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Math Help - finding lines

  1. #1
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    finding lines

    Here is the question. Find two lines L1, L2 that intersect at point (a,b). So how do yo go about it? Any help is appreciated, thanks.
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    Quote Originally Posted by Keep View Post
    Here is the question. Find two lines L1, L2 that intersect at point (a,b). So how do yo go about it? Any help is appreciated, thanks.
    Let m_1 denote the slope of L_1, m_2 the slope of L_2.

    Then the equations of the lines are

    L_1: y-b=m_1(x-a) and

    L_2: y-b=m_2(x-a)
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    Quote Originally Posted by earboth View Post
    Let m_1 denote the slope of L_1, m_2 the slope of L_2.

    Then the equations of the lines are

    L_1: y-b=m_1(x-a) and

    L_2: y-b=m_2(x-a)
    But you need to find out at least the slope of the line to get it, don't you? And since you don't have the y intercept of the line, that would also be difficult to get, isn't it? So you have two lines intersecting at point (6,18) and you don't have the gradient (slope) or the y intercept. You are just given the point of intersection of x and y where x = 6 and y = 18 as I said above.
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    Can anyone help? I just have ONE point, i.e. the point of intersection of the two lines ONLY (no slope or y intercept given) and nothing else. SO how do I get their equations?
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  5. #5
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    L_1:~y-b=2(x-a)
    L_2:~y-b=3(x-a)

    There. You now have two lines.
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  6. #6
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    Quote Originally Posted by Plato View Post
    L_1:~y-b=2(x-a)
    L_2:~y-b=3(x-a)

    There. You now have two lines.
    Thanks but I just want to understand more. 2 and 3 are the slopes for L1 and L2 respectively? If so, can you please tell me how you got because I want to know. So the equation will be y =2x-2a+b for L1 and y=3x-3a+b for L2? And what are a and b? Lastly, s it it not possible for us to get the equation in the form y=mx+b? Thank you and sorry for asking too many questions.
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  7. #7
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    Quote Originally Posted by Keep View Post
    Thanks but I just want to understand more. 2 and 3 are the slopes for L1 and L2 respectively? If so, can you please tell me how you got because I want to know.
    I just made them up.
    You can use any two numbers.
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    Quote Originally Posted by Plato View Post
    I just made them up.
    You can use any two numbers.
    So 2 and 3 are the slopes of the lines that you get by choosing any two numbers, right? OK, I understand that, thanks. But what is a and b? b is the y intercept? If b is y intercept, then what is a?
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    Quote Originally Posted by Keep View Post
    So 2 and 3 are the slopes of the lines that you get by choosing any two numbers, right? OK, I understand that, thanks. But what is a and b? b is the y intercept? If b is y intercept, then what is a?
    If y-b=m(x-a) then y=mx+(\color{blue}b-ma).
    So \color{blue}b-ma is the y-intercept.
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    Quote Originally Posted by Plato View Post
    If y-b=m(x-a) then y=mx+(\color{blue}b-ma).
    So \color{blue}b-ma is the y-intercept.
    Ok but the equatio says that y intrecept is b minus the product of the slope and a. My question is, what do these two variables (a and b) stand for. I mean we know what x, y and m are but what are a and b? Lastly is it not possible for me to give the lines in the usual form of y=mx+b without b being equal to (b-ma)?
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  11. #11
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    Quote Originally Posted by Keep View Post
    Ok but the equatio says that y intrecept is b minus the product of the slope and a. My question is, what do these two variables (a and b) stand for. I mean we know what x, y and m are but what are a and b? Lastly is it not possible for me to give the lines in the usual form of y=mx+b without b being equal to (b-ma)?
    Or maybe a and b are the points of intersection which I gave as 6 and 18?
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  12. #12
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    Quote Originally Posted by Keep View Post
    My question is, what do these two variables (a and b) stand for. ?
    I have no idea what they stand for other that the coordinates of a point in the plane.
    They simply were part of the given in the problem.
    y=mx+(b-ma) is in the form y=mx+B, where B=b-ma.
    Last edited by Plato; March 24th 2009 at 12:56 PM.
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    Quote Originally Posted by Plato View Post
    I have no idea what they stand for other that the coordinates of a point in the plane.
    They simply were part of the given in the problem.
    y=mx+(b-ma) is in the form y=mx+B, where B=b+mx.
    Oh OK I get it now. I see. I know the first time I gave the coordinates as (a,b) but later on I put them as (6,18). So let us say we have the intersection point as (6,18). Using the equation which you gave which was:
    L_1:~y-b=2(x-a)
    L_2:~y-b=3(x-a) we get:

    L1: y=2x+6
    L2: y=3x

    Am I right?
    Last edited by Keep; March 24th 2009 at 01:23 PM.
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    Do you understand that there are an infinite number of lines passing through the given point? You only asked for two of them. Any line passing through point (x_0, y_0) can be written in the form y= m(x- x_0)+ y_0. Pick any two numbers for m and you get two different lines through (x_0,y_0).
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  15. #15
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    Quote Originally Posted by HallsofIvy View Post
    Do you understand that there are an infinite number of lines passing through the given point? You only asked for two of them. Any line passing through point (x_0, y_0) can be written in the form y= m(x- x_0)+ y_0. Pick any two numbers for m and you get two different lines through (x_0,y_0).
    Ok, thanks. So if I have a point of intersection of the two lines (L1 and L2) as (6,18) and I pick m=2 for L1 and m=3 for L2, the equation will be y=2(x-6)+18 for L1 and y=3(x-6)+18 for L2, and I will get y=2x+6 for L1 and y=3x for L2.
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