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Homework Help: Find the equation

  1. Dec 29, 2011 #1
    Find the equation of the straight line joining points (-1,1) and (2,5)



    m = y1 - y2 / x1 - x2



    5 - 1 / 2 - (-1) = 4/3
    y - 1 = 4/3( x + 1 )
    y = 4/3x + 4/3 + 1
    y = 4/3x + 5/3< --- this is my attempt solution but I think is wrong! because on the back of my maths book the answer is 3y = 4x + 7 so I don't know what formula I have to use.
     
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  3. Dec 29, 2011 #2

    HallsofIvy

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    I think you mean (y1- y2)/(x1- x2). What you wrote is y1- (y2/x1)- x2.



    What do you get if you multiply both sides of y= (4/3)x+ 5/3 by 3?
     
  4. Dec 29, 2011 #3
    Why do I have to multiply by 3 ?? I dont remember exactly how to do that somebody teach me that like 3 years ago but is like y - 1 = 4/3 ( x + 1) and I think I have to multiply by 3 each number so I can get rid of the denominator, but I not sure ... y - 1 = 12 ( 3x + 3) <--- like that?
     
    Last edited: Dec 29, 2011
  5. Dec 29, 2011 #4
    Any ideas?
     
  6. Dec 29, 2011 #5

    SammyS

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    You have made an error in adding [itex]\displaystyle \frac{4}{3}+1\,.[/itex]

    The number one is how many thirds?
     
  7. Dec 29, 2011 #6
    Oh yes! 4/3 + 1/1 (3) = 4/3 + 3/3 = 7/3 ... But I still having the wrong answer because the answer is 3y = 4x + 7.. and I figure out only the slope formula ( y = mx + b )
     
  8. Dec 29, 2011 #7

    SammyS

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    Two ways to see that the following two equations are equivalent:
    [itex]\displaystyle y=\frac{4}{3}x+\frac{7}{3}\quad\quad \text{(Equation 1)}[/itex]

    [itex]\displaystyle 3y=4x+7\quad\quad\quad\text{(Equation 2)}[/itex]​

    One way: (You've been told this before.) Multiply both sides of Equation 1 by 3 . Remember, use the Distributive Law

    Second way: Divide [STRIKE]Multiply[/STRIKE] both sides of Equation 2 by 3 .
    (Mod note: removed extraneous operation above.)
     
    Last edited by a moderator: Dec 29, 2011
  9. Dec 30, 2011 #8
    luigihs,

    it should be 4/3 + 1/1(3/3) = 4/3 + 3/3 = 7/3, or


    [itex]\dfrac{4}{3} \ = \ \dfrac{1}{1}\bigg(\dfrac{3}{3}\bigg) \ = \ \dfrac{4}{3} + \dfrac{3}{3} \ = \dfrac{7}{3}[/itex]



    What you had is equivalent to [itex]\dfrac{4}{3} + 3.[/itex]
     
  10. Dec 30, 2011 #9

    Redbelly98

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    In case it wasn't clear from what others have said: you have the correct answer, it is equivalent to the answer given in the book.
     
  11. Dec 30, 2011 #10

    Mark44

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    The work above has a typo (marked). Here is the correction:
    [itex]\dfrac{4}{3} \ + \ \dfrac{1}{1}\bigg(\dfrac{3}{3}\bigg) \ = \ \dfrac{4}{3} + \dfrac{3}{3} \ = \dfrac{7}{3}[/itex]
     
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