Isosceles triangle $ABC$
$AB = AC = d_1$
$BC = d_2$
$A = (x_1, y_1)$
$B = (x_2, y_2)$
$C = (x_3, y_3)$
$\angle BAC = \phi$
$\angle ABC =\angle ACB = \theta$
I want an equation for $x_3$ and $y_3$ (and I know there will be two values)
[Math] How to get the third point coordinates in isosceles triangle
calculuscomputational geometrygeometrytrianglestrigonometry
Related Solutions
There are many solutions for $B$:
Draw a circle trough the points $A$ and $C$, with diameter $|AC|$, then all points on that circle except $A$ and $C$ are solutions
This is called Thales' theorem
We can find the points on this circle by first finding the equation of that circle with center
$$O=\dfrac{A+C}{2}$$
Taking your example: $A=(4,3)$ and $C=(2,1)$ we find that
$$O=\dfrac{(4,3)+(2,1)}{2}=\dfrac{(6,4)}{2}=(3,2)$$
And the radius of the circle is $|AO|=\sqrt{1^2+1^2}=\sqrt{2}$
So the equation of that circle is $$(x-3)^2+(y-2)^2=2$$
All points $B=(x,y)$ that satisfy this equation, except $A$ and $C$ make a right triangle with your given points.
Let's solve the equation for $y$:
$$y=\pm \sqrt{2-(x-3)^2}+2$$
So choose a value for $x$ but make sure the part under the square root will not be negative, and this will give you two valid values for $y$!
Example: choose $x=3$ then the formula gives $y=\pm \sqrt{2}+2$ so $$B=(3,\sqrt{2}+2)$$
Is one of many solutions.
Hint. Let $$f(u,v,w):=(v-w)^2(p-q)(p-r)+(w-u)^2(q-r)(q-p)+(u-v)^2(r-p)(r-q)\,.$$ Observe that $$\begin{align}f(u,v,w)&=\big((v-w)p\big)^2+\big((w-u)q\big)^2+\big((u-v)r\big)^2\\&\phantom{aaaaa}+2\,\big((w-u)q\big)\,\big((u-v)r\big)+2\,\big((u-v)r\big)\,\big((v-w)p\big)+2\,\big((v-w)p\big)\,\big((w-u)q\big)\,.\end{align}$$ Therefore, $$f(u,v,w)=\Big(\det\big(M(u,v,w)\big)\Big)^2\,,$$ where $$M(u,v,w):=\begin{bmatrix}1&p&u\\1&q&v\\1&r&w\end{bmatrix}\,.$$
Best Answer
I could solve this. The solution key was the conversion between the Cartesian Coordinate System and the Polar Coordinate System.
$x_3 = AB * Cos(ϕ) + x_1$
$y_3 = AB * Sin(ϕ) + y_1$