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Math Help - An analytic Geometry Riddle about Locus.

  1. #1
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    An analytic Geometry Riddle about Locus.

    OC starts from (0,0) AB is a line segment that has a midpoint C(because OC is touching it at C) Slope of OC is Tan60.Coordinates of A is (SQRT3,1)Find the equation of locus of point B.
    Last edited by serhanbener; March 23rd 2012 at 10:57 AM.
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  2. #2
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    Re: An analytic Geometry Riddle about Locus.

    Quote Originally Posted by serhanbener View Post
    OC starts from (0,0) AB is a line segment that has a midpoint C(because OC is touching it at C) Slope of OC is Tan60.Coordinates of A is (SQRT3,1)Find the equation of locus of point B.
    1. The point C is located on the straight line l:

    l:\vec c = \langle 1, \sqrt{3} \rangle \cdot t

    2. Let \vec b = \langle b_1, b_2 \rangle denotes the stationary vector of point B. C is the midpoint of \overline{AB}.

    Therefore:

    \displaystyle{\frac{\vec b + \vec a}2 = \vec c} with \vec a = \langle \sqrt{3} , 1 \rangle

    3. That means:

    \left| \begin{array}{rcl}\frac{b_1+\sqrt{3}}2 &=& t \\ \frac{b_2+1}2 &=& \sqrt{3} \cdot t \end{array} \right. .... \implies .... \left| \begin{array}{rcl} b_1 &=& 2t - \sqrt{3} \\ b_2 &=& 2 \sqrt{3} \cdot t - 1 \end{array} \right.

    4. The stationary vector of point B is:

    \vec b = \langle 2t - \sqrt{3}, 2 \sqrt{3} \cdot t - 1 \rangle~\implies~ \vec b = 2 \langle 1, \sqrt{3} \rangle \cdot t - \langle \sqrt{3} , 1 \rangle

    This is the equation of a parallel of line c.
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    Re: An analytic Geometry Riddle about Locus.

    Many Thanks for your answer. I managed to solve it like this:
    A(SQRT3,1) is given. B is (x,SQRT3) and C is C(A,B).I applied the
    midpoint formula.
    First of all I found A as 2x- SQRT3 and B=2SQRT3x-1.I replaced A in B.
    B=(A+SQRT3)SQRT3-1.Then B=SQRT3A+2.If we replace A and B with X and Y,
    it becomes y= SQRT3x+2
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    Re: An analytic Geometry Riddle about Locus.

    Quote Originally Posted by serhanbener View Post
    Many Thanks for your answer. I managed to solve it like this:
    A(SQRT3,1) is given. B is (x,SQRT3) and C is C(A,B).I applied the
    midpoint formula.
    First of all I found A as 2x- SQRT3 and B=2SQRT3x-1.I replaced A in B.
    B=(A+SQRT3)SQRT3-1.Then B=SQRT3A+2.If we replace A and B with X and Y,
    it becomes y= SQRT3x+2
    1. That's nearly perfect
    (You must use brackets so that the equation is unambiguous: y= SQRT(3) *x + 2)

    2. I was trapped by your headline: Analytic Geometry means in German vector geometry.
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    Re: An analytic Geometry Riddle about Locus.

    Vectors are a different subject of analytic geometry in Turkey. Analytic Geometry subjects are: Analytic geometry:Lines ,tangents,symmetry vs..,Analytic of the circle,Vectors,Parabolia,Hyperbolia and Elips. Thanks again for you help.
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