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Math Help - Find equation of the line

  1. #1
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    Find equation of the line

    Hi everyone,

    I would to check with someone if I didn't make a mistake with the problem below

    Positions are given with reference to a Cartesian coordinate system whose x- and y-axes point due East and due North, respectively. Distance is measured in kilometres.


    An aeroplane flies in a straight line from city A, at (300,-200), to city B, at (-100,600).


    ((i) Find the equation of the line of flight of the aeroplane.
    [HTML][/HTML]
    A(300,-200) B(-100,600)

    First find the slope

    Run is: -100-300 = -400
    Rise is: 600-(-200) = 800

    The slope is

    M = rise/run = 800/-400 = -2

    We can apply equation y-y1 = m(x-x1) with m= -2 and either (x1,y1) = (300,-200) or (x1,y1)= (-100,600).

    First possibility:
    y(-200)= -2(x-1)
    y= -2x+202
    [HTML][/HTML]

    Thanks again for helping me
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  2. #2
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    Quote Originally Posted by valerie-petit View Post
    Hi everyone,

    I would to check with someone if I didn't make a mistake with the problem below

    Positions are given with reference to a Cartesian coordinate system whose x- and y-axes point due East and due North, respectively. Distance is measured in kilometres.


    An aeroplane flies in a straight line from city A, at (300,-200), to city B, at (-100,600).


    ((i) Find the equation of the line of flight of the aeroplane.

    A(300,-200) B(-100,600)

    First find the slope

    Run is: -100-300 = -400
    Rise is: 600-(-200) = 800

    The slope is

    M = rise/run = 800/-400 = -2 Mr F says: Correct.

    We can apply equation y-y1 = m(x-x1) with m= -2 and either (x1,y1) = (300,-200) or (x1,y1)= (-100,600). Mr F says: Correct.

    First possibility:
    y(-200)= -2(x-1) Mr F says: Wrong! How on Earth did you get it? y - (-200) = -2 (x - 300) => y = .....

    y= -2x+202


    Thanks again for helping me
    ..
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  3. #3
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    Thumbs up Thanks Mr Fantastic

    y-200= -2(x - 300)
    y= -2x+302

    Sorry about that,
    Thanks again for helping me out
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  4. #4
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    Post Direction of travel

    The second question of that exercise I have no idea what is the steps to achieve it.

    ''Find the direction of travel of the aeroplane, as a bearing, with the angle in degrees correct to one decimal place.''


    Can someone Explain me what they looking for, It's the degrees part which confuse me. How do I find the angle if I don't have the distance between A and B.(distance between A & B is asked in the next question)

    Thanks again
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  5. #5
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    Quote Originally Posted by valerie-petit View Post
    The second question of that exercise I have no idea what is the steps to achieve it.

    ''Find the direction of travel of the aeroplane, as a bearing, with the angle in degrees correct to one decimal place.''


    Can someone Explain me what they looking for, It's the degrees part which confuse me. How do I find the angle if I don't have the distance between A and B.(distance between A & B is asked in the next question)

    Thanks again
    Plot the two points. Draw the obvious big right-angle triangle (it has a base length of 400 and a height of 800). Use tan to get one of the angles inside this triangle. Use this angle to get the bearing of B from A.
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  6. #6
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    Smile Re: direction of travel

    If I didn't do a big mess, I find the angle of B is 30.9 degree. So that should be S 30.9 W.
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  7. #7
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    Quote Originally Posted by valerie-petit View Post
    If I didn't do a big mess, I find the angle of B is 30.9 degree. So that should be S 30.9 W.
    1. \tan^{-1} 2 = 63.4^0.
    2. A is in the fourth quadrant and B is in the second quadrant so the bearing of B from A will definitely not be S 30.9 W.

    Did you draw a set of xy-axes, plot the points and draw the triangle?
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  8. #8
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    Smile Re: direction of aeroplane

    Sorry I did tan B= b/a instead of a/b, and I didn't draw the triangle with xy-axes which made my triangle look completely wrong.

    So the direction from B to A is S63.4E degree.

    I'm really bad in maths, the course I have give a book the size of a magazine, no much exercises and examples, but this is my last exam and I can't back up
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  9. #9
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    Re: direction of an aeroplane

    I start again from the begining, attached I put my drawing of the triangle.

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  10. #10
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    Thumbs up Thanks again

    Sorry about that completely confuse myself. I got it right this time.

    Thanks again Mr Fantastic
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  11. #11
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    Quote Originally Posted by valerie-petit View Post
    y-200= -2(x - 300)
    y= -2x+302

    Sorry about that,
    Thanks again for helping me out
    Not sure I agree with the final line equation. I get y = -2x + 400. What do you think?

    http://www.mathcelebrity.com/slope.php
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  12. #12
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    [quote=valerie-petit;180131]Hi everyone,

    I would to check with someone if I didn't make a mistake with the problem below

    Positions are given with reference to a Cartesian coordinate system whose x- and y-axes point due East and due North, respectively. Distance is measured in kilometres.


    An aeroplane flies in a straight line from city A, at (300,-200), to city B, at (-100,600).


    ((i) Find the equation of the line of flight of the aeroplane.
    A(300,-200) B(-100,600)

    First find the slope

    Run is: -100-300 = -400
    Rise is: 600-(-200) = 800

    The slope is

    M = rise/run = 800/-400 = -2

    We can apply equation y-y1 = m(x-x1) with m= -2 and either (x1,y1) = (300,-200) or (x1,y1)= (-100,600).

    First possibility:

    You are right upto here. Now see, the equation:

    y - y_1 = m(x - x_1)

    y - (- 200) = - 2 (x - 300)

    y + 200 = - 2x + 600

    y = - 2x + 600 - 200

    y = - 2x + 400.
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  13. #13
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    Thumbs up Thanks again

    Thanks mathceleb
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  14. #14
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    Thumbs up Thanks again

    Yes, at the end I did find my mistake
    Sad to know that question was just the beginning of a long exercise
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  15. #15
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    Quote Originally Posted by valerie-petit View Post
    Yes, at the end I did find my mistake
    Sad to know that question was just the beginning of a long exercise
    No worries. Keep working at it and you will do great.
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