Find Third point in right triangle given two points and a length I am trying to find the third point in a right triangle. I am given Points A and B, and the distance from B to C, but I don't know how to find the coordinates of point C.
An Image of the problem described (Excuse my sub-Par GeoGebra skills)
 A: Outline:
1) We know the coordinates of points $A=(u,v)$ and $B=(w,z)$.
So Q1) What is the slope, $m$, of the line $\overline{AB}$?
Answer: 

 If $A = (u,v)$ and $B= (w,z)$ then the slope is $m = \frac {z-v}{w-u}$.

2) We know the $\overline{BC}$ is a right angle to $\overline{AB}$.
So Q2) what is the slope, $m_2$, of the line $\overline{BC}$?
Answer:

 If the slope of $\overline{AB}=m$.  Then the slope of $\overline {BC} = -\frac 1m$.  Call it $m_2 = -\frac 1m = -\frac {w-u}{z-v}$.

3) Given $m_2$ is the slope of $\overline{BC}$ and that $B = (w,z)$ 
So Q3) What is the formula for the line $\overline{BC}$ in terms of variables $x$ and $y$?

 The slope of $\overline{BC}$ is $m_2=\frac {rise}{run}$ is if $x$ is $x-u$ away from $u$ then if $y$ is $y-v$ away from $y$ then $m_2 =\frac {y-v}{x-u}$ or in other words $y-v = m_2(x-u)$.  Or in slope intercept form $y = m_2x + (v-m_2u)$.  But we don't need slope intercept form.

4) So if $B= (u,v)$ and $C$ is an unknown $(x,y)$ then 
Q4) what is formula for the distance $d$ between $(u,v)$ and $(x,y)$.
Answer:

 $d = \sqrt{(x-u)^2 +(y-v)}$

5) So if $d$ is a known value for the distance between $B=(u,v)$ and $C$ the unknown $(x,y)$ and we have the line formula for the line $\overline{BC}$,
Q5) How do we solve for $x$ and how do we solve for $y$
Answer for $x$:

 $d = \sqrt{(x-u)^2 + (y-v)^2}$ and $y-v= m_2(x-u)$.  So $d =\sqrt{(x-u)^2 + m_2^2(x-u)^2}=\sqrt{(1+m_2^2)(x-u)^2}=|x-u|\sqrt{1+m_2^2}$.  If we assume $u > x$ (from the diagram) we have $d= (x-u)\sqrt{1+m_2^2}$.  So $x-u = \frac d{\sqrt{1+m_2^2}}$ and $x =\frac d{\sqrt{1+m_2^2}}+u$.

Answer for $y$:

 $x-u = \frac d{\sqrt{1+m_2^2}}$ and $y-v = m_2(x-u)$ so $y-v=\frac {m_2d}{\sqrt{1+m_2^2}}$ and $y = \frac {m_2d}{\sqrt{1+m_2^2}}+v$.

6)  So what is $C=(x,y)$?

 $C = (x,y) = (\frac d{\sqrt{1+m_2^2}}+u,\frac {m_2d}{\sqrt{1+m_2^2}}+v)=$ 

.

 $(\frac d{\sqrt{1 + (-\frac 1{m})^2}} +u, -\frac dm\frac 1{\sqrt{1+(-\frac 1m)^2}} + v) =$

.

 $(\frac d{\sqrt{1 + (\frac{z-v}{w-u})^2}}+ u, -\frac {d(w-u)}{(z-v)\sqrt{1 + (\frac{z-v}{w-u})^2}}+v)$

A: Dot product of $C-B$ with $A-B$ is zero.
Square of distance $BC$ is known.
Write down these two equations and solve for the coordinates of $C$.
A: Let us assume that the coordinates of A and B are :
$$ A \equiv (x_1,y_1)  ; B \equiv (x_2,y_2)$$
The the slope $m_1$ of line $\overline {AB}$ is :
$$ m_1 = \frac{y_2 - y_1}{x_2 - x_1}$$
Let as assume that :
$$ C \equiv (x,y) $$
$$ \overline{AB} \perp \overline{BC} $$
Slope of $\overline{BC}$ is $m_2$
Since as stated above $ \overline{AB} \perp \overline{BC} $ therefore $m_1\cdot m_2 = -1$
Therefore, $$ m_2 = - \frac{x_2 - x_1}{y_2 - y_1}= \tan \theta       --------(i)$$ 
Now let the distance BC be r , then applying parametric form of line :
$$ \frac{x - x_2}{\cos\theta} = \frac{y - y_2}{\sin\theta} = r $$
Now you can just plug in $x_2 , y_2, \cos\theta , \sin\theta $ (from eq (i)) and then find x and y.
Hope this helps. If you are unclear with the formulas of forms of line used you can refer to any good book regarding co-ordinate geometry
A: Let $A(a,1,a_2), B(b_1,b_2), d(B,C)=:d$, and $C(x,y)$.
1)Slope of $\overline {AB}$:
$m=\dfrac{b_2-a_2}{b_1-a_1}$.
2) Slope of $\overline {BC}$: $-1/m$ .
3) Equation of line $\overline{BC}$:
$y-b_2=-1/m(x-b_1)$.
4) Distance $d= \sqrt{(y-b_2)^2+(x-b_1)^2}$.
5)$d^2=(y-b_2)^2+m^2(y-b_2)^2$.
6) Solve 5) for $y$:
$(y-b_2)^2=\dfrac{d^2}{1+m^2}$.
$y_{1,2} = b_2 \pm \sqrt{\dfrac{d^2}{1+m^2}}$.
7) Combine with 3) to find corresponding $x_{1,2}$.
$x_{1,2}= m(b_2-y_{1,2})+b_1$.
