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VERY SIMPLE Coordinate Geometry

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Using the concept of equal slopes, state whether the following sets of points are collinear or not
    A(5, -6) B(0, -1) C(-4, 3)


    2. Relevant equations
    Slope = vertical rise/horizontal run


    3. The attempt at a solution
    I only got as far as...
    5-0/-6+1 = 5/-5
    I think what I just did was already incorrect. I actually have a few of these questions but I'm sure that if someone points me in the right direction with this one, I can figure out the rest.

    Thanks in advance. (yes, I know this is very basic maths)
     
  2. jcsd
  3. Mar 10, 2010 #2

    HallsofIvy

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    First, please use parentheses. What you wrote is really 5- 0+ 1= 6! You mean, of course, (5-0)/(-6+1)= 5/(-5)= -1.

    But also important is to say why you are doing calculations. I think you trying to do is calculate the slope of the line from (5, -6) to (0, -1). And yes, you are doing it wrong. Slope is defined as "change in y divided by change in x". Going from (0, -1) to (5, -6), y changes from -1 to -6 so the change is -6-(-1)= -5 and that is the numerator of the fraction. x changes from 0 to 5 so the change is 5- 0= 5 and that is the denominator of the fraction. The slope is (-5)/5= -1. (That is the same as before but only because this is a special case.)

    Now, what is the slope of the line from (0, -1) to (-4, 3)? What does that tell you?
     
  4. Mar 10, 2010 #3

    tiny-tim

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    Hi BL4CKCR4Y0NS! :smile:
    No, it's fine … the slope is -1.

    Now do B and C the same way. :smile:

    (remember, if you want to check your answer, you can easily draw it on a graph! :wink:)
     
  5. Mar 10, 2010 #4
    First you have to think, how you can use slope's to find out that lines are collinear. You have the concept that two lines are parallel, if they have their slope equal, now find out the slope of the line AB and BC, if you'll get the slope equal that means that AB and BC are parallel lines, but are they really parallel? They have 'B' point common, so they are not parallel lines, so only thing we can say about it is that they are collinear means AB and BC are the same line. Now to find the slope you have to use equation, [tex]\frac{y_2 - y_1}{x_2 - x_1}[/tex]
     
    Last edited by a moderator: Mar 10, 2010
  6. Mar 10, 2010 #5
    I messed up with latex code, so can anybody here can correct it.
     
  7. Mar 10, 2010 #6

    HallsofIvy

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    Yes, but only because 1/(-1)= -1. He actually calculated the reciprocal of the slope.

     
  8. Mar 10, 2010 #7
    Thanks HallsofIvy for correcting my latex code. I know that it isn't a right place to ask this question but I can't find any better place to ask. Can you give me the link to the page where the instructions to properly write latex code is given. I've seen it once but now I've lost that link and unable to find it using search functionality.
     
  9. Mar 10, 2010 #8

    tiny-tim

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    ah, but the question didn't ask for the slope …

    so long as he uses the same method for each line, the procedure is perfectly valid :smile:

    (and snshusat161, a good place to start is http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000" [Broken] :wink:)
     
    Last edited by a moderator: May 4, 2017
  10. Mar 10, 2010 #9
    I did what tiny-tim said, and I ended up with 5/(-4)
    But I'm not sure if I'm correct. What I did was (just in case I got it wrong and you needed working out to see where I went wrong)

    5/(-5) = 1/-1
    (1+4)/(-1)-3
    = 5/(-4)

    Thanks for all the replies guys... I appreciate it. =]
     
  11. Mar 10, 2010 #10
    What did you mean by 5/-4. Is it collinear or not. What i told earlier use it.
     
  12. Mar 10, 2010 #11
    Okay, I did as you said and...
    A and B came back with -1
    B and C came back with -1

    I'm not sure what I am supposed to tell from that ... does "-1" on both mean that they are collinear?
     
  13. Mar 11, 2010 #12
    no, it means that both are parallel but do you ever heard parallel lines intersecting at any point, therefore both lines AB and BC are not parallel but the same line and since points A, B and C lies on the same line they are collinear.

    And "-1" doesn't mean that they are collinear. Any number will do well if you get it for both AB and BC.
     
  14. Mar 11, 2010 #13

    tiny-tim

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    Yes, "-1" on both means they have the same slope, and since they also share a point, they must be on the same line. :smile:

    (incidentally, of course, if you check A and C, you should find the same :wink:)
     
  15. Mar 11, 2010 #14
    Thanks tiny-tim =D
    I didn't actually understand snshusat161's last post and was a little afraid to ask again because I'm so intellectually inferior and probably seem annoying... >_>

    Thanks again guys. I got the rest of the questions =D
     
  16. Mar 11, 2010 #15
    I wanted to clarify in my last post that equal slope doesn't mean that two lines are collinear but if they have some common point then they are collinear.
     
  17. Mar 11, 2010 #16
    Or simply use this equation:

    [tex]\frac{y_{2} - y_{1}}{x_{2} - x_{1}}[/tex] = [tex]\frac{y_{3} - y_{2}}{x_{3} - x_{2}}[/tex]
     
  18. Mar 11, 2010 #17

    Mentallic

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    There's no point in using the equation if the OP doesn't understand how and why it's being used.
     
  19. Mar 11, 2010 #18
    So the common point in this case would be the "-1" ?
     
  20. Mar 11, 2010 #19
    No, I think you are totally confused and you don't know how we represent any point in coordinate geometry. Let me search some good tutorial page for you.
     
  21. Mar 11, 2010 #20

    Mentallic

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    You need to go back and study coordinate geometry from the very start.
    Simply put, as snshusat161 has said, you seem to be complete oblivious as to how coordinate geometry works and putting this added pressure of trying to solve problems is completely pointless to you.
     
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