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Very HARD maths question

  1. Mar 13, 2005 #1

    aek

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    1).Find the co-ordinates of the point A on the line x= -3 such that the line joining A to B(5,3) is perpendicular to the line 2x+5y =12.

    2). Using the vertices A(2a,2b), B(-2c, 0), C(2c,0).
    (i) prove that the medians of a triangle are concurrent. (a median is a line joining a vertex to a midpoint of the opposite sides; the lines are concurrent if they have a common point of intersection);
    (ii) show that the centroid is 2a/3, 2b/3. (a centroid is the point of intersection of the medians).
     
  2. jcsd
  3. Mar 13, 2005 #2
    1) I will just give you hints.

    Can you find the slope of the line through A and B, in terms of the y-coordinate of A?

    Can you tell me anything else about what this slope must be, in order for the connecting line to be perpendicular to [tex]2x + 5y = 12[/tex]?

    2)

    (i) Again, just hints.

    What are the midpoints of the three sides of this triangle? Can you find the equations of the medians based on the vertex and midpoint coordinates?

    How do you find the intersection of two lines?

    (ii) If you can figure out how to find the intersection of two lines, you should be able to do this part on your own.
     
    Last edited: Mar 13, 2005
  4. Mar 15, 2005 #3
    2x+5y=12 is in another way y= (-2/5)x +12/5 so the slope of this line is -2/5.
    If this line is perpendicular to line (AB) then, the slope of (AB) is 5/2.

    the equation of line (AB) is: Y - Yb = slope (X - Xb) because B is a point of the line
    thus y - 3 = 5/2(x - 5)
    and if A is a point of x= -3 then it's x-coordinate is -3
    let's find its y-coordinate by replacing its x-coordinate in the equation of (AB)

    y - 3 = 5/2(-3 -5) which gives you y= -17

    and A(-3; -17)
     
    Last edited: Mar 15, 2005
  5. Mar 15, 2005 #4

    Integral

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    AI has made algebra errors in the first part of this problem, his method is close but his answer is not.
     
  6. Mar 15, 2005 #5
    you can draw ABC triangle and mark I as midpoint of [AB], J midpoint of [BC], K midpoint of [AC].
    Now you can obtain the coordinates of I, J and K
    I (a-c; b)
    J (0 ; 0)
    K (a+c; b)

    we can now say that [CI) and [BK) and [AJ) are the medians in the following triangle
    the next step is to find their equations and to proove that that intersection of for example (CI) and (BK) is the same as the intersection of (AJ) and (BK)

    let's find first the equation of (CI):

    the coordinates of vector CI are (a-3c; b)
    thus the slope of line (CI) is b/(a-3c)
    knowing that C is a point of (CI)
    we have (CI) :

    y - 0 = b/(a-3c)[x - 2c]

    we can next following the same methods find the other equations, i'll give them directly:

    (BK) : y = b/(a+3c)[x +2c]

    (AJ) : y = (b/a)x

    The following step is to take the intersection of two lines:
    for example (AJ) and (BK)

    for y(AJ) = y(BK)
    we have:

    (b/a)x = b/(a+3c)[x +2c]

    ok now it's a matter of calculus to arrive to the result that H is the intersection point
    with coordinates H(2a/3 ; 2b/3)

    we do the same for the intersection of another couple of lines
    you will obtain the same result

    thus the point with coordinates (2a/3; 2b/3) is the centroid of this triangle.



    As for part i of the question.
    i didn't understand exactly what do they want. because what they're asking is theoretical.
    Medians of a triangle should intersect

    Joe
     
  7. Mar 15, 2005 #6
    thanks Integral, i felt there was an error
    i edited my post :)
     
  8. Mar 16, 2005 #7

    Integral

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    AI,
    Perhaps you are not aware, but it is the general policy not to simply do the homework problems for the OP. We prefer to give hints to guide the OP to a solution. It is essential that you learn to solve problems on your own. Hints are best, complete solutions not so good.

    With that said you have still got problems with your solution to the first part.

    Have you tried piloting your lines to see if everything works?

    Here is how I would approach the problem.

