# Third coordinate of triangle (not right angled)

• Jan 31st 2010, 12:50 AM
cycleinmars
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

Attachment 15126

http://www.mathhelpforum.com/math-he...1&d=1264931248
• Jan 31st 2010, 01:21 AM
earboth
Quote:

Originally Posted by cycleinmars
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:

$\left| \begin {array}{l}(x-x_1)^2+(y-y_1)^2 = C^2 \\ (x-x_2)^2+(y-y_2)^2 = A^2 \end{array} \right.$
• Jan 31st 2010, 01:27 AM
cycleinmars
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
• Jan 31st 2010, 01:28 AM
cycleinmars
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
• Jan 31st 2010, 02:26 AM
HallsofIvy
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 $X_2$.