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Math Help - Find cordinates of the point of intersection.

  1. #1
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    Find cordinates of the point of intersection.

    Please help me find the cordinates of the point of the intersection of the medians.





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    Length A right triangle is formed in the first quadrant by the x- and y-axes and the line through the point (3,2). write the length L of the hypotenuse as a function of x.

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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by bilalownsu View Post
    Please help me find the cordinates of the point of the intersection of the medians.

    one median is the line segment joining (b,c) and the midpt of (-a,0) and (a,0) which is (0,0).. so, get the equation on the line joining (b,c) and (0,0)..

    another median is the line segment joining (a,0) and the midpt of (-a,0) and (b,c).. get the midpt first then get the equation of the line joining that midpt and (a,0)..

    so you have two linear equations, just get their intersection..


    Quote Originally Posted by bilalownsu View Post
    Length A right triangle is formed in the first quadrant by the x- and y-axes and the line through the point (3,2). write the length L of the hypotenuse as a function of x.

    so, L=\sqrt{x^2+y^2}

    get the equation of the line joining (x,0) and (3,2): Y = \frac{2}{3-x}(X-x).. since (0,y) is in this line, it must be satisfied.. thus y = \frac{2}{3-x}(-x) = \frac{2x}{x-3} (we just sabstitute 0 to X and y to Y)

    therefore, the length of the hypotenuse as a function of L is L(x) = \sqrt{x^2+\left(\frac{2x}{x-3}\right)^2}
    Last edited by kalagota; August 29th 2008 at 05:36 AM.
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  3. #3
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    Hello, bilalownsu!

    A right triangle is formed in the first quadrant by the x- and y-axes
    and the line through the point P(3,2).
    Write the length L of the hypotenuse as a function of its x-intercept.
    Code:
          |
          |B
    (0,b) o
          |   * (3,2) 
          |       o
          |       P   *
          |               *   A
      - - + - - - - - - - - - o - -
          |                 (a,0)

    The line has intercepts: A(a,0)\,\text{ and }\,B(0,b)
    . . Note that: a > 3

    The slope of the line through P and A is: . m \:=\:\frac{-2}{a-3}


    The equation of the line through A with slope m is:

    . . y \:=\:\frac{-2}{a-3}(x-a) \quad\Rightarrow\quad y \:=\:\frac{-2}{a-3}x + \frac{2a}{a-3}


    Hence: line L has intercepts: . (a, 0)\,\text{ and }\,\left(0,\frac{2a}{a-3}\right)


    Therefore, the length of L is: . \sqrt{a^2 + \left(\frac{2a}{a-3}\right)^2} \;=\;\boxed{\frac{a\sqrt{a^2-6a+13}}{a-3}}

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