Analytic Geometry: Point coordinates, same distance from two points Given are two points, $P1(x_1, y_2)$ and $P2(x_2, y_2)$, and distance $a$. Now I want to find the two points $T1$ and $T2$.
$$d(P1,T1) = d(P2, T1) = a = d(P1,T2) = d(P2, T2)$$
Eg: (T1 and T2 are my two points I want to find)
           T1
          /|\
 len = a / | \ len = a
        /  |L \
      P1 ------ P2
        \  |  /
 len = a \ | / len = a
          \|/
           T2

len = a The distance between the points is given (they are all the same ($a$))
L Angle of 90 degrees.
I've been trying to solve this equation for an hour. I need this for an application, so I need an abstract formulla, in which I can insert the given $x_1, y_1, x_2, y_2$.
Thanks
 A: The point at the centre of the diagram has coordinates $(x_1+x_2)/2$ and $(y_1+y_2)/2$. The distance $d$ from each of $P_1$ and $P_2$ to the centre is half the distance between $P_1$ and $P_2$, so $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}/2$. Then the distance $h$ from each of $T_1$ and $T_2$ to the centre follows by Pythagoras: $h=\sqrt{a^2-d^2}$. The vector with components $y_2-y_1,x_1-x_2$ is orthogonal to the line $P_1P_2$ and thus points along the line $T_1T_2$. Thus you just have to add $\pm h$ times a normalized version of that vector to the centre to get $T_1$ and $T_2$:
$$
\begin{eqnarray}
T_{1,2}
&=&
\frac12\left({x_1+x_2\atop y_1+y_2}\right)\pm\frac{\sqrt{a^2-\frac14((x_2-x_1)^2+(y_2-y_1)^2)}}{\sqrt{(x_1-x_2)^2+(y_2-y_1)^2}}\left(y_2-y_1\atop x_1-x_2\right)
\\
&=&
\frac12\left({x_1+x_2\atop y_1+y_2}\right)\pm{\sqrt{\frac{a^2}{(x_2-x_1)^2+(y_2-y_1)^2}-\frac14}}\left(y_2-y_1\atop x_1-x_2\right)
\end{eqnarray}
$$
A: Another way of looking at it:


*

*Knowing the coordinates $(x_1,y_1)$ and $(x_2,y_2)$, it is easy to construct the equation for the perpendicular bisector of the segment joining those two points:


$$\frac{y-\frac{y_1+y_2}{2}}{x-\frac{x_1+x_2}{2}}=-\frac{x_2-x_1}{y_2-y_1}$$


*

*The perpendicular bisector also intersects the circles of radius $a$ centered at the two given points, $(x-x_1)^2+(y-y_1)^2=a^2$ (and similarly for the other circle).


If you solve these two simultaneous equations (you only need to pick one of the two circles), you should obtain the same results as in joriki's answer.
