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Math Help - Analytic Geometry Help!

  1. #1
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    Analytic Geometry Help!

    Show that (1/3 [2x1 + x2], 1/3[2y1+y2]) is one of the points of trisection of the line segment joining (x1,y1) and (x2,y2). Then find the second point of trisection by finding the midpoint of the segment joining

    (1/3[2x1 + x2], 1/3[2y1 + y2]) and (x2, y2)

    Use exercise 28 to find the points of trisection of the line segment joining the given points.

    a (1 , -2) (4, -1)
    b. (-2, -3) (0,0)
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  2. #2
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    Quote Originally Posted by Nimmy View Post
    Show that (1/3 [2x1 + x2], 1/3[2y1+y2]) is one of the points of trisection of the line segment joining (x1,y1) and (x2,y2). Then find the second point of trisection by finding the midpoint of the segment joining

    (1/3[2x1 + x2], 1/3[2y1 + y2]) and (x2, y2)

    Use exercise 28 to find the points of trisection of the line segment joining the given points.

    a (1 , -2) (4, -1)
    b. (-2, -3) (0,0)
    To do the first part I would imagine you need to show that the distance between (x_1, y_1) and \left ( \frac{1}{3}(2x_1 + x_2), \frac{1}{3}(2y_1 + y_2) \right ) is 1/3 the distance between (x_1, y_1) and (x_2, y_2).

    So the distance between (x_1, y_1) and \left ( \frac{1}{3}(2x_1 + x_2), \frac{1}{3}(2y_1 + y_2) \right ) is:
    d = \sqrt{\left ( x_1 - \frac{1}{3}(2x_1 + x_2) \right ) ^2 + \left ( y_1 - \frac{1}{3}(2y_1 + y_2) \right ) ^2 }

    d = \sqrt{\left ( \frac{1}{3}x_1 - \frac{1}{3}x_2 \right ) ^2 + \left ( \frac{1}{3}y_1 - \frac{1}{3}y_2 \right ) ^2 }

    d = \sqrt{\frac{1}{9}(x_1 - x_2)^2 + \frac{1}{9}(y_1 - y_2)^2 }

    d = \sqrt{\frac{1}{9}}\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 }

    d = \frac{1}{9} \cdot \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 }

    Note that \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 } is the distance between (x_1, y_1) and (x_2, y_2) so we have proven what we have set out to show.

    For the second part we need to find the midpoint of the two points \left ( \frac{1}{3}(2x_1 + x_2), \frac{1}{3}(2y_1 + y_2) \right ) and (x_2, y_2).

    M \left (\left ( \frac{1}{3}(2x_1 + x_2), \frac{1}{3}(2y_1 + y_2) \right ) , (x_2, y_2) \right ) = \left ( \frac{1}{2} \left [ \frac{1}{3}(2x_1 + x_2) + x_2 \right ] , \frac{1}{2} \left [ \frac{1}{3}(2y_1 + y_2) + y_2 \right ] \right )

     = \left ( \frac{1}{2} \left [ \frac{2}{3}x_1 + \frac{1}{3}x_2 + x_2 \right ] , \frac{1}{2} \left [ \frac{2}{3}y_1 + \frac{1}{3}y_2 + y_2 \right ] \right )

     = \left ( \frac{1}{2} \left [ \frac{2}{3}x_1 + \frac{4}{3}x_2 \right ] , \frac{1}{2} \left [ \frac{2}{3}y_1 + \frac{4}{3}y_2 \right ] \right )

     = \left ( \frac{1}{3}x_1 + \frac{2}{3}x_2 , \frac{1}{3}y_1 + \frac{2}{3}y_2 \right )

     = \left ( \frac{1}{3} \left [ x_1 + 2x_2 \right ] , \frac{1}{3} \left [ y_1 + 2y_2 \right ] \right )

    This is your second point of trisection.

    I'll do a) as an example. We are given the points (1 , -2) and (4, -1). The first point of trisection is:
    \left ( \frac{1}{3} \left [ 2x_1 + x_2 \right ] , \frac{1}{3} \left [ 2y_1 + y_2 \right ] \right )

     = \left ( \frac{1}{3} \left [ 2(1) + (4) \right ] , \frac{1}{3} \left [ 2(-2) + (-1) \right ] \right )

     = \left ( \frac{1}{3} \left [ 6 \right ] , \frac{1}{3} \left [ -5 \right ] \right )

     = \left ( 2 , -\frac{5}{3} \right )

    The second point of trisection is:
    \left ( \frac{1}{3} \left [ x_1 + 2x_2 \right ] , \frac{1}{3} \left [ y_1 + 2y_2 \right ] \right )

     = \left ( \frac{1}{3} \left [ (1) + 2(4) \right ] , \frac{1}{3} \left [ (-2) + 2(-1) \right ] \right )

     = \left ( \frac{1}{3} \left [ 9 \right ] , \frac{1}{3} \left [ -4 \right ] \right )

     = \left ( 3 , -\frac{4}{3} \right )

    -Dan
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  3. #3
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    Hello, Nimmy!

    Show that \left(\frac{1}{3}[2x_1 + x_2],\:\frac{1}{3}[2y_1+y_2]\right)
    is one of the points of trisection of the line segment joining P(x_1,y_1) and Q(x_2,y_2).

    Make a sketch . . .
    Code:
                                                    * Q(x2,y2)
                                                *   | :
                                            *       | :
                                        *           | :
                                    *               | y2 - y1
                            T   *                   | :
                            *                       | :
                        *   |                       | :
                    *       |                       | :
      (x1,y1) P *-----------*-----------------------* R(x2,y1)
                            U
              : - - - - - - - x2 - x1 - - - - - - - :

    We have: . \Delta PTU \sim \Delta PQR

    Since PT = \frac{1}{3}PQ, then: . \begin{array}{cc}PU \:=\:\frac{1}{3}PR\:=\:\frac{1}{3}(x_2-x_1) \\ TU \:= \:\frac{1}{3}QR\:=\:\frac{1}{3}(y_2-y_1)\end{array}

    The coordinates of T are: . \begin{array}{cc}x \:= \: x_1 + \frac{1}{3}(x_2-x_1) \: =\: \frac{1}{3}(2x_1 + x_2) \\<br />
y \:= \: y_1 + \frac{1}{3}(y_2-y_1) \: =\: \frac{1}{3}(2y_1+y_2)<br />
\end{array}

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