Results 1 to 5 of 5

Math Help - Lines in a Plane 10

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    425

    Lines in a Plane 10

    Prove that the shortest distance from a point to a line is the distance measured along the perpendicular from the point to the line.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by skeske1234 View Post
    Prove that the shortest distance from a point to a line is the distance measured along the perpendicular from the point to the line.
    Assume it isn't -- take a point p which is on the line and not perpendicular to the point and creates the shortest distance. Sketch this and try seeing what you can do with the right triangle formed by the original point and p
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    May 2009
    Posts
    471
    Let (p,q) be a point in the plane and y=ax+b be a line in the plane

    Then the distance from (p,q) to any point on the line is \sqrt{(x^2-p^2)+((ax+b)^2-q^2)} here we took x to be x and y to be ax+b

    So we need to minimize the distance

    It should be obvious that to minimize the distance we can minimize the square of the distance, or in other words, we can ignore the radical sign and minimize what is under it

    From calculus, to minimize something look at the derivative when it equals zero

    Let E=the square of the distance, then

    E'=2x+2(ax+b)*a=0

    2x+2a^2x+2ab=0

    x(1+2a)=-ab

    x=-\frac{ab}{2a+1}

    Now use the second derivative test to confirm this is a minimum and not a maximum

    E''=2+2a^2 this quantity is always positive, as a^2 is always positive for all a, so we do have a local min

    This means that the closest point to our point (p,q) on the line is (-\frac{ab}{2a+1}, a(-\frac{ab}{2a+1})+b)=(-\frac{ab}{2a+1}, -\frac{a^2b}{2a+1}+b)

    Now all you have to do is come up with the line perpindicular to our line that goes through point (p,q) and show that it contains the point we just found



    *there is probably an easier, shorter way that I can't come up with at the moment, but whenever I see prove things about min and max, my brain jumps to calculus... sorry
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    Posts
    425
    Quote Originally Posted by Defunkt View Post
    Assume it isn't -- take a point p which is on the line and not perpendicular to the point and creates the shortest distance. Sketch this and try seeing what you can do with the right triangle formed by the original point and p
    ok, so this is what I have.. can you or someone check to see if my method of "proving" this is valid?

    assuming it isn't true: (like defunkt says)
    Let point be (5,4)
    A be (0,0)
    B be (-1,0)
    let theta be 30 degrees
    cos(theta)=a/h
    h=a/(costheta)
    h=sqrt1/cos30
    =1.15

    now assuming it is true: (what the question asks to prove)

    Let point be (5,4)
    A be (0,0)
    B be (1,0)
    let theta be 30 degrees
    tan(theta)=o/a
    atan(theta)=o
    sqrt1 (tan30)=o
    0.577=o

    so the shortest distance from a point to a line is the distance measured along the perpendicular from the point to the line (which is the second method above)

    So let me know if this is a valid method of proving this question.
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    May 2009
    Posts
    471
    if you haven't taken a proofs class yet then proofs can be very challenging but you cannot prove a general statement with examples... you've only shown it holds for a very specific case and not for all lines ax+b and points (p,q),

    if you are going to assume it isn't you still have to be general and use variables not numbers
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Lines in a Plane 13
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 8th 2009, 11:47 AM
  2. Lines in a Plane 7
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 6th 2009, 04:40 AM
  3. Lines in a Plane 13
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 5th 2009, 11:22 PM
  4. Lines in a Plane 14
    Posted in the Calculus Forum
    Replies: 0
    Last Post: September 5th 2009, 06:14 PM
  5. n lines in a plane
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: June 21st 2008, 10:59 AM

Search Tags


/mathhelpforum @mathhelpforum