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Beg Algebra: Finding equations Given Two Points

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About this Lesson

  • Type: Video Tutorial
  • Length: 6:31
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 70 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)
College Algebra: Equations of a Line (5 lessons, $7.92)

Using the slope-intercept equation of a line, Professor Burger teaches you how to write the equation of a line if you are given two points on that line. Given two points, you can find the slope. Once you have found the slope of the line, you can input any point on that line into the equation with the slope to solve for b. Once you have found the slope and b, you have the slope-intercept equation (y = mx + b).

Learn how to find the slope of a line: http://www.mindbites.com/lesson/920-beg-algebra-finding-the-slope-given-two-points

Learn about slope-intercept equations:
http://www.mindbites.com/lesson/921-beg-algebra-equations-in-slope-intercept-form

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 http://www.thinkwell.com/student/product/beginningalgebra. 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

Thinkwell
Thinkwell
2174 lessons
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11/13/2008

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 www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.

Thinkwell lessons feature a star-studded cast of outstanding university professors: Edward Burger (Pre-Algebra through...

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Recent Reviews

Nopic_orng
Awesome
09/02/2009
~ Sashkya

I've been racking my brain forever trying to figure this out...Prof Berger helped me in a matter of minutes! Thank you so much!!!!!

Nopic_orng
Awesome
09/02/2009
~ Sashkya

I've been racking my brain forever trying to figure this out...Prof Berger helped me in a matter of minutes! Thank you so much!!!!!

Page 1 of 1 www.thinkwell.com © Thinkwell Corp.
WRITING AN EQUATION GIVEN TWO POINTS
So now we see the equation of a line can be expressed using this slope intercept notation where you just have the slope and you have the y intercept, and then you have y = mx + b. This is one of my favorite ways of writing lines, by the way. So m represents the slope and b represents the y intercept. Fantastic. But now, suppose that I asked you to find the equation of the line that actually passes through two points and that’s all you know. I don’t tell you the slope, I just tell you the two points. So let’s do a first example here. So I want you to find the equation of the line that passes through 0, 3 and also 2, 2. Okay, now, let’s think about that. Well, first of all, I want to find the slope. I need the slope, so what do I do? Well, that’s just the change in y--I’m going to write in in shorthand notation--over the change in x. So I’ve got to subtract the y from the y and the x from the x. And always remember, subtract in the same order: y from y, x from x. If you don’t like that you can do it this way. Now, watch this. This is different. y from y, x from x. Either one is correct, but what’s bad would be y from y, x from x. That doesn’t even feel right if you jut do it physically. You want to go y from y, x from x. Okay, let’s do it that way. So let’s go y--that minus that. So 2 minus 3, and divide that by 2 - 0. So that equals -1 divided by 2, which equals -½. So the slope we immediately see is is -½. That was pretty easy. So that goes in there.
Now, what about the y intercept? Well, that’s where the line is going to cross the y-axis. That means that x would be 0 there. So that’s where x is 0 because that’s the y-axis, and if you notice, look at that--0, 3. That means we go zero in the x direction and just up 3. So, in fact, that’s the y intercept. The y intercept is hidden right there. So, in fact, since this is a 0, something, that’s the y intercept, so I can just immediately say y = -½x + 3. So that was really easy because, in fact, they gave us the y intercept. We just had to realize that that was the y intercept, because x is 0. So if x is 0 that means that 0, 3, that’s the y intercept.
Okay, let’s try one where, in fact, the poser of the question wasn’t so friendly. So let’s find the equation for the line, and let’s do it in slope intercept form--I’ll remind you what that is--if the line contains the following two points: one of them is 1, 2, and the other one is -1, -4. Well, of course, we’re going to need the slope again, so let’s compute that. So slope equals the change in y over the change in x. So remember, we can subtract this from this, or this from this. Let’s do the following: Let’s take this and subtract this. So that’s going to be 2 - -4, and a I divide that now by 1 - -1. So you have to be careful of all those negative signs. Those negative signs actually combine to give a positive, so in fact, that’s 2 + 4, which is 6, divided by 1 + 1. You might think zero here, but be careful. It’s 1 + 1, which is 2. So this has slope 3. So that tells us what the slope is. So that’s pretty cool.
Okay, but now what I want to do is I want to find out the y intercept. But you’ll notice, neither of these points are an intercept. This is 1, 2 and this is -1, -4, so it looks like I’m in trouble here. However, if you think about it, two points should determine a line uniquely, so if I know two points I should be able to find the equation of the line. I already found the slope. I’m missing this value here. Well, the way to proceed is to actually put in what you have. So I have y = 3x + b. And the question is, what’s b? That’s the question. What is this? That’s the question. But what do I know? I know that both of these points lie on the line. That means that both of these points must satisfy this equation. If I plug in 1 for x and 2 for y, this must be satisfied. So all I have to do is pick one of these points and plug it in for x and y. Let’s do that. Let’s take this point here and what I’m going to do is, I’m going to take that 2 and I’m going to plug it in for x, and I’m going to take this 1 and plug it in--oh, my goodness, I made a huge typo. Sorry. This 1 is the x value, so I’m going to plug this in for x, and then this 2 I’m going to plug in for y. Okay? So I’m going to take this point that I know has to satisfy this equation, and I’m going to put this in for x, and that value in for y, and that has to be true. But what would it look like? Well, here I’d see a 2 = 3(1) + b. I still don’t know what b is. But look, now I have an equation and I can solve for b. So by taking one of the points that I know must lie on the line, and substituting it in for x and y appropriately, I will have an equation that just has b and everything else is known to be a number, and so I can solve for b. In fact, in this case, I see that 2 = 3 + b, so if I bring the 3 over as a -3, I see that b must equal -1. So I actually can find the y intercept even if I’m not given it, by just taking one of the points and putting it in. By the way, if you would have taken this point and plugged it into here, you would actually get the same answer for b, and you can try that. You can take -1, -4, plug in the -4 here, plug in the -1 here, and solve, and you’ll still get -1 for b. You can try it.
Anyway, armed with that, I now see that the equation of the line is y = the slope, which is 3x and then + b, and b here is -1, so I’ll just put down plus, minus, or just -1. And there’s the equation of the line. So just given two points, I can now find the equation of the line, and I’m writing it in the slope intercept form. Neato! All right. You try some of these. See what you think.

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