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Math Help - Find Orthocenter of Triangle

  1. #1
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    Smile Find Orthocenter of Triangle

    I need to find the orthocenter of a triangle with coordinates:

    G(-2,5)
    H(6,5)
    J(4,-1)

    AND...

    A(4,-3)
    B(8,5)
    C(8,-8)


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  2. #2
    Lord of certain Rings
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    Let O(a,b) be the orthocenter of triangle GHJ. Since O is the orthocenter, OG is perpendicular to HJ and OH is perpendicular to GJ.

    Thus (slope of OG) x (slope of HJ) = -1. Also (slope of OH) x (slope of GJ) = -1. Those two slope equations will you give you two simultaneous equations in a and b. Solve them and get O.
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  3. #3
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    ok........

    Quote Originally Posted by Isomorphism View Post
    Let O(a,b) be the orthocenter of triangle GHJ. Since O is the orthocenter, OG is perpendicular to HJ and OH is perpendicular to GJ.

    Thus (slope of OG) x (slope of HJ) = -1. Also (slope of OH) x (slope of GJ) = -1. Those two slope equations will you give you two simultaneous equations in a and b. Solve them and get O.
    Thanks for the tips.
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  4. #4
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    Hello, magentarita!

    The orthocenter is the intersection of the altitudes.
    Did you make a sketch ??


    Find the orthocenter of a triangle with vertices:
    G(-2,5), H(6,5), J(4,-1)
    Code:
                |
      (-2,5)    |          (6,5)
        A o  *  *  *  *  *  o B
            *   |     :    *
              * |     :   *
                *     :  *
        - - - - + * - : * - - - - - -
                |   * :*
                |     o C
                |  (4,-1)
                |

    Recall that an altitude is perpendicular to the opposite side.

    Since AB is horizontal,
    . . we see the altitude from C is the vertical line x = 4

    The slope of AC is: . m_1 \:=\:\frac{\text{-}1-5}{4-(\text{-}2)} \:=\:-1

    Hence, the altitude from B has slope: . m_2 \:=\:1

    The equation of altitude from B(6,5) with slope 1
    . . y - 5 \:=\:1(x-6) \quad\Rightarrow\quad y \:=\:x-1

    The intersection of this altitude with the vertical line x = 4

    . . is: . \begin{Bmatrix}y \:=\:x-1 \\ x \:=\:4\end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}x \:=\:4\\y\:=\:3 \end{Bmatrix}


    Therefore, the orthocenter is (4,3)

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  5. #5
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    great...

    Quote Originally Posted by Soroban View Post
    Hello, magentarita!

    The orthocenter is the intersection of the altitudes.
    Did you make a sketch ??

    Code:
                |
      (-2,5)    |          (6,5)
        A o  *  *  *  *  *  o B
            *   |     :    *
              * |     :   *
                *     :  *
        - - - - + * - : * - - - - - -
                |   * :*
                |     o C
                |  (4,-1)
                |
    Recall that an altitude is perpendicular to the opposite side.

    Since AB is horizontal,
    . . we see the altitude from C is the vertical line x = 4

    The slope of AC is: . m_1 \:=\:\frac{\text{-}1-5}{4-(\text{-}2)} \:=\:-1

    Hence, the altitude from B has slope: . m_2 \:=\:1

    The equation of altitude from B(6,5) with slope 1
    . . y - 5 \:=\:1(x-6) \quad\Rightarrow\quad y \:=\:x-1

    The intersection of this altitude with the vertical line x = 4

    . . is: . \begin{Bmatrix}y \:=\:x-1 \\ x \:=\:4\end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}x \:=\:4\\y\:=\:3 \end{Bmatrix}


    Therefore, the orthocenter is (4,3)
    Another great job. Happy New Year, Soroban.
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