Finding a point within a 2D triangle I'm not sure how to approach the following problem and would love some help, thanks!
I have a two-dimensional triangle ABC for which I know the cartesian coordinates of points $A$, $B$ and $C$.
I am trying to find the cartesian coordinates of a point $P$. I know the lengths (distances) $PA$, $PB$ and $PC$.
How can I find the coordinates for $P$?
Thanks.

Let $(x,y)$ be the coordinate of $P$ and $(x_i,y_i)$ be the coordinates of $A,B,C$ respec. $(i=1,2,3)$ Since you know the lengths of $PA, PB,PC$, you'll get three equations like $$(x-x_i)^2+(y-y_i)^2=d_{i}^2,\ i=1,2,3$$ Then, subtract the equation $i$ from equation $j$ to get something like $$x(x_i-x_j)+y(y_i-y_j)+x_i^2-x_j^2+y_i^2-y_j^2=d_i^2-d_j^2$$ From these equations solve for $(x,y)$ Actually only two lengths are enough to find out $P$. This method, for $3D$ is called triangulation.
The equation that you'll get from point $A,B$ is $$x(x_1-x_2)+y(y_1-y_2)=-x_1^2-x_2^2+y_1^2-y_2^2+d_1^2-d_2^2\tag{1}$$
Similarly, equation that you'll get from $B,C$ is $$x(x_2-x_3)+y(y_2-y_3)=-x_2^2-x_3^2+y_2^2-y_3^2+d_2^2-d_3^2\tag{2}$$
So the equations are now in the form $$a_1x+b_1y=c_1\\
a_2x+b_2y=c_2$$
which have the solution $$x=\frac{b_2c_1-b_1c_2}{a_1b_2-a_2b_1}\\
y=\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}$$
