Determine whether the three points are collinear. (-7,4),(6,-2...

Question

Determine whether the three points are collinear. (-7,4),(6,-2),(-1,1)

Accepted Answer

Welcome back. I am so glad you're here. We are given three points and were asked to identify the three points are co linear. So what I'm going to do first is I'm going to label the three points. So the first point will be A that's negative 10. And similarly recognizing that negative nine is the X value for point A. So I'm going to label it X sub A and that 10 is the Y value for point A. So I'm going to label it Y sub A, the next point I'm going to label point B that is negative 38. And recognizing that negative three is the X value for B so X sub B and that eight is the Y value for B. So why sub B and the last one here will be point C that will be at and two is the X value for C so X sub C and four is the Y value for C. So Y sub C. And what do we do next? How do we figure out if these are co linear while recalling from previous lessons, we have the distance formula method to determine whether these points are co linear. We're going to measure the distance between each of these points. So from A to B will be one distance we measure from B to C will be another distance that we measure. And then from A to C will be the final distance that we measure. And once we've measured those distances, then we take the two shortest ones add them together and see if they equal the longest one. If they do, then these are co linear. If they don't equal, then these are not co linear. Alright. So the next thing we need to recall from previous lessons is the distance formula itself. The distance formula is equal to the square root of, we take X sub two minus X sub one in parentheses and then square that then add that to new parentheses. Y sub two minus Y sub one and then square that. Well, we've got excess and wise that have A S and B's and C's not ones and twos. So I'll show you how that's going to work. So for the measuring of A B, the line segment A B, we're going to have A B R first and be, be our second. So when we're plugging this in, we take the square root of and then we have X sub two. Well, that's going to be X sub B because B is the second one and X sub B is negative three. And then that will be minus our X sub one, R X sub one is going to be our X sub A. So that's minus negative nine and then we are going to square that and add that to. Now taking our Y sub two, that's going to be our Y sub B which is eight, then minus R Y sub one, which is going to be our Y sub A which is 10 and then closing those parentheses and squaring. Now it's just a matter of simplification. So doing the first thing here, negative three minus negative nine, that's the same as negative three plus nine, bringing down the rest. Then we have the addition and subtraction negative three plus nine that equals a positive six, still square that plus eight minus 10. That's a negative two and still squaring that. Now we can do the exponents. So we have six square, that's 36 plus negative two squared that equals positive four. And now doing this edition 36-plus 4 that equals 40. So we have the square root of 40. We could further simplify that, but I'm going to leave it like that for now. Now let's find the measurement of B C. So B C equals the square root of and I might need a little more room from that A C scoot over. Alright, so B C we are taking X sub two R B is going to be R one R C is going to be our two. So our X sub two is our X sub C which is two and then minus our X sub one which is our X sub B so minus negative three and then closing those parentheses and squaring it. Adding that to now are Y sub two, R Y sub two is our Y sub C which is four and then minus our Y sub one, which is our Y sub B which is eight, closing those parentheses and squaring it now to simplify two minus negative three is the same as two plus three. Bringing down the rest. Now on the next line doing our addition and subtraction two plus three is five. So we have five still squared plus four minus eight, that's negative four and still squared. Now we can do the exponents, five squared is 25 plus negative four squared is positive 16. And now doing the addition 25-plus equals 41. So we have the square root of 41 for BC. Now let's do A C R A is going to be our first and C is going to be our second. So inside the radical, we have our X sub two which is going to be our X sub C which is two minus our X R X sub one, which is our X sub A. So that's negative nine and then closing those parentheses and squaring it plus our Y sub two, which is our Y sub C which is four minus our Why? Sub one, which is our y sub a which is 10, closing those parentheses and squaring it. Now we can do some simplification. Two minus negative nine is the same as two plus nine. And then bringing down the rest. Now doing the addition and subtraction in the parentheses, two plus nine equals 11. Still squared plus four minus 10 equals negative six still squared. Now doing our exponents, 11 squared is 121 and negative six squared is positive 36. Now doing the addition 121 plus equals 157. So this one is the square root of 157. Okay, we found the distances, the lengths for all three of those line segments. Now we need to figure out which two are the shortest. So I'm going to pop these in a calculator. The square to 40 that comes out to approximately 6.32. The square root of 41 is approximately 6.40 and the square root of is approximately 12.52. Now, we take the two shortest ones, the two shortest ones are going to be our 6.32 and our 6.40. We add our two shortest ones together 6.32 plus 6.40 And they should equal our longest one if these are all co linear points. So do they equal 12.52 while estimating they're very close. But actually doing the addition, they're not equal to each other. So actually these are not co linear points which is answer choice B well done. We'll catch you on the next one.

Determine whether the three points are collinear. (-7,4),(6,-2),(-1,1)

Key Concepts

Collinearity of Points

Slope Formula

Using Area to Test Collinearity

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