Perpendicular Segments

Discussion in 'Algebra' started by nycmathguy, Jun 28, 2021.

  1. nycmathguy

    nycmathguy

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    See attachment.

    Let me see.

    To find d_1 & d_2, use the distance formula.
    I then must use the distance again to find the length of the hypotenuse. Finally, express my lengths in terms of the Pythagorean Theorem.

    Yes?

    20210628_092208.jpg
     
    nycmathguy, Jun 28, 2021
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  2. nycmathguy

    MathLover1

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    d1=sqrt((1-0)^2+(m1-0)^2)
    d1=sqrt(1+m1^2)



    d2=sqrt((1-0)^2+(m2-0)^2)
    d2=sqrt(1+m2^2)

    hypotenuse= (d1)^2+(d2)^2

    hypotenuse= (sqrt(1+m1^2))^2+(sqrt(1+m2^2))^2

    hypotenuse=1+m1^2+1+m2^2
    hypotenuse=2+m1^2+m2^2.............eq.1


    distance between endpoints of hypotenuse is

    hypotenuse=sqrt((1-1)^2+(m2-m1)^2)

    hypotenuse=sqrt(0+(m2-m1)^2) ............eq.2

    from eq.1 and eq.2 we have

    2+m1^2+m2^2=sqrt((m2-m1)^2)...square both sides

    (2+m1^2+m2^2)^2=(m2-m1)^2.......take sqrt of both sides


    2+m1^2+m2^2=m2-m1

    2+m1^2+m2^2-m2+m1 =0.....complete squares

    (m1 + 1/2)^2 + (m2 - 1/2)^2 + 3/2 = 0..........solve for m1

    (m1 + 1/2)^2 =- (m2 - 1/2)^2 - 3/2

    m1 + 1/2 =sqrt(- (m2 - 1/2)^2 - 3/2)

    m1 + 1/2 =(1/2) sqrt(-4 (m2 - 1) m2 - 7)

    m1 =(1/2) sqrt(-4 (m2 - 1) m2 - 7)-1/2
     
    MathLover1, Jun 28, 2021
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  3. nycmathguy

    nycmathguy

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    Great math work as always. What about my reasoning for this thread? Am I wrong in my discussion?
     
    nycmathguy, Jun 28, 2021
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  4. nycmathguy

    MathLover1

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    your reasoning for this thread is perfect
     
    MathLover1, Jun 28, 2021
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  5. nycmathguy

    nycmathguy

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    Thanks. More questions tomorrow. By the way, I think we are the only two members actively involved here. What do you think? Good site? Not so good?
     
    nycmathguy, Jun 28, 2021
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