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Thread: Proving that there is no such point on a line

  1. #1
    Junior Member gusztav's Avatar
    Jan 2008

    Proving that there is no such point on a line


    I'd greatly appreciate any help with the following problem:

    Let $\displaystyle l$ be a line and $\displaystyle P$ a point not on the line.
    Prove that there is no such point $\displaystyle T_0 \in l $ such that

    $\displaystyle (\forall T \in l) $_____$\displaystyle d(P,T) \leq d(P, T_0)$.

    In other words, prove that there is no point $\displaystyle T_0$ on the line, such that the distance from $\displaystyle T_0$ to $\displaystyle P$ is greater than the distances from all the other points of the line to the point $\displaystyle P$.

    Just to illustrate the problem, we may take $\displaystyle T_0 \in l$, and in fact $\displaystyle d(T_1,P)\leq d(T_2,P) \leq d(T_3,P) \leq d(T_0,P)$.

    However, that is not always true for $\displaystyle T_0$, for there is a point $\displaystyle T_4$ such that $\displaystyle d(T_4,P) \nleq d(T_0,P) $:


    I have tried to prove this problem by supposing that the opposite is true and then arriving at contradiction, and tried to use the triangle inequality $\displaystyle (d(A,B) \leq d(A,C) + d(C,B))$ somewhere on the way, but so far none of my attempts have been successful.

    Thanks a lot!
    Last edited by gusztav; Jan 12th 2010 at 03:58 PM.
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  2. #2
    Senior Member nikhil's Avatar
    Jun 2008

    my thinking

    drop a perpendicular from P on the line (let point of intersection be A).
    (this is the shortest distance between P and line)
    now from P draw a line parallel to the given line and take an arbitrary point B on it.(take initial point of line as P)
    now angle APB=90 degrees.
    if this drawn line want to touch the other line then angle APB must be smaller then 90 degrees.
    suppose we draw a line at the angle which is just smaller than 90 degrees.
    let this angle be x degrees.
    but since an angle greater than x but smaller than 90 can still be obtained which may be given by
    hence a point of intersection (T) belonging to l can always be obtained such that PT>PT(o)
    no point T(o) exist such that PT(o)>PT.

    2) let the equation of line be
    let coordinates of P be (r,t)
    now let the coordinates of perpendicular from P on line be (m,n)
    suppose there is a point T(o) whose distance from T is maximum.
    Let its coordinates be (M,N)
    but another point T can be obtained on l whose coordinates can be given by
    (m+M+c/a,n+N) or (m+M,n+N+c/b) whose distance from P i.e PT will be greater than PT(o)
    there is no point T(o) belonging to l such that PT(o)>PT
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