analytic geometrydeterminanteuclidean-geometrytrigonometry
In geometry the in-center is the point of intersection/concurrence of angle bisectors;
If I have a triangle with three vertices
$$A=(x_a,y_a), B=(x_b,y_b), C=(x_c,y_c),$$
known to be related by
$$\left( \frac{ax_a + bx_b + cx_c}{a+b+c};\frac{ay_a + by_b + cy_c}{a+b+c} \right) \tag 1$$
how can I find/derive the above formula $(1)$ using the determinant of a matrix or by any other means?
$\def\A{{\bf A}} \def\B{{\bf B}} \def\C{{\bf C}} \def\R{{\bf R}} \def\D{{\bf D}} \def\f{\phi}$As others have mentioned, if the task is to find the coordinates of $B$ and $C$, this problem is underdetermined.
Let's first consider the triangle with point $A$ located at the origin and point $C$ lying along the positive $x$-axis. Then the coordinates of the points can be found with simple trigonometry, $$\begin{eqnarray*} \A_0 &=& (0,0) \\ \B_0 &=& (c \cos A, c\sin A) \\ \C_0 &=& (b,0). \end{eqnarray*}$$ The angle $A$, and the sides $b$ and $c$ have been given. This is an SAS triangle. The triangle can be solved by finding $a$ with the law of cosines and one of the other angles with the law of sines.
(Added: For completeness, $A = 72^\circ$, $b = 2.61r$, and $c=r$. The law of cosines gives $a = \sqrt{b^2+c^2-2bc\cos A} = 2.49r$. Then $\frac{\sin A}{a} = \frac{\sin B}{b}$ implies $B = 86^\circ$. Lastly, $A+B+C = 180^\circ$ implies $C = 22^\circ$.)
The collection of triangles you are interested in have vertices of the form $$\begin{equation*} \D = \A + \R(\f)\D_0 \tag{1} \end{equation*}$$ where $\A = (A_x,A_y)$ is the given location of $A$, and where $\R(\f)$ is a rotation matrix. The transformation (1) is a counterclockwise rotation by the angle $\f$, followed by a shift so the point $A$ has the given coordinates. In components, $$\begin{eqnarray*} A_x &=& A_x \\ A_y &=& A_y \\ B_x &=& A_x + c\cos A\cos\f - c\sin A\sin\f \\ &=& A_x + c\cos(A+\f) \\ B_y &=& A_y + c\cos A\sin\f + c\sin A\cos\f \\ &=& A_y + c\sin(A+\f) \\ C_x &=& A_x + b\cos \f \\ C_y &=& A_y + b\sin \f. \end{eqnarray*}$$
Below we plot the triangle before rotation and translation in black.
(We set $r = 1$ in the figure.)
The dotted triangle has been rotated counterclockwise by $\f = 30^\circ$, and then translated so the new location of point $A$ is $(2,1)$.
For reference,
$$\begin{eqnarray*}
A_x &=& 2 \\
A_y &=& 1 \\
c &=& r = 1 \\
b &=& 2.61 \\
A &=& 72^\circ \\
\f &=& 30^\circ.
\end{eqnarray*}$$
Hint:
using elemental geometrical properties of equilateral triangles you can see that:
$$ PC=PA=PB=\frac{10\sqrt{3}}{3} $$
and the distance of $P$ from the sides is half this value.
Can you do from this adding or subtracting such values from the coordinates of $P$ to find the vertices?
Best Answer
This may be similar enough to what you're looking for.
Represent the points as vectors $A=(x_a,y_a)$, $B=(x_b,y_b)$, $C=(x_c,y_c)$. Let a point $(p:q:r)$ stand for $$\frac{pA+qB+rC}{p+q+r},$$ so that $A=(1:0:0)$, $B=(0:1:0)$, $C=(0:0:1)$; we want to show that the incenter is $(a:b:c)$. (These are called barycentric coordinates, and the ``extra dimension,'' using three coordinates instead of just two, comes from the fact that if you scale $p$, $q$, and $r$ each by the same amount, the point it represents doesn't change.)
It shouldn't be too hard to convince yourself that lines can, in this system, be represented by a triple $(\alpha,\beta,\gamma)$, and consist of the points $(p:q:r)$ with $\alpha p+\beta q+\gamma r=0$. You can see this by considering the places in which a line intersects each of the sides of the triangle.
Now, let's find the equation of the bisector of $\angle BAC$. We know this line contains $A$, and it also contains the point $D$ on $BC$ with $\frac{BD}{DC}=\frac cb$. This point can be represented as $(0:b:c)$. So, the bisector of $\angle BAC$ is the line $0p+cq-br=0$, or equivalently $\frac qb=\frac rc$.
By symmetry, the three angle bisectors intersect at the point $(p:q:r)$ with $\frac pa=\frac qb=\frac rc$, i.e. the point $(a:b:c)$.
A side-note: You mention looking for a way to show this using determinants. While this isn't as direct, one can show that area in barycentric coordinates is closely related to determinants, in that if you have three points $(p_1:q_1:r_1)$, $(p_2:q_2:r_2)$, and $(p_3:q_3:r_3)$, normalized such that $$p+1+q_1+r_1=p_2+q_2+r_2=p_3+q_3+r_3=1,$$ then the area of the triangle represented by these three points is $$\begin{vmatrix}p_1&q_1&r_1\\p_2&q_2&r_2\\p_3&q_3&r_3\end{vmatrix}$$ times the area of $ABC$. One way to define the incenter is as the unique point $I$ inside $ABC$ for which $$\frac{\operatorname{Area}(IBC)}a=\frac{\operatorname{Area}(ICA)}b=\frac{\operatorname{Area}(IAB)}c.$$ Applying this barycentric area formula gives you the same result as before, i.e. that $I=(a:b:c)$. I believe this proof can, with enough work, be translated out of barycentric coordinates and into Cartesian, using the determinant version of the shoelace formula.