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Equation of bisector of line

  1. Apr 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Equations of the bisectors of the lines ## 3x - 4y + 7 =0 ## and ## 12x + 5y - 2 = 0##
    are ?

    2. Relevant equations
    Line equation is y = mx + c
    where m is slope and c is y intercept


    3. The attempt at a solution
    Bisector is a line cutting equal angles between those 2 lines.
    But what to do or approach that problem?
     
  2. jcsd
  3. Apr 27, 2015 #2

    BvU

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    Make a drawing.
    What do you know of the line you are looking for ? 1. some point it has to go through 2. something about the slope. That should be sufficient !
     
  4. Apr 27, 2015 #3
    Drawing seems difficult to me in which slopes are in fraction and y intercept is not 0.
    I found the point of bisector, by solving that 2 equations which is (-3/7,10/7).
    How to find slope?
     
  5. Apr 27, 2015 #4

    BvU

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    You should really learn how to draw such a line. Here's a recipe:
    1. take x = 0 and calculate what y is
    2. take y = 0 and calculate what x is
    you now have two points, so you can draw the line.
     
  6. Apr 27, 2015 #5
    Okay, I will do that but first what is the use of that when we can solve algebraically the points ?
     
  7. Apr 27, 2015 #6

    BvU

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    You can visually check your answer, develop a feeling, etc. etc. Comes in very handy when things become more complicated and more abstract.

    Now for the slope of the bisectors: if line 1 has slope ##\alpha## and line 2 has slope ##\beta##, what is the slope of one of the the bisectors ?
     
  8. Apr 27, 2015 #7
    Did the drawing and it gives me bad feeling,
    image.jpg
     
  9. Apr 27, 2015 #8

    BvU

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    Gives me a bad feeling too:

    "I found the point of bisector, by solving that 2 equations which is (-3/7,10/7)"
    -- but that point is clearly not where these two lines intersect!​

    but not all that bad: small error only in ##12x - 2 = 0 \Rightarrow x = 1/6##, not ##-1/6##.

    Now for the slope of the bisectors: if line 1 has slope ## \alpha## and line 2 has slope ## \beta## , what is the slope of one of the the bisectors ?
     
  10. Apr 27, 2015 #9
    Don't know.
    If I know that, then I can easily find the equation of bisector.
    So can you give a hint?
     
  11. Apr 27, 2015 #10

    Mark44

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    I agree completely with BvU here. My understanding of how the brain works is that one hemisphere is primarily analytical, and the other is more visual. The best way to tackle mathematics problems is to come at them with both sides of your brain. Too often, inexperienced students become enamoured with manipulating symbols of equations, and due to an error in their logic, come up with an incorrect solution. If they had also engaged the visual part of their brains, they would have been able to see that they had made a mistake.
     
  12. Apr 27, 2015 #11
    But my point in this question is that if we know the diagram also, then also I think we cannot solve diagrammatically. ( Moreover we not get accurate diagrams).
    We have to use algebra to solve.
    Why to make a diagram? It is not giving me any info.
    Without making drawing we can think that lines are intersecting.
     
  13. Apr 27, 2015 #12

    BvU

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    RE your post #11: well, then we can still check whether our ultimate answer agrees with the picture we made.

    Re your post #9: Don't know is fine with me, but not enough for you to bring the exercise to a succesful conclusion. I could do the exercise for you, or give the answer to my own question, but that doesn't help you. That's why I asked in post #8.

    Well, now we rummage through our toolkit of available relevant equations and come up with nothing at all, eh ? y = mx + c isn't good enough, so we'll have to develop a new tool somehow. To answer the question "if line 1 has slope ## \alpha## and line 2 has slope ## \beta## , what is the slope of one of the bisectors ?"

