1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Symmetric point of line

  1. May 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the coordinates of the symmetric point of the point M(2,1,3) of the line


    2. Relevant equations

    3. The attempt at a solution

    Out from here:



    [tex]M_1(-2,-1,1) ; M_2(-1,1,0)[/tex]

    I got two conditions

    lets say that the point we need to find is N.




    How will I find the 3-rd condition? I tried also with normal distance from M to the line to be equal with the normal distance of N to the line... Please help... Thank you.
  2. jcsd
  3. May 18, 2008 #2
    Ok I solved this, using the 3-rd condition MN=2*distance from the point to line... But I have another task:

    Find point at equal distance from the points A(3,11,4) and B(-5,-13,-2) at the line
    x+2y-z-1=0 & \\
    3x-y+4z-29=0 &

    I find the line using x=0.

    The equation of the line is:


    Also I got:


    And I put the conditions in one system:
    \sqrt{(x-3)^2+(y-11)^2+(z-4)^2}=\sqrt{(x+5)^2+(y+13)^2+(z+2)^2} & \\
    \frac{x}{7}=\frac{y-\frac{33}{7}}{-7}=\frac{z-\frac{59}{7}}{-7} &

    I get that point [tex](\frac{664}{77} ; \frac{-43}{11} ; \frac{-15}{77}[/tex]

    And in my text book they got: [tex](2,-3,5[/tex]

    Is my way correct?
  4. May 19, 2008 #3
    tiny-tim, HallsofIvy, can you confirm me?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?