Results 1 to 7 of 7

Math Help - Absolute value (similar to triangle ineq.)

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    64

    Mid-coordinate value

    How can I show that (0, \frac{1}{2}) is the equidistant point from the 3 points: A(2,2), B(2,-1) and C(-2,-1). I know that (0, \frac{1}{2}) is the midpoint of AC but how can you conclude it's equidistant to B? (is it because of the midpoint of AB?)
    Last edited by dannyc; March 24th 2010 at 01:42 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,423
    Thanks
    1331
    Quote Originally Posted by dannyc View Post
    How can I show that (0, \frac{1}{2}) is the equidistant point from the 3 points: A(2,2), B(2,-1) and C(-2,-1). I know that (0, \frac{1}{2}) is the midpoint of AC but how can you conclude it's equidistant to B? (is it because of the midpoint of AB?)
    You can't- it's not equidistant from the three points. You can show that by calculating the three distances from that point to A, B, and C. The distance from (0,1/2) to A and B is 3/2 for both but the distance from (0, 1/2) to C is NOT 3/2. To find the point equidistant from three point, average the coordinates of all three points- ((2+ 2+ 1)/3, (2- 1- 1)/3) ("averaging" does not work for more than three points). To show that is the equidistant point, again- calculate the three distances.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2010
    Posts
    64
    I'm still confused Ivy..I'm not sure how to proceed? So I should calculate the distances? The actual question is what point is equidistant to the 3 points (we aren't actually supposed to know it's (0, 1/2) but I was working backwards before..)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    Quote Originally Posted by dannyc View Post
    I'm still confused Ivy..I'm not sure how to proceed? So I should calculate the distances? The actual question is what point is equidistant to the 3 points (we aren't actually supposed to know it's (0, 1/2) but I was working backwards before..)
    If you show
    AB^2 + BC^2 =AC^2,
    then the mid point of AC is equidistant from A, B and C.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    905
    Thanks
    27

    mid coordinate value

    Quote Originally Posted by dannyc View Post
    How can I show that (0, \frac{1}{2}) is the equidistant point from the 3 points: A(2,2), B(2,-1) and C(-2,-1). I know that (0, \frac{1}{2}) is the midpoint of AC but how can you conclude it's equidistant to B? (is it because of the midpoint of AB?)
    The three points form a 3-4-5 right triangle and the midpoint of the hypothenuse is equidistant to each point.You prove it

    bjh
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,423
    Thanks
    1331
    Quote Originally Posted by dannyc View Post
    I'm still confused Ivy..I'm not sure how to proceed? So I should calculate the distances? The actual question is what point is equidistant to the 3 points (we aren't actually supposed to know it's (0, 1/2) but I was working backwards before..)
    I recommend you go back and read my answer again. I told you exactly how to find the point equidistant from three given points. And it is not a good idea to "work backwards" from a wrong answer.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    Quote Originally Posted by HallsofIvy View Post
    You can't- it's not equidistant from the three points. You can show that by calculating the three distances from that point to A, B, and C. The distance from (0,1/2) to A and B is 3/2 for both but the distance from (0, 1/2) to C is NOT 3/2. To find the point equidistant from three point, average the coordinates of all three points- ((2+ 2+ 1)/3, (2- 1- 1)/3) ("averaging" does not work for more than three points). To show that is the equidistant point, again- calculate the three distances.
    Distance between (0,1/2) and A(2,2) is [2^2 + (3/2)^2]^1/2 = 5/2.
    Similarly the distances between the point (0,1/2) and B and C are also 5/2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cheby Ineq and Markov
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: December 13th 2009, 12:35 PM
  2. Similar triangle problem
    Posted in the Geometry Forum
    Replies: 2
    Last Post: November 23rd 2009, 03:35 PM
  3. similar triangle
    Posted in the Geometry Forum
    Replies: 10
    Last Post: May 3rd 2009, 08:15 AM
  4. Linear Algebra.Similar to a triangle matrix.pls help!
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: March 29th 2009, 01:27 AM
  5. Similar Triangle
    Posted in the Geometry Forum
    Replies: 10
    Last Post: January 24th 2009, 03:03 PM

Search Tags


/mathhelpforum @mathhelpforum