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Math Help - Finding The Distance Between Parallel Lines...?

  1. #1
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    Finding The Distance Between Parallel Lines...?

    How do you find the distance between the two parallel lines:

    y = 4x+6
    y = 4x-9

    A lot of people think it's 15, which sounds plausible, but the answer REALLY is 3.64. My teacher says so. My textbook says so. But I don't know how to arrive at that answer in a step by step process...

    Is anyone smart enough to do this question?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by AlphaRock View Post
    How do you find the distance between the two parallel lines:

    y = 4x+6
    y = 4x-9

    A lot of people think it's 15, which sounds plausible, but the answer REALLY is 3.64. My teacher says so. My textbook says so. But I don't know how to arrive at that answer in a step by step process...

    Is anyone smart enough to do this question?
    0=4x+6\Rightarrow{x=\frac{-3}{2}}
    0=4x-9\Rightarrow{x=\frac{9}{2}}
    \frac{3}{2}+\frac{9}{2}=3.75
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    0=4x+6\Rightarrow{x=\frac{-3}{2}}
    0=4x-9\Rightarrow{x=\frac{9}{2}}
    \frac{3}{2}+\frac{9}{2}=3.75
    I also thought it'd be 3.75, but suprisingly, it's 3.64.

    My friend explained this to me hour ago, but I accidentally exited the conversation.

    Thanks for your help though. I thanked you for your effort.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by AlphaRock View Post
    I also thought it'd be 3.75, but suprisingly, it's 3.64.

    My friend explained this to me hour ago, but I accidentally exited the conversation.

    Thanks for your help though. I thanked you for your effort.
    It is 3.75...I think your book may be wrong
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    It is 3.75...I think your book may be wrong
    Probably, but if it helps... it included tangent that opp is 4 times bigger than four or something. And that the tangent = slope.

    and x^2 + #x^2 = 225
    17x^2 = 225
    4.12x = 15
    x = 3.64


    These are all that I can remember. I can't remember the steps or understanding in between. Perhaps you can.
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  6. #6
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    I REALLY don't know where the X^2 or 17 came from. If somebody could explain everything clearly... I would appreciate it.
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  7. #7
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    What Mathstud found was the horizontal distance between two points of the two lines which isn't necessarily the shortest distance.

    Recall the distance formula from a point to the line (presumably the shortest):
    d = \frac{|Ax_{0} + By_{0} + C|}{\sqrt{A^{2} + B^{2}}}

    where (x_{0},y_{0}) is your point and Ax + By + C = 0 is your line.

    Now, let's pick a point from y = 4x + 6 ... let's say (0,6). This will be our point.

    From our other line y = 4x - 9, we must but it in standard form (Ax + By + C = 0):
    y = 4x - 9 \: \Rightarrow \: 0 = 4x - y - 9

    Now apply the formula:
    d = \frac{|4(0) + (-1)(6) + (-9)|}{\sqrt{(4)^{2} + (-1)^{2}}} \approx 3.638
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  8. #8
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    Also, note that the shortest distance between two lines is a perpendicular line that intersects both.

    Start with y = 4x + 6. Let's use the point (0,6). We want to find the equation of a line perpendicular to y = 4x + 6 that goes through (0,6). Perpendicular lines have negative reciprocal slopes, so our line will have a slope of -1/4 - note that it has the same y-intercept.

    So our equation is y = (-1/4)x + 6. To see where that line intersects our OTHER line (in other words, to find the point on the other line closest to (0,6), set the two equal to one another.

    (-1/4)x + 6 = 4x - 9. Solve for x. It isn't pretty, but you get x = 60/17. Plug that in to either our perpendicular line's equation or the second equation to get y, which is 87/17. So now we have these two points, (0,6) and (60/17, 87/17), and we know, since they're on a perpendicular line that intersects our two original lines, that they're as close as possible. But how close?

    The distance formula:

    {\sqrt{\frac{60}{17}^2 + \frac{15}{17}^2}\approx 3.638}

    Admittedly, this is a longer method than the one o_O proposed, but I never remember the equation for the distance between a point and a line - this particular method relies on a little more "basic" geometrical concepts and formulas. And, having multiple methods to solve a problem is never a bad thing.
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  9. #9
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    Finding The Distance Between Parallel Lines...?-parallel-lines.jpg

    The triangles in my diagram are similar.

    Using this fact you can label one triangle with sides 15, 4x and x and then use Pythagoras' theorem.

    x^2+(4x)^2=15^2

    17x^2=15^2

    x=\frac{15}{\sqrt{17}}

    x=3.638..
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  10. #10
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    Does this apply to any dimension.if we have a1*x1+a2*x2+...an*xn=b1 and a1*x1+a2*x2+...an*xn=b2.
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  11. #11
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    Yes

    Yes, the formula works for higher dimensions as well

    If we have a hyperplane wx=b, then the distance from point x1 (which is a vector) to the line is given by:

    |wx-b|/||w||^2

    Remember w is also a vector of weights so ||w||= w1^2+w2^2+.....
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