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Math Help - Orthocenter

  1. #1
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    Orthocenter

    Heres the problem : Given triangle RST, with R= (-3,2), S=(4,5) , T= (7,-2), find the cords of the orthocenter.

    I have no idea what to do, my teacher gave this to us and im confused..

    thanks (:
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  2. #2
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    Quote Originally Posted by tehc0ffee View Post
    Heres the problem : Given triangle RST, with R= (-3,2), S=(4,5) , T= (7,-2), find the cords of the orthocenter.

    I have no idea what to do, my teacher gave this to us and im confused..

    thanks (:
    Woah, I am on the exact same problem right now.

    Page 664, no? O_o

    Find the length of each side, and then find the altitudes to those sides from the vertex of the opposite sides. The point where the altitudes concur is the orthocenter.
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  3. #3
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    LOL.

    Yes, page 644 (: thanks a lot :P
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    664***
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  5. #5
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    Hello,

    Quote Originally Posted by tehc0ffee View Post
    Heres the problem : Given triangle RST, with R= (-3,2), S=(4,5) , T= (7,-2), find the cords of the orthocenter.

    I have no idea what to do, my teacher gave this to us and im confused..

    thanks (:
    The orthocenter is the intersection between two of the altitudes in the triangle.
    So the thing will be to find the equations of the lines representing the altitudes.
    For example, we'll look for the altitudes coming from R (to ST) and S (to RT).

    1st step
    Equations of the lines ST and RT.

    We know that to a line of equation ax+by+c=0 corresponds a direction vector (-b,a) and an orthogonal vector (a,b).


    -------------------
    \vec{ST}~:~(x_T-x_S ~,~y_T-y_S)=(3~,~-7).
    So here, -b=3 and a=-7.
    Therefore, according to the red part, the equation of line ST is : -7x-3y+c=0.

    But we're interested in the orthogonal line to ST, passing through R.
    Its vector director is, according to the red part above : (-7,-3).
    So here again, to find out the equation of this line, we use the red part, with -b=-7 (b=7) and a=-3

    The equation of a line orthogonal to ST is : -3x+7y+c=0.

    i do these steps because I can never remember the equations...


    We want this line passing through R(-3,2).
    By substituting in the equation :

    -3(-3)+7(2)+c=0 \implies c=-14


    Therefore, the equation of the altitude coming from R to ST is :
    \boxed{-3x+7y-14=0}

    -------------------

    Now, for RT, we work the same way :

    \vec{RT}~:~(10~,~-4)

    Equation of line RT : -4x-10y+c=0 (thanks to the red part)

    Vector director of a line orthogonal to RT : (-4~,~-10) (thanks to the blue part)

    Equation of a line orthogonal to RT : -10x+4y+c=0 (thanks to the red part)

    We know the altitude to RT goes through S(4,5).
    By substituting :

    -10(4)+4(5)+c=0 \implies c=20


    Therefore, the equation of the altitude coming from S to RT is :
    \boxed{-10x+4y+20=0}

    ~~~~~~~~~~~~~~~~~~~~~~
    2nd sted
    H is on these two altitudes.

    So solve the system :

    \left\{ \begin{array}{ccc} -3x_H+7y_H-14&=&0 \\ -10x_H+4y_H+20&=&0 \end{array} \right.


    -----------------------
    If there are any mistakes, don't hesitate ! I'm likely to have made typos, but the reasoning may be correct.
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