Page 1 of 2 12 LastLast
Results 1 to 10 of 12

Thread: finding distances between pairs of parallel lines

  1. #1
    Junior Member
    Join Date
    Mar 2006
    Posts
    65

    finding distances between pairs of parallel lines

    Ok well, I don't have a very good math teacher at all, so I find myself basically teaching myself math by reading in the book (which we aren't allowed to bring home because we don't have enough). So anyway, please help me on this and any help at all will be greatly appreciated:

    1) Find the distance between each pair of parallel lines:

    y = x + 9
    y = x + 3

    2) Find the distance between each pair of parallel lines:

    y = -5x
    y = -5x + 26

    I have other problems, but once I understand how to do these two, I think I can figure out how to do them...Thanks in advance for any help.
    there are three kinds of people in the world, those who can count, and those who cant

  2. #2
    Full Member
    Join Date
    Jun 2006
    Posts
    553
    Hi Cole92!

    A good way to do this is to graph this equations and improvise a way to solve it.

    Here's the graph:



    So count how many units up the lower line is from the upper line (look at the y - axis). Or you could use the distance formula knowing your points since the slope and y - intercept is constant.

    You can derive the points [tex]\{(0,3)\,,\,(0,9)\}[/tex] because they are you y - intercepts.

    So now use the distance formula:[tex]\L \;d\,=\,\sqrt{{(x_1\,-\,x_2)}^2\,+\,{(y_1\,-\,y_2)}^2}[/tex]

    ....To get [tex]\L \;d\,=\,6[/tex]

    Think you can do the second one now?
    Try your best in everything and don't be scared to do something different.

  3. #3
    Elite Member
    Join Date
    Apr 2005
    Location
    USA
    Posts
    8,905
    Except that distance normally would be defined at PERPENDICULAR distance.

    You might be shooting for [tex]3\sqrt{2}[/tex] and [tex]\sqrt{26}[/tex].

    Read your definition of "Distance" and proceed according to that definition.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  4. #4
    Full Member
    Join Date
    Jun 2006
    Posts
    553
    Quote Originally Posted by tkhunny
    Except that distance normally would be defined at PERPENDICULAR distance.

    You might be shooting for [tex]3\sqrt{2}[/tex] and [tex]\sqrt{26}[/tex].

    Read your definition of "Distance" and proceed according to that definition.
    How did you get those answers?
    Try your best in everything and don't be scared to do something different.

  5. #5
    Junior Member
    Join Date
    Mar 2006
    Posts
    65
    Yes, I believe it is supposed to be perpendicular.

    Ok, from what I am understanding, this is how I did number two...

    y = -5x ----->slope is -5
    y(2) = -5x + 26 ------> made slope 1/5

    then...

    I made the equation y(2) = 1/5x + 26 ...I get the points (0,26)

    Then I plugged in:

    -5x = 1/5x + 26
    solve...
    x = -5 [gives me points (-5,?)]

    Finally I plugged in -5 (my answer for x) into the first equation and got 25 [making my second set of points (-5,25)]

    Then I did the distance formula using points (0,26) and (-5,25)...and my answer was:

    Square Root of 26?

    (I'm sorry if this is wrong. I'm really confused by my teacher and am trying to put my notes with what you guys are showing me).
    there are three kinds of people in the world, those who can count, and those who cant

  6. #6
    Elite Member
    Join Date
    Jan 2005
    Posts
    7,400
    Given [tex]T:(p,q)\quad \& \quad Ax + By + C = 0[/tex], a point and a line, then the distance from the point to the line is given by: [tex]D = \frac{{\left| {Ap + Bq + C} \right|}}{{\sqrt {A^2 + B^2 } }}.[/tex]

    Now with two parallel lines, take any point on one of the lines then find its distance to the other line.
    “A professor is someone who talks in someone else’s sleep”
    W.H. Auden

  7. #7
    Elite Member
    Join Date
    Apr 2005
    Location
    USA
    Posts
    8,905
    y = x + 9
    y = x + 3 ==> x - y + 3 = 0

    Pick ANY point on the first, (0,9).
    Calculate the distance to the other [tex]\frac{|1(0)-1(9)+3|}{\sqrt{(1)^{2}+(-1)^{2}}}\;=\;\frac{6}{\sqrt{2}}\;=\;3\sqrt{2}[/tex]

    y = -5x
    y = -5x + 26 ==> 5x + y - 26 = 0

    This one's even easier, since it contains the Origin. Pick (0,0)
    Calculate the distance to the other [tex]\frac{|5(0)+1(0)-26|}{\sqrt{(5)^{2}+(1)^{2}}}\;=\;\frac{26}{\sqrt{2 6}}\;=\;\sqrt{26}[/tex]

    That's about it.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  8. #8
    Junior Member
    Join Date
    Mar 2006
    Posts
    65
    Great so then my answer to number two is correct. Ok guys thanks for all of your help...I think I get it now.
    there are three kinds of people in the world, those who can count, and those who cant

  9. #9
    Junior Member
    Join Date
    Mar 2006
    Posts
    65
    One last question, I think I totally screwed this one up:

    y = -x - 7
    7 = -x - 9
    there are three kinds of people in the world, those who can count, and those who cant

  10. #10
    Senior Member skeeter's Avatar
    Join Date
    Dec 2005
    Location
    Fort Worth, TX
    Posts
    2,416
    Quote Originally Posted by cole92
    One last question, I think I totally screwed this one up:

    y = -x - 7
    7 = -x - 9
    did you mean for the second equation to be y = -x - 9 ?

    if so, the perpendicular distance between the lines is [tex]\L \sqrt{2}[/tex].

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •