analytic geometrygeometrytriangles
I have a simple right-angle triangle $\triangle ABC$. I know the coordinates of $B$ and $C$ and the distances $AB = c$, $AC = b$ and $BC = a$. How can I find the coordinates of point $A$. For example, if $B = (4,3)$ and $C = (4,0)$ and the distance from A to B is $5$, how can I get the coordinates of $A$, which in this case $A = (0,0)$.
Edit:
The $AC$ and $BC$ Wont always be parallel to the $x$ and $y$ axis respectively
You need to find the coordinates of a point $D(x,y,z)$. Then, you need to make three equations (in order to find the three variables $x$, $y$ and $z$. Solve the system of equations and you´ll get the value of $x$, $y$ and $z$.
Construct each equation by the information given, for instance:
$$\tag {1} d_{AD}=\sqrt {(x-x_A)^2+(y-y_A)^2+(z-z_A)^2}$$
If $d_{AD}$ is the distance from $A(x_A,y_A,z_A)$ to $D(x,y,z)$
The same thing with:
$$\tag {2} d_{BD}=\sqrt {(x-x_B)^2+(y-y_B)^2+(z-z_B)^2}$$ $$\tag {3} d_{CD}=\sqrt {(x-x_C)^2+(y-y_C)^2+(z-z_C)^2}$$
Solve the system of the equations (1), (2) and (3) and find the values of $x$, $y$ and $z$
Working with complex numbers, and assimilating a point with its associated complex number, we can write :
$$A=D+d \exp(i\theta_D), \ \ B=E+e \exp(i \theta_E), \ \ C=F+f \exp(i \theta_F)\tag{1}$$
for certain angles, giving the 3 constraints :
$$\begin{cases}|E-D+e \exp(i \theta_E)-\exp(i\theta_D)|&=&c\\|F-E+e \exp(i \theta_F)-\exp(i\theta_E)|&=&a\\ |D-F+e \exp(i \theta_D)-\exp(i\theta_F)|&=&b\end{cases}\tag{2} $$
If you want an approximate solution, you can use brute force : divide for example each circle into 360°, and consider the $360^3$ possibilities for $(\theta_E,\theta_F,\theta_G)$ (three embedded "for" loops) and test for each possibility whether the LHS of the 3 equations is close or not to its RHS (of course, this must be turned into a more effective formulation with a certain threshold, etc...). Hopefully, one of the combinations (sometimes more) will be in agreement with a at most one degree tolerance.
If you want arbitrary precision, you must turn to iterative methods (multidimensional Newton). If this case, one should work on a system of 3 real equations with 3 unknowns that are obtained by squaring all constraints in (3),
No hope (IMHO) that there is an explicit solution.
Remark : a particular case : there is at least a case where the conditions given provide an infinite number of solutions : it is when ABC and DEF have identical lengthes and distances $d=e=f$ (see diagram below) :
Fig. 1 : Triangle DEF in black. An infinite number of positions of triangle ABC are possible.
Another case where there are (at least) 2 solutions
$D(2,0), E(6,4), F(5,1)$ with $AB=4\sqrt{2}, AC=BC=4, d=e=2, f=\sqrt{2}$
You can have $A(0,0),B(4,4),C(4,0)$ or $A(2,2),B(6,6),C(6,2).$
Best Answer
Hint:
You can compute the distance $BC = \sqrt{\left(x_c - x_b\right)^2 + \left(y_c - y_b\right)^2}$, where $B = \left(x_b, y_b\right)$, and $C = \left(x_c, y_c\right)$. Because you now know $AB$ and $BC$, you can find the third length $AC$ with the help of the Pythagora's Theorem, i.e., $AC = \sqrt{AB^2 - BC^2}$. Finally, let $A = (x_a, y_a)$. Then, $$\begin{align}AC &= \sqrt{\left(x_c - x_a\right)^2 + \left(y_c - y_a\right)^2}\tag{1}\\AB &= \sqrt{\left(x_b - x_a\right)^2 + \left(y_b - y_a\right)^2}\tag{2}\end{align}$$ Solve the equations simultaneously to get the values for variables $x_a$ and $y_a$.