Results 1 to 5 of 5

Math Help - Third coordinate of triangle (not right angled)

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    3

    Talking Third coordinate of triangle (not right angled)

    I have a triangle (as per below figure)
    the Known points are all the triangle distances ie A,B,C
    The two vertices are known,x1,y1, x2,y2.

    I require the third vertice formula / equation for x3 and y3

    the angle are not known.(i do not need them now)
    this is not a right angle triangle
    i need this for my programming purpose

    Please help me as soon as possible
    Third coordinate of triangle (not right angled)-drawing1-model.jpg

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by cycleinmars View Post
    I have a triangle (as per below figure)
    the Known points are all the triangle distances ie A,B,C
    The two vertices are known,x1,y1, x2,y2.

    I require the third vertice formula / equation for x3 and y3

    the angle are not known.(i do not need them now)
    this is not a right angle triangle
    i need this for my programming purpose

    ...
    The 3rd point is the point of intersection between the 2 circles around the other 2 points:

    \left| \begin {array}{l}(x-x_1)^2+(y-y_1)^2 = C^2 \\ (x-x_2)^2+(y-y_2)^2 = A^2 \end{array} \right.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    3
    i know this but i do not know how to solve this and how to het x3 and y3
    i have tried lot of times still i am getting a wrong answer
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2010
    Posts
    3

    thanks for formula

    i know this but i do not know how to solve this and how to get x3 and y3
    i have tried lot of times still i am getting a wrong answer
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,602
    Thanks
    1421
    The circle with center (x_1,y_1) and radius C is given by (x-x_1)^2+ (y-y_1)^2= C^2 and the circle with center [tex](x_2, y_2) and radius A is [tex](x-x_2)^2+ (y-y_2)^2= A^2, just as earboth said.

    Multiply both of those out:
    x^2- 2x_1 x+ x_1^2+ y^2- 2y_1 y+ y_1^2= C^2
    x^2- 2x_2 x+ x_2^2+ y^2- 2y_2 y+ y_2^2= A^2

    Subtract the two equations and the squares of both x_1 and x_2 cancel out:
    2(x_2- x_1)x+ (x_1^2- x_2^2)+ 2(y_2- y_1)y+ (y_1^2- y_2^2= C^2- A^2

    You can solve that for y in terms of x:
    y= \frac{C^2- A^2- (y_1^2- y_2^2)- (x_1^2- x_2^2)- 2(x_2-x_1)x}{2(y_2- y_1)}.

    Put that into the equation of either circle and you have one equation to solve for x. Of course, there will be two solutions, one on either side of the line between X_1 and X_2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 23rd 2011, 08:10 AM
  2. in a right angled triangle ABC
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: August 12th 2010, 07:32 AM
  3. Finding coordinate length from angled datum
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: March 5th 2010, 09:30 PM
  4. Replies: 2
    Last Post: December 15th 2009, 12:38 PM
  5. right angled triangle
    Posted in the Geometry Forum
    Replies: 2
    Last Post: October 16th 2009, 07:56 AM

Search Tags


/mathhelpforum @mathhelpforum