I am trying to discover another type of geometry. I want to transform a geometry problem to a coordinate geometry problem as a coordinate geometry problem that is easier than a pure geometry (Euclidean geometry) problem. A geometry problem needs deep thinking (I am telling about hard problems i.e. Olympiad level problems), but a coordinate geometry problem doesn't need deep thinking ,it just needs calculations. So, let us go to the original problem.
The vertices of a triangle $ABC$ are $(a_1,b_1),(a_2,b_2),(a_3,b_3)$. Find the following coordinates:
$1)$ Incentre
$2)$ Circumcentre
$3)$ Orthocentre
$4)$ Centre of the nine point circle
Please tell me with good explanation. Thank you.
There are well known formula for the triangle centers in trilinear coordinates $x : y : z$,
$$\begin{array}{rc:c} \text{name} & & x : y : z \\ \hline \text{incenter} & I & 1 : 1 : 1\\ \text{circumcenter} & O & \cos A : \cos B : \cos C\\ \text{orthocenter} & H & \sec A : \sec B :\sec C\\ \text{nine-point center} & N & \cos(B-C) : \cos(C-A) : \cos(A-B) \end{array} $$ You can convert them to Cartesian coordinates by the recipe: $$P = (x,y,z) \quad \leftrightarrow \quad \vec{P} = \frac{ax \vec{A} + by\vec{B} + cz\vec{C}}{ax + by + cz}$$
In terms of coordinates, this means
$$\vec{P} = (p_1,p_2) = \left( \frac{ax a_1 + by a_2 + cz a_3}{ax + by + cz}, \frac{ax b_1 + by b_2 + cz b_3}{ax + by + cz} \right) $$ The parameters $a,b,c$ and $A,B,C$ are given by the formula: $$\begin{cases} a &= \sqrt{(a_2-a_3)^2 + (b_2 - b_3)^2}\\ b &= \sqrt{(a_3-a_1)^2 + (b_3 - b_1)^2}\\ c &= \sqrt{(a_1-a_2)^2 + (b_1 - b_2)^2} \end{cases} \quad\text{ and }\quad \begin{cases} A &= \cos^{-1}\frac{b^2+c^2-a^2}{2bc}\\ B &= \cos^{-1}\frac{c^2+a^2-b^2}{2ca}\\ C &= \cos^{-1}\frac{a^2+b^2-c^2}{2ab} \end{cases} $$ The parameters $x, y, z$ can be obtained by plugging these expressions of $a,b,c, A,B,C$ into corresponding formula in above table.