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Math Help - Please Help Me!

  1. #1
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    Please Help Me!

    Hi, these are questions in grade 12 geometry course i'm taking. Please Help me out!

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  2. #2
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    can u see it this time?

    Please Help Me!-mod_1_act_6.jpg




    If u can't read it, the question says "Supply the coordinates for each figure without introducing any new coordinates.

    question (a) says: Triangle ABC has a right angle at A and angle B=Pie over 3.

    question (b) says: BCi s a quarter circle with centre at (0,0). Line AD has equation y=x.
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  3. #3
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    In part a, the slope of line BC is \tan \left( {\frac{{2\pi }}{3}} \right).
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Raiden_11 View Post
    Click image for larger version. 

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    If u can't read it, the question says "Supply the coordinates for each figure without introducing any new coordinates.

    question (a) says: Triangle ABC has a right angle at A and angle B=Pie over 3.
    The length of the base of the triangle is a.

    Now, using the trig ratio for tangent, we see that

    \tan B = \frac {|AC|}{a} ........ |AC| means the length of the line connecting A and C

    \Rightarrow \tan \left( \frac { \pi}{3} \right) = \frac {|AC|}{a}

    \Rightarrow \sqrt {3} = \frac {|AC|}{a}

    \Rightarrow a = \frac {|AC|}{ \sqrt {3}} = \frac { \sqrt {3} |AC|}{3}

    So we have B \left( \frac { \sqrt {3} |AC|}{3}, 0 \right)


    question (b) says: BCi s a quarter circle with centre at (0,0). Line AD has equation y=x.
    Note that r = |AD| = |AB|

    Also note that the angle the line y = x makes with the x-axis is 45^{ \circ}, or \frac { \pi}{4}

    Using the trig ratio for sine we see that

    \sin \left( \frac { \pi}{4} \right) = \frac {|DE|}{|AD|} = \frac {|DE|}{r} .........E is the point where a vertical line from D cuts the x-axis, |DE| represents the length of that line

    \Rightarrow r = \frac {|DE|}{ \frac { \sqrt {2}}{2}} = \frac {2 |DE|}{ \sqrt {2}}

    So B \left( 0, \frac {2 |DE|}{ \sqrt {2}} \right)


    These questions are a bit vague to me. I guess we could do the last one without introducing a new point, if we used the cosine rule or something like that
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  5. #5
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    It appears to me as if the question (a) is asking for the coordinates of C.
    Because line BC has slope \tan \left( {\frac{{2\pi }}{3}} \right) =  - \sqrt 3 then say C(0,c) so \frac{{c - 0}}{{0 - a}} =  - \sqrt 3 \quad  \Rightarrow \quad c = a\sqrt 3 \quad or\quad C:\left( {0,a\sqrt 3 } \right).
    Thus we have written the coordinates of C in terms of the given.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    It appears to me as if the question (a) is asking for the coordinates of C.
    Because line BC has slope \tan \left( {\frac{{2\pi }}{3}} \right) =  - \sqrt 3 then say C(0,c) so \frac{{c - 0}}{{0 - a}} =  - \sqrt 3 \quad  \Rightarrow \quad c = a\sqrt 3 \quad or\quad C:\left( {0,a\sqrt 3 } \right).
    Thus we have written the coordinates of C in terms of the given.
    oh yeah, silly me. and i suppose question (b) is asking for coordinate D right...or C? or both, yeah, both

    well, i'll leave my post up there, just in case anybody wonders, "well, how does a and r relate to all of this?"
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  7. #7
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    Here's another question guys....it seems simple but i think theres more to it.

    Thanks for the help earlier can u help me answer this?
    Please Help Me!-mod_1_act_6_2.jpg
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  8. #8
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    I really need help!

    These are questions from the grade 12 geometry course i am taking.

    1. Triangle ABC has medians CE and BD as shown in the diagram. Use a coordinate proof and a vector proof to prove BF : FD = 2 : 1

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  9. #9
    MHF Contributor red_dog's Avatar
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    Coordinates prove:
    We can assume that A\in Oy,B,C\in Ox and A(0,2a),B(2b,0),C(2c,0), for the simplicity of calculus.
    Now D(c,a),E(b,a). Then you can write the equations of BD and CE, solve the system and you will find the coordinates of the point F. Then you can calculate the length of the segments BF,FD.
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  10. #10
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    thanks....but

    Thanks but can you show me how to solve it just so i can have a clear idea.
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  11. #11
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    Also

    Also,how did you get those points in the first place...can u show me in detail please. Much appreciated.
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  12. #12
    MHF Contributor red_dog's Avatar
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    Given a triangle ABC in plane we can get the Ox and Oy axis anywhere in that plane. So we can get Ox axis passing through B and C and Oy axis passing through A.
    Now, if there are two points M_1(x_1,y_1),M_2(x_2,y_2), then the midpoint M of the segment M_1M_2 has the coordinates \displaystyle \frac{x_1+x_2}{2} and \displaystyle \frac{y_1+y_2}{2}. Then D(c,a), \ E(b,a).
    The equation of the line passing through M_1(x_1,y_1),M_2(x_2,y_2) is \displaystyle \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1} or
    \left|\begin{array}{lll}<br />
x & y & 1\\<br />
x_1 & y_1 & 1\\<br />
x_2 & y_2 & 1<br />
\end{array}\right|=0
    Using one of these equations you can now write the equations of BD and CE, then solve the system formed by these equations to find the coordinates of F.
    Then you can use the distance formula between two points in plane:
    M_1M_2=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
    to calculate BF and FD
    It is more clear now?
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  13. #13
    MHF Contributor red_dog's Avatar
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    Vector prove:
    Let \displaystyle \frac{BF}{FD}=k
    Then \displaystyle \overrightarrow{CF}=\frac{1}{1+k}\overrightarrow{C  B}+\frac{k}{1+k}\overrightarrow{CD} and
    \displaystyle \overrightarrow{CE}=\frac{1}{2}(\overrightarrow{CB  }+\overrightarrow{CA})=\displaystyle \frac{1}{2}\left(\overrightarrow{CB}+2\overrightar  row{CD}\right)=\frac{1}{2}\overrightarrow{CB}+\ove  rrightarrow{CD}.
    The vectors \overrightarrow{CF} and \overrightarrow{CE} are colinear, so
    \displaystyle \frac{\frac{1}{1+k}}{\frac{1}{2}}=\frac{\frac{k}{1  +k}}{1}\Rightarrow k=2, so \displaystyle \frac{BF}{FD}=2
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  14. #14
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    I have another problem

    Thanks man, i really appreciate the help but i have two more problems.

    Please Help Me!-4-8-2.jpg

    Please Help Me!-4-8-3.jpg
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