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anonymous

  • 5 years ago

Consider two straight lines L1 (y = 2x +1) and L2 (y=x). a) What is the slope of line L1 in respect to line L2? b) What is the equation of line L1 in respect to line L2? NOTE. Think that the original xy co-ordinate system has been rotated in respect to the origin so that line L2 defines the new x-axis x´ and y´ is the new y-axis!

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  1. anonymous
    • 5 years ago
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    I need a smart person to answer this, please, I REALLY need to understand this

  2. amistre64
    • 5 years ago
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    a smart person eh..... hmmm

  3. anonymous
    • 5 years ago
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    yeah

  4. amistre64
    • 5 years ago
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    the slope of L1 with respect to L2 suggests to me that it is a relative answer. If L2 were flat, what would the slope of L1 be with it..

  5. amistre64
    • 5 years ago
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    y=x has a slope of 1, which is a tan(45)

  6. anonymous
    • 5 years ago
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    ok, I know this, but where do I go from here?

  7. amistre64
    • 5 years ago
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    y = 2x... has a slope of 2....

  8. anonymous
    • 5 years ago
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    so it is 2 :)

  9. amistre64
    • 5 years ago
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    i aint determined that yet :)

  10. amistre64
    • 5 years ago
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    it is the angle of slope 2 - slope1 but you use the "angles" and not the slopes

  11. amistre64
    • 5 years ago
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    what angle has a tan^-1(2)?

  12. anonymous
    • 5 years ago
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    oh, so, basically, the slope of L1 in relation to slope L2 will be 1, right?

  13. anonymous
    • 5 years ago
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    I am really bad at this sorry... forgot all the high school stuff

  14. amistre64
    • 5 years ago
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    tan(63.43) = 2 not really, since we have rotated the axis by a certain degree, we need to determine the degrees involved, then you trig stuff to re dress the problem

  15. amistre64
    • 5 years ago
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    .... we dont need the angles perse, just their ration....

  16. anonymous
    • 5 years ago
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    isn't there a simple way?

  17. amistre64
    • 5 years ago
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    cos(a-b) = cos(a)cos(b)-sin(a)sin(b)

  18. amistre64
    • 5 years ago
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    cos(a) = 1/sqrt(2) cos(b) = 1/sqrt(5) sin(a) = 1/sqrt(2) sin(a) = 2/sqrt(5)

  19. amistre64
    • 5 years ago
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    cos(a-b)=1/sqrt(2) * 1/sqrt(5) - 1/sqrt(2) * 2/sqrt(5)

  20. amistre64
    • 5 years ago
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    1/10 - 2/10 = -1/10

  21. amistre64
    • 5 years ago
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    might be better to work it in tans :) that way we get a new "slope"....tan = slope by the way

  22. amistre64
    • 5 years ago
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    tan(a-b) = tan(a)-tan(b) ----------- if I recall correctly 1+tan(a)tan(b)

  23. amistre64
    • 5 years ago
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    tan(a) = 1/1 tan(b)=2/1

  24. amistre64
    • 5 years ago
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    hold up, make a the bigger one... tan(a) = 2; tan(b)=1 2-1 ---- = 1/3 right? 1+2

  25. anonymous
    • 5 years ago
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    yeah

  26. amistre64
    • 5 years ago
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    the new slope is 1/3 :) what was the second part?

  27. anonymous
    • 5 years ago
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    equation of the line L1 in relation to eq of the line of L2

  28. amistre64
    • 5 years ago
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    y=(1/3)x + b just gotta determine where the lines intersect and use that as a parameter for the new system....

  29. amistre64
    • 5 years ago
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    they originally intersect at x=-1; y=-1

  30. amistre64
    • 5 years ago
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    the new point of intersection is the length of -sqrt(2)....

  31. amistre64
    • 5 years ago
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    so the new coord to fill in is (-sqrt(2),0)

  32. anonymous
    • 5 years ago
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    why sqrt(2)?

  33. amistre64
    • 5 years ago
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    0 = (1/3)(-sqrt(2)) + b b= sqrt(2)/3 y = (1/3)x - (sqrt(2)/3) should be it

  34. amistre64
    • 5 years ago
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    why sqrt(2)? the original lines cross at (-1,-1) when we rotate the axis to accomodate the new axises.... that point is still there, but we use the distance of it from the origin as the x intercept

  35. anonymous
    • 5 years ago
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    oh, ok, got it

  36. amistre64
    • 5 years ago
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    1-1-sqrt(2) is the triangle of a 45 degree..

  37. anonymous
    • 5 years ago
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    how did you get so good at maths?

  38. amistre64
    • 5 years ago
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    by messing up alot lol

  39. amistre64
    • 5 years ago
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    my son is nagging me for the laptop....ciao :)

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