    A first step could be to make it a bit simpler: what is a bisector of y= 0 and y = x ? Both through the origin, one slope zero, the other slope 1. Don't fall into the trap and claim "the line with slope 1/2!" because that's a little too easy and plain wrong. (Easily disproved: the bisector of y=0 and x=0 doesn't have slope ##\infty/2## but slope ##1##)

    Since the "bisector is a line cutting equal angles between those 2 lines" (should have had a place in the toolkit!) and one angle is 0 and the other is ##\pi/4## the line through the origin with a slope corresponding to an angle of ##\pi/8## seems a good candidate. So half the angle. Slope calculation awkward! Something with solving ##\tan 2\phi = 1\rightarrow \tan\phi = ?##
     
    Last edited: Apr 27, 2015
  14. Apr 27, 2015 #13

    Mark44

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    In post 1 you didn't say anything about the problem giving you a diagram. Also, your diagram doesn't have to be 100% accurate. It can give you a rough idea of what the solution looks like.
    As long as your diagram is reasonably accurate, it can give you lots of information. For example, in the diagram you show in post #7, I can see that one bisector will have a slope of about 1. The other bisector will have a slope of about -1. I can also get a reasonable idea of where the two lines intersect,

    So from the drawing alone I can use the point of intersection and the two slopes, and write equations for the two bisectors. These won't be exactly correct, but they should be reasonably close, and should agree with the equations you get by using algebra alone.

    If you make a mistake in your algebra calculation, you won't know that your work is incorrect. With a drawing, you have something to check against.
     
  15. Apr 27, 2015 #14
    I have written that in attempt but not in relevant equation. :smile:
    ##\tan 2\phi = 1 ##
    $$ ⇒ \frac{2tan\phi}{1-tan^2\phi} = 1 $$
    Taking tanΦ = x

    We get ## 2x = 1- x^2 ##

    ## ⇒ x^2 + 2x - 1 = 0 ##

    ## ⇒ x = \frac{-2 ± \sqrt{8}}{2} ##

    ## ⇒ tan\phi = -1 ± √2 ##
    Now what is the next part?
     
  16. Apr 27, 2015 #15

    BvU

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    Well done. So you know how to find the tangent of a halved angle.
    If we go back to the exercise: first line has slope ##m_1 = 3/4##, the second has slope ##m_2 = -12/5## .

    (the notation ##\alpha## and ##\beta## for the slopes wasn't such a good idea , so I change to ##m_1## and ##m_2##)​

    If the angle of the first one is ##\phi_1## and the angle of the second one is ##\phi_2##, what do you think you can expect for the angle of a bisector ?
     
  17. Apr 28, 2015 #16
    It would be ##\frac{\phi_1 + \phi_2}{2} ## ? But I think it is not valid for all cases.
    For this diagram it is true
    image.jpg We have only one dotted bi sector here?

    For this it is not true
    image.jpg
    Sorry in second diagram, the angle ##\phi_2## should be on other side
    and slope angle and between angles are different.
     
    Last edited: Apr 28, 2015
  18. Apr 28, 2015 #17

    BvU

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    Bisec.jpg
    In other words: since the correct angle is indeed ##180^\circ - \phi_2## (your ##\phi_2##) the expression ##\phi = {\phi_1+\phi_2\over 2}## (##\phi_2## in this picture) is correct also for your second picture.

    Why do you draw arrows ?

    And: now we are at it, and have pictures at hand (how useful :wink:!) once you have one bisector, what about the other ?
     
  19. Apr 28, 2015 #18
    We are supposed to draw arrows on line as they are lines but not line segments.
    Otherwise our fraction of marks are cut for not showing arrows.

    Don't know the angle phi of other
    See this,

    image.jpg
     
  20. Apr 28, 2015 #19

    BvU

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    OK, I understand the arrows. And two arrows on each line is a bit much, so apparently you have settled on one per line :smile: .

    Well, try out a few more pictures, preferably a bit more accurate ! There's a simple discovery waiting for you !
     
  21. Apr 28, 2015 #20
    Sorry for that will take care in future.o:)
    Getting ## \phi=\frac{\phi_1 + \phi_2 - 180}{2} ##

    image.jpg
     
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