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Math Help - Orthocenter...what is it with this triangle

  1. #1
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    Orthocenter...what is it with this triangle

    Whats the orthocenter of a triangle with coordinates R (-1 , -4) , P (-1 , 5) , Q (7 , 2).

    Please show how you did it as simple as possible

    Thanks
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  2. #2
    Junior Member lanzailan's Avatar
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    Quote Originally Posted by squat4speed View Post
    Whats the orthocenter of a triangle with coordinates R (-1 , -4) , P (-1 , 5) , Q (7 , 2).

    Please show how you did it as simple as possible

    Thanks
    i think you need to find the midpoint of the 2 sides of the triangle. from these new points you will come up with 2 equation (linear equation) and find the coordinates of the point of intersection. that would be your orthocenter..
    i hope this is correct...
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  3. #3
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    how do i find coordinates of point of intersection
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  4. #4
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    Alright I got a head ache..and im confusing myself somehow, can someone please give me a step by step show of how you find the coordinates of the orthocenter.
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  5. #5
    Junior Member lanzailan's Avatar
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    Quote Originally Posted by squat4speed View Post
    how do i find coordinates of point of intersection
    when you got 2 equation what you need to do is solving using simultaneous or substitution method.

    first draw your triangle.
    ok i got midpoint of RQ=(3,-1) <<---let this point be point A
    and PR=(-1,1/2)<<-------- this point be point B

    now find the equation of line that passes through point B and point Q (first equation)
    and equation of line that passes through A and P(second equation).
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  6. #6
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    Hello, squat4speed!

    What's the orthocenter of a triangle with coordinates: R (-1 , -4) ,\:P (-1 , 5) ,\: Q (7 , 2).

    The orthocenter is the intersection of the three altitudes of the triangle.

    Find the equation of two of the altitudes and locate their intersection.

    If you make a sketch, the problem is much simpler.
    Code:
              P   |
        (-1,5)*   |
              * * |
              *   *
              *   | *
              *   |   *
              *   |     *
          - - * - + - - - *Q
              *   |     * (7,2)
              *   |   *
        ------*---+-*----------
              *   *
              * * |
       (-1,-4)*   |
              R   |
                  |

    Side PR is vertical.
    Hence, the altitude from Q is the horizontal line, y = 2. .[1]

    Side QR has slope: . m \:=\:\frac{2-(\text{-}4)}{7-(\text{-}1)} \:=\:\frac{3}{4}

    Hence, the altitude from P(\text{-}1,5) has slope \text{-}\tfrac{4}{3}
    Its equation is: . y - 5 \:=\:\text{-}\tfrac{4}{3}(x+1) \quad\Rightarrow\quad y \:=\:\text{-}\tfrac{4}{3}x + \tfrac{11}{3} .[2]

    The intersection of [1] and [2]: . 2 \:=\:\text{-}\tfrac{4}{3}x + \tfrac{11}{3} \quad\Rightarrow\quad \tfrac{4}{3}x \:=\:\tfrac{5}{3}


    Therefore: . x \:=\:\tfrac{5}{4},\;y \;=\:2

    The orthocenter is: . \left(\tfrac{5}{4},\:2\right)

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  7. #7
    Junior Member lanzailan's Avatar
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    if you draw on the graph paper you will get (1.6,1)
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