)
Beginning Algebra Review (19 lessons, $37.62)
Int Algebra: Applications of Radicals (3 lessons, $4.95)
College Algebra: Relationships between Two Points (2 lessons, $2.97)

Professor Burger proves the distance formula (distance = square root [(x1-x2)^2 + (y1-y2)^2] ) using the pythagorean theorum. Using this formula, you can find the distance between two points on a line. Professor Burger goes on to prove the midpoint formula ( [x1 + x2]/2, [y1 + y2]/2). The midpoint formula is in the form of a point on a line and is the average of the points on the x and y axes.

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, Beginning Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers linear equations, inequalities, polynomials, rational expressions, relations and functions, roots and radicals, quadratic equations and systems of equations.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

About this Author

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Founded in 1997, Thinkwell has succeeded in creating "next-generation" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technology-based textbooks. For more information about Thinkwell, please visit or visit Thinkwell's Video Lesson Store at

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Page 1 of 2 © Thinkwell Corp.
Suppose I give you two points on the plane and I want you to find the exact distance between them. How would you do it? Well, there’s actually a thing called “the distance formula,” and I hope you will not memorize it—you shouldn't memorize it. Let’s think through exactly how one could figure out what that formula would be. So let’s suppose I give you two points. Maybe one is way over here. Now, I won’t know the coordinates of the point just for this example because I want to actually generate this formula for you. So suppose that this x value is x1, and the y value would be y1. Then I have another point—let’s say this point’s way over here, and that value, the x value—I go over x2, but I go up to y2. So I want you to understand what this means. This is just a point. The coordinates of the point are x1, y1. Don’t get nervous by that, it’s just some number. It could be like, maybe, (3,2). I just don’t know what those values are right now and I want to use general values to actually create those formulas for you. This second point has some different coordinates, so maybe this is like (10,8). So I’m calling x2 that value and I go up y2.
Okay, now I want to find the distance between them, so what I want to do is connect them with a straight line. So the question is how long is this line? Well, you know, you can memorize, like I said, the formula for this, or you could just think about it for a second and realize that that’s actually the hypotenuse of a right triangle. Let me draw in the right triangle so you can actually see it. In fact, let me draw it in in my right triangle color, which I just decided is going to be green. So if I just draw these two lines—one horizontal and one vertical—you see I have a nice right triangle here. Well, this is great, because I can use now the Pythagorean theorem to find the hypotenuse if I know the lengths of each of these sides. But I do, because this length is going to be what? Well, it’s the distance, basically, between here and here. So that’s going to be x2, and I subtract off that distance, which is x1. So this is actually x2 - x1. And this distance right here, well, that’s just going to be all of y2, but then I subtract off y1. So this is actually y2 - y1.
So what does the Pythagorean theorem say? Well, this is the distance I want. Let me call this “dist”. Anyway, if I called this distance between these two points “dist”, then what do I know? I know from the Pythagorean theorem that this length squared plus that length squared will equal this squared. So what that would mean is that dist, the hypotenuse squared equals, well, this squared, which is (x2 - x1) 2, which, by the way, doesn’t equal x2 squared minus x1 squared. That is the number one classic mistake—the squaring mistake—remember to foil. Plus, that length squared, which is (y2 - y1) 2. So if now want to solve for distance alone, I have to take square roots. Now, technically, I have to take plus or minus the square root, but since this is a distance we know that there are no negative distances, so it has to be the positive root. So now, today, I can actually take the one square root and say that the distance is equal to just the regular square root of (x2 - x1)2 + (y2 - y1)2. So, in fact, that is the formula for distance, or sometimes called the distance formula.
So the distance between two points is given by this. You subtract the x values and square it, subtract the y values, square it, add those two numbers together, take the square root. That’s the distance. But I hope you won’t memorize that. I hope instead you just think of this picture. That’s what I always do.