    The original expression for a line yields:

    [tex] y = - \frac 2 5 x + \frac {12} 5 [/tex]

    The line perpendicular to this is

    [tex] y= \frac 5 2 x + b [/tex]

    Now you are given that this line must pass through the point (5,3) so use that point to find b.

    solve

    [tex] 3= \frac 5 2 5 + b [/tex]

    for b.

    Now simply find y when x = -3.

    I get y=2
     
  9. Mar 16, 2005 #8
    Sorry, I am not going to oppose you. But,
    I'd like my consultant to give me the answers, rather than a solution or hints for maths.
    Since you guys are very bright and successful, your archieve on the problem I asked can let me know it is correct. Then I will solve it myself again.
     
  10. Mar 16, 2005 #9

    aek

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    I totally agree, i rather have worked out steps following an answer in which i can guide myself through it. Hints can be rather too little to grasp the question your trying to answer. So Integral, i believe your theory is wrong and yes, i would help other young students finding trouble by giving them a long, worked out solution in which they can follow through.

    King regards.
     
  11. Mar 16, 2005 #10

    Integral

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    I have stated the general policy of these forums. It is to your advantage to put some effort and thought into these problems. Having the solution handed to you on a silver platter is not going to help you. YOU must but in the effort to arrive at a solution. YOU are the one taking a class to learn how to solve the problem. We will guide you and we WILL continue to discourage complete solutions, regardless of what YOU want.
     
  12. Mar 16, 2005 #11

    arildno

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    Now, I would just add to what Integral has said:
    It is, in fact, quite common that once it is clear that an OP (I'm not referring to the OP of this thread, just a general name) has struggled a bit, and still faces difficulties, some member on the forum chooses to give a detailed solution.

    Note that this is completely different from just flicking out the solution on the first opportunity:
    Basically, what you gain by that as an OP is the chance to fail on your own (but be helped into the solution) or solve it on your own purely by the hints (i.e, success!), i.e, learning opportunities denied to you if we simply gave off the solution straight-away.
     
  13. Mar 16, 2005 #12
    I'll usually post a solution if someone else has posted one with clear mistakes in it, to avoid confusing the original poster. Alternatively, if someone else has already posted a correct solution and I see a better way to do it I'll usually post it. Otherwise I agree completely.
     
  14. Mar 17, 2005 #13
    I want to clarify that I just want someone help me to check out whether the answer in the book is wrong.lol
     
  15. Mar 18, 2005 #14

    aek

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    nah i think you lost the plot however, i got no grounds to argue upon soo yeah. But what i wanna say is that some people just need to be give a work solution, because hints are to vague sometimes for them to figure out. My job is not here to help the kid succeed at school but rather answer his question. If you think it's wrong to answer it, then ignore it.

    AEK
     
  16. Mar 18, 2005 #15
    Sorry, i've been absent for a while.
    Thanks for your remark Integral, and i will surely work by the rule.
    But I noticed from your correction that you gave the constant b in the following equation:

    3 = 5*(5/2) +b therefore it gives b = 3 - (25/2)
    which gives b= (-19/2)

    y = (5/2)x -19/2
    if we replace x with -3

    we get:

    y = (-15/2) -19/2 = -34/2 = -17.

    Do I have to start doubting about myself??
    Anyway, thanks all for your understanding
     
  17. Mar 18, 2005 #16

    Integral

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    Opps! Yes you are correct. x= -17
     
  18. Mar 18, 2005 #17

    aek

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    See integral, even YOU make mistakes. I just like to thank AI personally for helping me. I really appreciate it.
     
  19. Mar 18, 2005 #18

    Integral

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    If all you got out of it was a problem done, then you have accomplished nothing. If you learned to do this sort of problem on your own then it was worth the effort. The proof in the pudding will be test time.


    It is not the first mistake I have made and probably not the last either.
     
  20. Mar 19, 2005 #19
    maybe we should conclude this topic by learning that everyone is prompt to do mistakes.
    You r welcome aek :)
    As for Integral, you are right, hints are sometimes better than giving the answer
     
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