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Math Help - Check my work please?

  1. #1
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    Check my work please?

    I have this problem for orthocenter and circumcenter, but I'll talk about circumcenter for now.

    I have these 3 equations of lines:

    y = -63/33X +118.364
    y= -31/5x +576.8
    y= -94/38x +178.684

    I found that all 3 intersect at (106.8389, -85.60117). Since I need the exact coordinates and solve it algebraically, I did some graphing and found that the first and second lines would intersect at the same coordinates as all 3 lines, thus giving me the exact intersection. I did teh math, but I got (106.9237288, -86.12711856)

    Did I go wrong? How should I do this? I can supply all previous work done to find these 3 lines, please help.
    Last edited by Dyscrete; January 10th 2010 at 02:28 PM.
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  2. #2
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    Hi

    I graphed it att www.coolmath.com/graphit and then zoomed in on your intersection, if you keep zooming in and zooming in you se that all 3 lines dont intersect perfectly at the same point. Could it be because, too some line the values are more precise?? for example :+118.364; then we have: 576.8.....? is it. 576.854?? or 576.751?

    You have 3 intersects: Att these points two of your three lines cross.
    106.83797;-85.599395
    106.83890;-85.601177
    106.83905;-85.602016

    Hope this was somewhat helpfull.
    Last edited by Henryt999; January 9th 2010 at 01:26 PM.
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  3. #3
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    Slope of line YZ-

    74 69 = 5
    97 66 = 31

    Slope of Perpendicular bisector XP-

    -31
    5

    Midpoint of YZ is ( 81.5 , 71.5)

    Equation of Perpendicular bisector XP-

    Y 71.5 = -31 (x 81.5)
    5





    That's how I got y= -31/5x +576.8

    =/

    Can anyone help me out here? Should I post all of my work to see what went wrong?
    Last edited by mr fantastic; January 10th 2010 at 03:21 AM.
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  4. #4
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    Quote Originally Posted by Dyscrete View Post
    Can anyone help me out here? Should I post all of my work to see what went wrong?
    Maybe that would help...
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  5. #5
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    My initial coordinates:

    X (3, 36)
    Y (66, 69)
    Calculations Next Page

    Z (97, 74)


    Here is what I did to get the 3 equations of lines:

    P midpoint of ZY,

    P 97 + 66 , 74 + 69
    2 2



    P 163 , 143
    2 2




    P ( 81.5 , 71.5 )

    Q- midpoint of XZ,



    Q 3 + 97 , 36 + 74
    2 2



    Q 100 , 110
    2 2



    Q (50 , 55)




    R midpoint of XY,



    R 3 + 66 , 36 + 69
    2 2

    R 69 , 105
    2 2


    R (34.5 , 52.5)





    Slope of XY-

    69 36 = 33
    66 3 = 63

    Slope of Perpendicular bisector RZ-

    -63
    33

    Midpoint of XY is (34.5 , 52.5)

    Equation of line RZ-

    Y-52.5 = -63 (x 34.5)
    33
    X + 118.364




    Y = -63
    33



    Slope of line YZ-

    74 69 = 5
    97 66 = 31

    Slope of Perpendicular bisector XP-

    -31
    5

    Midpoint of YZ is ( 81.5 , 71.5)

    Equation of Perpendicular bisector XP-

    Y 71.5 = -31 (x 81.5)
    5

    X + 576.8

    Y = -31
    5

    Slope of line XZ-

    74 36 = 38
    97 3 94

    Slope of Perpendicular bisector QY-

    -94
    38

    Midpoint of XZ (50, 55)

    Equation of Perpendicular bisector QY-

    Y 55 = -94 (x 50)
    38
    X + 178.684


    Y = -94
    38



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  6. #6
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    Thanks for your help guys!

    Also I think it got messed up, but the final equations I found were:


    y = -63/33X +118.364
    y= -31/5x +576.8
    y= -94/38x +178.684
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  7. #7
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    Almost perfect

    -31/5*x+576.7960000000001 this gives you an almost perfect intersect
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  8. #8
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    Quote Originally Posted by Henryt999 View Post
    -31/5*x+576.7960000000001 this gives you an almost perfect intersect
    It still doesn't solve my question but I'll give you thanks anyway.
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  9. #9
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    To find the circumcentre, you do not need the equations of the triangle sides,
    when you are given the points of the vertices.
    Just calculate the perpendicular slopes from the side slopes,
    get the side midpoints and write the equations of any two of the three perpendicular bisectors.