Let’s try an example. So suppose I want to find the distance between the points (6,4) and (-8,11). So what’s the distance? Well, what I do is I just think about it graphically and set up the triangle, and we’d see that the distance would equal the square root, and I’d take the differences of the x’s. Now, the important thing here is to always subtract this x from that x, so that would be (6 - -8) 2 + (4 - 11) 2. Noticed I’m subtracting these points from these points. So that equals the square root. Well, 6 - -8 is 6 + 8, and 6 + 8 is around 14, so I have 142, and then I have 4 - 11, and so 4 - 11 = (7)2, so what is this? Well, the square of 14 is 196 and the square of -7 is 49. And so what I see here is the square root of 245. And you can use a calculator, for example, to see that this equals 15.65 stuff—it goes on. So this is the actual answer, and this gives you a sense of what it is—it’s around 15.65, but the actual distance between these two points, if you put down a measuring tape, would be the square root of 245, or 15.6-something. So that’s how you find distances.
What if I give you two points and I want you to find the point that’s right in between them? Let’s think about that for a second. So now we can find the distance, but suppose I give you a point right here. Let me call this point, again, x1, and I’m going to call this y1. And I give you a point way over here—I call this x2, and its y value I’ll call y2. Well, you can draw the line that connects them—this is the shortest distance. We just figured out how to find that distance if we wanted to. But suppose instead of finding that distance I want to find the point right in between? So that point right in between would look just like this; it would be right there. How would you find that point? Well, let’s think about that for a second. What would I have to do? Well, it would sort of be in the middle of this, and it would be in the middle of this. So how would I find the middle of this? Well, the middle of this, in some sense, is like the average of these two numbers. It’s the thing
Page 2 of 2
right in the middle. So how do you find the average? What I would do is I would just add these two numbers up and divide by 2. So x1 + x2, and I’d divide the whole thing by two. That’s the middle—the thing that’s right in the middle here.
And where would this point be? Well, it would be in the middle of these two things, which would be taking the average, y2 + y1, divide by 2. That’s this point right here. So, in fact, the coordinates of the midpoint between two points are going to be 221xx+, and the y is going to be 212yy+. So, in fact, this is how you find the coordinates of the midpoint between two points. And again, I don't think you should memorize it. Just remember that you sort of want to find the average between the x’s and then the average between the y’s.
Let’s do a quick example together. Let’s find the midpoint between the points 1, -1 and the point 5, 5. So if I want to find the midpoint what I want to do is average the x’s and then average the y’s, so the midpoint would be—well, I’d average the x’s, so I’d add 1 + 5 and divide it by 2, then I’d average the y’s, -1 + 5 divided by 2, and that would equal 26 and 24, which would be the point 3, 2. That should be the midpoint. Let’s graph this and see if that looks okay. So I’m going to put that over there so you can see there’s the midpoint. Let’s see if this makes any sense at all. So the first point over there was (1,-1). So I go one unit over and then -1, so that’s one unit down. So there’s the point (1,-1). The other point is (5, 5)—1, 2, 3, 4, 5. This is, of course, very rough, as you can see—1, 2, 3, 4, 5. I’m literally eyeballing this. I’m not even trying to be accurate; just to get a sense if this answer makes sense or not. Sometimes it’s worth thinking about if an answer even makes sense. This is (5, 5). If I connect these two points with a straight line, that line would look like this. There’s the line. And where’s the midpoint? The midpoint looks like it’s sort of around here, and let’s see, does that midpoint look like it’s around the point 3, 2? Well, in fact, look at this picture. This is absolutely amazing. Here is 3… But look at it—it’s exactly over 3. And here is 2—it is exactly on 2. So even on this little sketchy, sketchy, lousy little drawing, you can see that this answer is certainly reasonable, and of course, from the mathematics we know it’s absolutely correct.
So pretty easy to find midpoints. We just average the x’s, add them up, divide by 2, average the y’s, add them up, divide by 2, that gives the midpoint. To find the distance formula I just use the Pythagorean theorem to create a right triangle and find the length of the hypotenuse.
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