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Co-ordinate geometry

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DevB Rising GMAT Star Default Avatar
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Co-ordinate geometry Post Fri Jul 25, 2014 10:01 am
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    Hi All,

    Please help in answering the attached question from GMAT Prep.

    Answer options are:

    1. 1/2
    2. 1
    3. square root 2
    4. square root 3
    5. (square root 2)/ 2
    Thanks!
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    abhasjha GMAT Destroyer! Default Avatar
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    Post Fri Jul 25, 2014 10:22 am
    slope of PO = a/b
    slope of QO = t/s
    PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
    In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1

    Alternatively

    from the figure we can see that OP=OQ = radius of the semicircle. Also Angle POQ = 90degree

    thus PQ^2 = OP^2 + OQ^2-------eqn1

    using distance formula (O is origin)
    OP = sqrt [(1-0)^2 + (sqrt -3 -0)^2] = sqrt [4] = 2 = OQ

    also OQ = sqrt [ (s-0)^2 + (t-0)^2] = sqrt[s^2 + t^2]
    OQ^2 = s^2+t^2 =4 ----eqn2

    now PQ = 2sqrt 2 when we put values of OP and OQ in eqn 1

    using distance formula length of PQ = sqrt [ (s+ sqrt3)^2 + (t-1)^2]
    PQ^2 = (s+sqrt3)^2 + (t-1)^2 = 8--------eqn3

    soving eqn 2 and 3 we get sqrt 3 s = t.Equating this in eqn 2 we get s=1

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    DevB Rising GMAT Star Default Avatar
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    Post Fri Jul 25, 2014 10:35 am
    Thanks a lot Abhas for the both the explanations!! Smile

    Post Fri Jul 25, 2014 10:38 am
    Quote:


    In the figure above, points P and Q lie on the circle with center O. What is the value of S?

    a) 1/2
    b) 1
    c) √2
    d) √3
    e) (√2)/2
    Here's one approach:


    So, s = 1
    Answer: B

    Cheers,
    Brent

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    DevB Rising GMAT Star Default Avatar
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    Post Fri Jul 25, 2014 10:42 am
    This is also a very nice approach Brent...Thanks a lot!!

    Post Fri Jul 25, 2014 11:42 am
    Hi DevB,

    When dealing with geometry/graphing questions, there are certain patterns you should be on the lookout for:

    1) A diagonal line on a graph IS the hypotenuse of a right triangle. You can draw that triangle and figure out its base and height. From there, you have a variety of triangle formulas that might be helpful involving the question.

    2) The GMAT has a variety of standard formulas and patterns that it will test you on. When you come across a triangle, chances are it fits some type of pattern. Look to use the Pythagorean Theorem, and look for the special right triangles (30/60/90, 45/45/90, 3/4/5, 5/12/13).

    3) Weird numbers almost always point to a pattern. Notice the (root3) in this question? I wonder what that might refer to in a right triangle….?

    GMAT assassins aren't born, they're made,
    Rich

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    Gurpreet singh Rising GMAT Star Default Avatar
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    Post Tue Jun 07, 2016 5:21 pm
    Please explain this part

    In this case, b = √3, a = -1



    abhasjha wrote:
    slope of PO = a/b
    slope of QO = t/s
    PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
    In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1

    Alternatively

    from the figure we can see that OP=OQ = radius of the semicircle. Also Angle POQ = 90degree

    thus PQ^2 = OP^2 + OQ^2-------eqn1

    using distance formula (O is origin)
    OP = sqrt [(1-0)^2 + (sqrt -3 -0)^2] = sqrt [4] = 2 = OQ

    also OQ = sqrt [ (s-0)^2 + (t-0)^2] = sqrt[s^2 + t^2]
    OQ^2 = s^2+t^2 =4 ----eqn2

    now PQ = 2sqrt 2 when we put values of OP and OQ in eqn 1

    using distance formula length of PQ = sqrt [ (s+ sqrt3)^2 + (t-1)^2]
    PQ^2 = (s+sqrt3)^2 + (t-1)^2 = 8--------eqn3

    soving eqn 2 and 3 we get sqrt 3 s = t.Equating this in eqn 2 we get s=1

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    Post Tue Jun 07, 2016 11:13 pm
    Gurpreet singh wrote:
    Please explain this part

    In this case, b = √3, a = -1
    He's using similar triangles here. Each triangle has sides of length 1 and √3, which, applied to coordinate plane, gives (-√3, 1) for the first and (1, √3) for the second.

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