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Math Help - Check my work please?

  1. #16
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    Well done!
    you corrected yourself,
    that's learning!
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  2. #17
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    x= 94.019

    y= -61.128

    I got it, thanks Archie ( I round to the nearest thousandth)

    Thanks a lot for your help.
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  3. #18
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    Uh, perpendicular bisectors had the property of equidistance, meaning we could use the midpoints. What would I use for y1 now? The point that the altitudes originate from to go through the line it is perpendicular to?
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  4. #19
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    You use the perpendicular slopes again.
    Now, instead of using side midpoints, you use the co-ordinates of the opposite vertex instead (the point that is not on the triangle side).

    Again, you only need 2 perpendiculars (altitudes).

    The perpendiculars may not even be touching the triangle side if the triangle has an obtuse angle.

    They will be parallel to the bisectors in the circumcentre case.

    It's my naptime so i'll check tomorrow to see how you're getting on.
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  5. #20
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    Okay, so I got these two lines:

    Y = -63/33x + 6185/33
    Y = -31/5x - 55/5

    According to my calc (no idea how to continue algebraically) the intersection is at (-46.24294, 275.70621)

    Were my lines correct?
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  6. #21
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    Quote Originally Posted by Dyscrete View Post
    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.
    Please post the actual question.

    CB
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  7. #22
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    Quote Originally Posted by CaptainBlack View Post
    Please post the actual question.

    CB

    Yeah..
    After the circumcenter I found the orthocenter, hopefully Archie would check if it's correct.
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  8. #23
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    Hi Dyscrete,

    recheck your equations for the 2 perpendicular lines (altitudes)
    that contain the opposite vertex.

    At least one is wrong.

    You say you have no idea how to continue with the algebra,
    at least you are honest but it means you should study why the line equations
    are needed, how you write them and how you find the point of intersection.

    I calculate (-47.67, 350.2) for the orthocentre.

    I will post the calculations in fraction form,
    but you should practice until you are very clear about your algebraic calculations and how they relate to the geometry.
    Last edited by Archie Meade; January 10th 2010 at 01:41 PM. Reason: mistype
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  9. #24
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    The 3 equations of the altitudes are...

    y=-\frac{31}{5}(x-3)+36

    y=-\frac{63}{33}(x-97)+74

    y=-\frac{47}{19}(x-66)+69

    Only 2 are required to find the orthocentre.
    The x and y values of these equations are the co-ordinates of points on the line. If we let x be the same on all 3 lines and also let y be the same on all 3 lines, then we are referring to the orthocentre, the common point of intersection of the 3 altitudes.

    Using the following 2 lines...

    y-36=-\frac{31}{5}(x-3)

    y-74=-\frac{63}{33}(x-97)

    we get.. y=-\frac{31}{5}x+\frac{3(31)+5(36)}{5}

    and...... y=-\frac{63}{33}x+\frac{97(63)+74(33)}{33}

    giving... -\frac{63}{33}x+\frac{8553}{33}=-\frac{31}{5}x+\frac{273}{5}

    \frac{31}{5}x-\frac{63}{33}x=-\frac{8553}{33}+\frac{273}{5}

    \frac{33(31)-5(63)}{5(33)}x=-\frac{5(8553)-33(273)}{5(33)}

    [33(31)-5(63)]x=-[5(8553)-33(273)]

    (1023-315)x=-(42765-9009)

    x=-\frac{33756}{708}=orthocentre x co-ordinate.

    Place this co-ordinate into any of the 3 altitudes to find the orthocentre y co-ordinate.

    All 3 should give the exact same y.
    Last edited by Archie Meade; January 10th 2010 at 01:36 PM.
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  10. #25
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    Thank you!!

    Now I understand what you did, I kept doing crazy things with my lineq equations which is why my answer was messed up. Thanks for helping!
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  11. #26
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    These examples encourage the student to co-ordinate their written work very carefully, due to all the points, slopes and equations.

    These problems are worth the patience.
    You showed good spirit.
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  12. #27
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    Thanks again, but wasn't the x coordinate of the intersection negative? when I divide it becomes a positive, obviously.
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  13. #28
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    Yes, I mistyped a sign, I'll correct it.
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  14. #29
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    Quote Originally Posted by Archie Meade View Post
    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{33}{5}x+\frac{33}{5}\frac{163}{2}

    y=-\frac{33}{5}x+\frac{33(163)+5(143)}{10}

    We get

    y=-\frac{66}{10}x+\frac{5594}{10}

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

    Therefore,

    \frac{66}{10}x+\frac{5594}{10}=\frac{42}{22}x+\fra  c{2604}{22}

    \frac{66}{10}x-\frac{42}{22}x=\frac{5594}{10}-\frac{2604}{22}

    \frac{66(22)-420}{220}x=\frac{22(5594)-26040}{220}

    1032x=97028

    x=\frac{97028}{1032}\ exactly.

    y=-6.6(\frac{97028}{1032})+559.4\ exactly.
    You got -33/5 there somewhere for the slope.. I got -31/5 is that an error as well or am I missing something?
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  15. #30
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    No, you're right!
    I'll clean that up.
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