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    A line with the origin of 300,400 exists. It travels at bearing of 45(degrees). Find the point at which this line and the line y = 100 intercept with one another.
    The question above has no answer.
    NEW QUESTION: How would you find the equation of the line that has its origins at (300,400) and has a bearing of 45degrees.
    P.S. need a solution that works for different degrees too, 45 is rather easy because the gradient will be 1
    I've been stuck on this question for a while. If anyone could help that would be great.
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    (Original post by bahjat93)
    A line with the origin of 300,400 exists. It travels at bearing of 45(degrees). Find the point at which this line and the line y = 100 intercept with one another.

    I've been stuck on this question for a while. If anyone could help that would be great.
    It doesn't. If the line begins from (300,400) and travels at 45 degrees then both x and y will be increasing above their starting position.

    If you simply consider a line that goes through (300,400) at a bearing of 45 then the equation of the line is y-400=tan(45)(x-300)
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    (Original post by RDKGames)
    It doesn't. If the line begins from (300,400) and travels at 45 degrees then both x and y will be increasing above their starting position.

    If you simply consider a line that goes through (300,400) at a bearing of 45 then the equation of the line is simple y-400=tan(45)(x-300)

    God damn it you're right! New question how do you find the equation of the line that has it's origins at (300,400) and bearing at 45degrees
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    (Original post by bahjat93)
    God damn it you're right! New question how do you find the equation of the line that has it's origins at (300,400) and bearing at 45degrees
    Use the general equation of a line y-y_1=m(x-x_1) with gradient m and going through the point (x_1,y_1). The gradient can be found by using tan of the angle. But you should notice that a bearing of 45 is the same as a gradient of 1.
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    (Original post by RDKGames)
    Use the general equation of a line y-y_1=m(x-x_1) with gradient m and going through the point (x_1,y_1). The gradient can be found by using tan of the angle. But you should notice that a bearing of 45 is the same as a gradient of 1.
    The point slope formula doesn't work if you don't know "m". I need an equation that will convert my degrees into m.
    BTW Just looked at your profile and it turns out we both went to the same secondary school :lol::lol:
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    (Original post by bahjat93)
    The point slope formula doesn't work if you don't know "m". I need an equation that will convert my degrees into m.
    BTW Just looked at your profile and it turns out we both went to the same secondary school :lol::lol:
    You can easily find m. There is no equation that will convert degrees to 'm'. If you think about tan, it's sin over cos which would be change in y over change in x which would give you the gradient. So tan(45) is your gradient.

    Haha nice. When did you finish TDA?
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    This is the idea im trying to explain but im too tired so here's a diagram. Its a unit circle.Name:  ImageUploadedByStudent Room1471215457.479362.jpg
Views: 94
Size:  136.6 KB


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    (Original post by RDKGames)
    You can easily find m. There is no equation that will convert degrees to 'm'. If you think about tan, it's sin over cos which would be change in y over change in x which would give you the gradient. So tan(45) is your gradient.

    Haha nice. When did you finish TDA?

    I'm gonna make some changes to my program and put your solution in, fingers crossed it works.

    I graduated from TDA 4 years ago. So probably before you started
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    (Original post by bahjat93)
    I'm gonna make some changes to my program and put your solution in, fingers crossed it works.

    I graduated from TDA 4 years ago. So probably before you started
    Yeah I was in Y9/Y10 at that time, I graduated this year xD

    Also, if you want the general gradient, it would be m=tan(90-\theta) where \theta is your bearing.
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    (Original post by RDKGames)
    Yeah I was in Y9/Y10 at that time, I graduated this year xD

    Also, if you want the general gradient, it would be m=tan(90-\theta) where \theta is your bearing.
    Nice man, hope you get into your first choice uni.

    Would that equation work for larger bearings such as 276?
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    (Original post by bahjat93)
    Nice man, hope you get into your first choice uni.

    Would that equation work for larger bearings such as 276?
    Thanks. And yes it should except for 0 or 180 as the value of tan doesn't exist at intervals of 90. If your bearing is 0 or 180 then it's simply a straight vertical line in the form of x=a. Test it in your 'program' or whatever it is you've got there. The 90 wouldn't be necessary if the angle was being measured from the positive x-axis but it is required since bearings are from the positive y-axis clockwise. Make sure it doesn't calculate in radians though.
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    (Original post by RDKGames)
    Thanks. And yes it should except for 0 or 180. Test it in your 'program' or whatever it is you've got there. The 90 wouldn't be necessary if the angle was being measured from the positive x-axis but it is required since bearings are from the positive y-axis clockwise. Make sure it doesn't calculate in radians though.
    Unfortunately every programming language works with radians. I just put a little conversion in to convert from degrees to radians before i use the inverse tan function. With the 45 degrees example it gave me the same answer (after i put in the conversion) so hopefully that won't be a problem
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    (Original post by bahjat93)
    Unfortunately every programming language works with radians. I just put a little conversion in to convert from degrees to radians before i use the inverse tan function. With the 45 degrees example it gave me the same answer (after i put in the conversion) so hopefully that won't be a problem
    Yeah that's not a problem, the gradient simply turns into m=tan(\frac{\pi}{2}-\frac{\theta \pi}{180}) where \theta is still your bearing in degrees.
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    (Original post by RDKGames)
    Yeah that's not a problem, the gradient simply turns into m=tan(\frac{\pi}{2}-\frac{\theta \pi}{180}) where \theta is still your bearing in degrees.
    I think this is the same (if not say let me know please) m = tan ((90-ray)*(pi/180))
    ray is representing theta
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    (Original post by bahjat93)
    I think this is the same (if not say let me know please) m = tan ((90-ray)*(pi/180))
    ray is representing theta
    Yeah that's right. It's the exact same thing; you're simply getting the angle from the horizontal axis first in degrees before applying the conversion so it's fine.
 
 
 
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