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Finding vertices given opposite vertices

  1. Mar 10, 2013 #1
    Hi everyone, is there a way of finding the other two vertices of a square on the coordinate system given that two opposite vertices are already there? I already attempted this problem with two methods: with Pythagoras and with circles. So, is there a general formula or not?
     
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  3. Mar 10, 2013 #2

    tiny-tim

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    hi eddybob123! :smile:

    hint: first find the centre

    then … :wink:
     
  4. Mar 10, 2013 #3
    I tried that too. I ended up with two equations with two variables including their squares, and I didn't know how to solve for them, even with the quadratic formula.
     
  5. Mar 10, 2013 #4

    tiny-tim

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    but once you've found the centre, all you need do is to find the line through it at right-angles to the given diagonal :confused:
     
  6. Mar 10, 2013 #5
    Wait, the center of what? And can you define what you mean more precisely, maybe provide a diagram? I think I have also tried that, but I might have overlooked it because it seemed so easy.
     
  7. Mar 10, 2013 #6

    tiny-tim

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    the center of the square (and of the given diagonal)!
    erm, i've a better idea :redface:

    you define what you mean more precisely, and maybe provide a diagram :wink:
     
  8. Mar 10, 2013 #7
    The two opposite vertices are labeled (a,b) and (c,d). The points labeled (e,f) and (g,h) are the points I wish to find (in terms of a, b, c, and d). Can you just give me your approach to this problem?
     

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  9. Mar 10, 2013 #8

    tiny-tim

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    ok, then the centre dot is at ( , ) ?

    and the slope of the second diagonal is ? :smile:
     
  10. Mar 11, 2013 #9
    The center is at the midpoint of the line connecting (a,b) and (c,d), that is, by the distance formula, (sqrt((c-a)2+(d-b)2))/2. The slope of the perpendicular line is (a-c)/(d-b). As I told you, I have attempted this method before, but the equations I got were terribly complicated to solve for.
     
  11. Mar 11, 2013 #10

    tiny-tim

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    so what are its coordinates? :smile:
     
  12. Mar 11, 2013 #11
    I am pretty sure it is ((c+a)/2,(d+b)/2), but how does that concern finding the other vertices?
     
  13. Mar 12, 2013 #12

    Curious3141

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    Have you learnt complex numbers? I find that the problem becomes almost trivially easy when you use these.

    You can also use vectors, but I just find the problem easier to do with complex numbers. They're basically equivalent anyway.

    Using coordinate geometry is quite the slog, though.

    If you're familiar with complex numbers, I can take you through the steps. So let us know.
     
  14. Mar 12, 2013 #13

    tiny-tim

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    (just got up :zzz:)
    that's right! :smile:

    isn't it obvious? the centre is the average of the two endpoints, so it must be half of one and half of the other

    here's the start of the proof for the position of the centre C of AB, for you to finish …

    vector AC = vector CB, so C - A = B - C, so C = … ? :smile:

    ok, now if the vector OP has coordinates (p,q), what are the coordinates of the perpendicular vector of the same length?
     
  15. Mar 12, 2013 #14

    HallsofIvy

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    Here's another way: the line through (a, b) and (c, d) has slope (d- b)/(c- a) so the line perpendicular to that has slope (a- c)/(d- b) (-1 times the reciprocal). The line with that slope through the center point, ((a+c)/2, (b+ d)/2), is given by y= (a-c)/(d-b)(x- (a+c)/2)+ (b+d)/2. The circle with center at ((a+c)/2, (b+d)/2) and radius [itex]\sqrt{(a-c)^2+(d- b)^2}/2[/itex], and so passing through all four vertices of the square, is given by [itex](x- (a+c)/2)^2+ (y- (b+ d)/2)^2= ((a- c)^2+ (d- b)^2)/4[/itex]

    If you replace y in that last equation with (a-c)/(b-d)(x- (a+c)/2)+ (b+d)/2, you get a quadratic equation in x to solve for the x coordinates of the other two vertices.
     
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