    If you want exact values of the x and y co-ordinates of the circumcentre, then
    work with fractions.

    The perpendicular slopes and midpoints were correctly solved for.

    Perp bisector of XY is y-\frac{105}{2}=-\frac{21}{11}(x-\frac{69}{2})

    y=-\frac{21}{11}x+\frac{21(69)+105(11)}{22}

    Perp bisector of YZ is y-\frac{143}{2}=-<br />
\frac{31}{5}x+\frac{31}{5}\frac{163}{2}

    y=-\frac{31}{5}x+\frac{31(163)+5(143)}{10}

    We get

    y=-\frac{62}{10}x+\frac{5768}{10}

    y=-\frac{42}{22}x+\frac{2604}{22}

    Therefore,

    \frac{62}{10}x+\frac{5768}{10}=\frac{42}{22}x+\fra  c{2604}{22}

    \frac{62}{10}x-\frac{42}{22}x=\frac{5768}{10}-\frac{2604}{22}

    \frac{62(22)-420}{220}x=\frac{22(5768)-26040}{220}

    944x=100856

    x=\frac{100856}{944}\ exactly.

    y=-6.2(\frac{100856}{944})+576.8\ exactly.
    Last edited by Archie Meade; January 10th 2010 at 02:28 PM.
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  10. #10
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    Thanks! But...

    Thanks Archie Meade, it looks like you used point slope form in the beginning..

    Now I'm a bit lost... I got -61.127 when solving for y, it's a bit confusing. What would I do next?
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  11. #11
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    I see that perpendicular bisector of XY...which I labeled RZ. Wouldn't it be the negative reciprocal? Now I'm really lost =/ (sorry)
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  12. #12
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    Wait a minute...I think I get this. My equations for the line were wrong, so they gave me iffy co-ordiantes. Now that the equations are correct, I got a new point of intersection, right?
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  13. #13
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    YES! WE DID IT!

    I put it in my calculator and it all works! Thanks Archie Meade!

    I'll try to fix my altitudes now.
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  14. #14
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    Yes, the y co-ordinate of the circumcentre is negative.

    The work may appear tedious, but this type of problem can make you very fluent with linear equations.
    That's the point of them.

    Yes, I found the slopes of the 3 triangle sides.
    Then I inverted and negated them to obtain the perpendicular slopes of the bisectors. Then I found the triangle side midpoints.

    Exactly as you yourself had done correctly.

    Now, you only need to use 2 perpendicular bisectors.

    You use the perpendicular slope and midpoint to write the equations of 2 bisectors.

    We solve their equations by setting the y's equal.
    We are then finding the x co-ordinate of the centre.
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  15. #15
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    Actually, I asked a teacher to review my altitudes(orthocenter) and he said they were correct. Here is the work I've done for them:

    Slope of line YZ-

    69 74 = -5 = 5
    66 97 = -31 = 31

    Slope of line XM -

    -6.2

    -31 =
    5 =

    Equation of line XM- (3, 36)

    Y= -6.2x + b

    36 = -6.2 (3) + b
    36 = -18.6 + b



    b = 54.6

    Equation of line XM-



    Y= -6.2x + 54.6

    Line YO is perpendicular to line XZ

    Slope of line XZ-

    74 36 = 38 = 19
    97 3 = 94 = 47



    Slope of line YO-
    2.474




    -47 =
    19

    Equation of line YO-

    Y= -2.474x + b (66, 69)

    69= -2.474 (66) + b

    69 = -163.284 + b



    b = 232.284



    Y = -2.44x + 232.284

    Line ZN is perpendicular to line XY

    Slope of line XY-

    69 36 = 33 = 11
    66 3 63 21

    Slope of line ZN-
    -1.909




    -21 =
    11 =

    Equation of line ZN

    Y = -1.909x + b ( 97, 74)

    74 = -1.909 (97) + b

    74 = -185.173 + b



    b = 259.173



    Y = -1.909x + 259.173






    -y + (-6.2x) = -54.6
    Y + 2.44x = 232.284

    -3.76x = 177.684
    -3.76 -3.76




    X= -47.256

    Y + 2.44 (-47.256) = 232.284

    Y + (-115.305) = 232.284



    Y = 347.589



    So Archie, redoing this with fractions instead of decimals would give me the correct result?
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