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Problem on co-ordinate geometry

  1. Sep 2, 2011 #1
    1. The problem statement, all variables and given/known data

    A(3,4) and [STRIKE]C(1,-2)[/STRIKE] C(1,-1) are the two opposite angular points of a square ABCD . Find the coordinates of the other two vertices .

    Here is the image : http://oi55.tinypic.com/2mq80i8.jpg

    2. Relevant equations

    Not sure what is most relevant . I tried applying distance formula , section formula and slopes but failed to crack the question .

    3. The attempt at a solution

    I tried applying distance formula of two diagonals and equated them but failed . Got four variable equation ! I tried slope formulas , but still failed .
    I tried combination of above got a very complicated quadratic equation which yielded wrong answer .

    Please help .

    (I am 14 years , 10th class )

    Thanks in advance .


    NOTE: The coordinates of C should be as shown in the image: C = (1,-1), not (1,-2)
     
    Last edited by a moderator: Sep 3, 2011
  2. jcsd
  3. Sep 2, 2011 #2

    Doc Al

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    Staff: Mentor

    Hints: Where's the center of the square? What's the slope of the given diagonal? What's the slope of the diagonal between the other two vertices? Write the equation for the diagonal.
     
  4. Sep 2, 2011 #3
    I have already tried these equations but equations become very complicated . Well thanks for replying Doc .

    Centre by section formula is

    (2, 3/2)

    Slope of diagonal AC : 5/2
    So of BD : -2/5
    Equation of AC : 2y-5x+7=0

    Now ?

    :confused:
     
  5. Sep 2, 2011 #4

    Doc Al

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    How far are the vertices from the center? Use that fact.
     
  6. Sep 2, 2011 #5
    Ok .

    Distance of each vertex from centre :

    sqrt(29/4)
    sqrt(29)/2

    Now ?
     
  7. Sep 2, 2011 #6

    Doc Al

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    Make use of that fact! Assume the vertices are at a point Δx, Δy away from the center. Use the slope and the distance to solve for Δx and Δy.
     
  8. Sep 2, 2011 #7
    Tally image post #1

    I got 1 equation :

    (b-3/2)/(a-2) = -2/5


    on applying m x m' = -1 of two vertices intersecting at 90 degrees .
     
  9. Sep 2, 2011 #8

    Doc Al

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    Good. Now get an equation for the distance to the center.
     
  10. Sep 2, 2011 #9
    Both the equations are coming same ie 2a + 5b = 23/2 !
     
  11. Sep 2, 2011 #10

    Doc Al

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    Show me your equation for the distance. This can't be it.
     
  12. Sep 2, 2011 #11
    (b-3/2)/(a-2) = -2/5 equation 1
    On solving you'll get ,
    Solution:
    4a+10b=23
    = 2a+5b=23/2

    Now

    1+25/4 + (a-2)2 + (b-3/2)2 = (4-b)2 + (3-a)2 equation 2

    On solving this you'll get ,
    2a+5b=23/2


    Now what can I do ?
     
  13. Sep 2, 2011 #12

    Doc Al

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    Not sure what you're doing here. The square of the distance to the center is given by:
    (a-2)2 + (b-3/2)2

    Set that equal to the square of the distance you found before.
     
  14. Sep 2, 2011 #13

    (a-2)2 + (b-3/2)2 = 29/4 ?


    a2-4 a+b2-3 b-1 = 0

    It is giving a degree 2 equation .
     
  15. Sep 2, 2011 #14

    Doc Al

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    OK. Now combine it with your other equation.
     
  16. Sep 2, 2011 #15

    verty

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    Problems like this become much easier if you plan how to tackle them. You should almost already know how to get the answer before you write the first equation. This is like in (classical) geometry when you have to first find the answer before you can write the proof.
     
  17. Sep 3, 2011 #16

    2a + 5b = 23/2
    => 4a + 10b = 23
    => a= (23 - 10b) / 4

    a2-4a+b2-3b-1 = 0

    Ok so on combining
    ((23 - 10b) / 4)2-4((23 - 10b) / 4)+b2-3b-1 = 0

    A very complicated quadratic equation indeed .

    Let me try solving it .

    29\16 (4 b^2-12 b+5) = 0
    b = 1\2
    so b = 1/2
    and d = 5/2

    In the same time ,
    a = 9/2
    C = -1/2

    Any easier method ?

    Instead of slope can we also do it like this ?
    Tally image post 1

    AB2 = BC2 ( where AB and BC are sides of square ) -> equation X

    So we get one equation in form of a and b .

    Now we can get

    AC2 ( diagonal ) = 29

    Now we put equation of AC2

    (3-a)2 + (4-b)2 + (a-1)2 + (b+1)2 = 29

    =2 a2-8 a+2 b2-6 b+27 = 29 --> equation Y

    Now we combine equation X and Y ??

    Well , anyways I got correct answer so thanks a lot for your help .

    :)




    Do you know any easier method ?

    Please explain this also : http://in.answers.yahoo.com/question/index?qid=20070306031100AASHBBs

    I just know searched it on the net after finishing my question on advice of Doc .
     
    Last edited: Sep 3, 2011
  18. Sep 3, 2011 #17

    HallsofIvy

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    No, that's incorrect. The given points are A(3, 4) and C(1, -2). The midpoint (center of the square) is ((3+1)/2, (4- 2)/2= (2, 1).

    No, that's incorrect. The slope is (4-(-2))/(3- 1)= 6/2= 3.

    No, that's incorrect. The distance between A and C is [itex]\sqrt{(3- 1)^2+ (4-(-2))^2}= \sqrt{4+ 36}= \sqrt{40}= 2\sqrt{10}[/itex]. The distance from the center to each vertex is [itex]\sqrt{10}[/itex]
     
  19. Sep 3, 2011 #18

    Doc Al

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    Thanks, Halls. (I didn't even bother to check those results as I was focused more on the process than the answers. :redface:)

    sankalpmittal, my apologies for not checking your work more carefully. In any case, with Halls' corrections, please redo your analysis following the method I outlined.
     
    Last edited: Sep 3, 2011
  20. Sep 3, 2011 #19
    Umm Sir Halls , those two coordinates are (3,4) and (1,-1) .
    Please tally that image in post 1.

    Please check my post 16 .


    No you checked my work carefully . By mistake sir hallsofivy typed wrong coordinates .
    Please check my post 16 .

    Oops I wrote wrong question . Sorry . Tally image that is correct question itself in post 1 .

    Edit post 1 : In place of (1,-2) edit it to (1,-1)
    (I apologize HallOfIvy )
     
    Last edited: Sep 3, 2011
  21. Sep 3, 2011 #20

    Doc Al

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    That's the problem: In post #1 you wrote C = (1,-2), but it should have been what you had in the image, C = (1,-1). Right?
     
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