Coordinates of specific points on a triangle, given the triangles coordinates

Question

I'm working on a programing project. For that project I have a triangle with points $A,B,C$, where $A(a_1,a_2,a_3);B(b_1,b_2,b_3);C(c_1,c_2,c_3)$. Given the coordinates of the points $A,B$ and $C$, I want to find the coordinates of the orthocenter, circumcenter,incenter and the points where the perpendicular bisectors ,altitudes and angle bisectors meet with the $AB,BC,CA$. I believe there are formulas for each of these things. I tried looking online, but I couldn't find anything, so I'm asking you - Are there formulas for these things, and if so what are they?

Answer 1

Of course there are formulas, but it is probably easier to derive them than to find them online. The derivations become much easier if you work with vectors and take point $A$ to be $(0,0,0)$ (that is, translating by subtracting (a_1, a_2, a_3)$ from all of the points until the very end when you add it back.

Let's take the meeting points of the altitudes with the sides first, and look for point $P$ where the altitude from $C$ meets $AB$ (the other two cases are easy switches of $A$, $B$ and $C$ once you know that case). Let $AB = \vec{b}$ and $AC = \vec{c}$ and $AP = \vec{p}$. Then because $P$ is on (extended) line $AB$, $$ \vec{p} = k\vec{b} $$ for some scalar $k$. And since $CP \perp AP$,
$$ \vec{c}-\vec{p} \perp \vec{b} \rightarrow (\vec{c} - k\vec{b})\cdot \vec{b} = 0 \rightarrow k = \frac{\vec{c}\cdot \vec{b}}{|b|^2} = \frac{b_1 c_1 + b_2 c_2 + b_3 c_3}{b_1^2 + b_2^2 + b_3^2} $$ $$\vec{p} = \frac{\vec{c}\cdot \vec{b}}{|b|^2}\vec{b} = \frac{b_1 c_1 + b_2 c_2 + b_3 c_3}{b_1^2 + b_2^2 + b_3^2}\left( b_1, b_2, b_3\right) $$ Translating back the the original coordinates this gives $$ \left( a_1 + k (b_1-a_1), a_2 + k (b_2-a_2), a_2 + k (b_2-a_2) \right) $$ with $$ k=\frac{(b_1-a_1) (c_1-a_1) + (b_2-a_2) (c_2-a_2) + (b_3-a_3) (c_3-a_3)}{(b_1-a_1)^2 + (b_2-a_2)^2 + (b_3-a_3)^2} $$ (You can see how to do the translation to original coordinates from this; from here forward I will only show the work in the coordinates with $A$ at the origin.)

As long as we are working with altitudes, let's do the orthocenter next: We start with $Q$, the foot of the altitude on $AC$ which by the same reasoning as above is at $$ \vec{q} = k_b\vec{c} $$ where $$ k_b = \frac{\vec{b}\cdot \vec{c}}{|c|^2} = $$ and for notational symmetry we write the $k$ given above as $$ k_c = \frac{\vec{c}\cdot \vec{b}}{|b|^2} = $$ Line $BQ$ is described by $\vec{b} + \alpha (\vec q - \vec{b}) $ and line $CP$ is described by $\vec{c} + \beta (\vec p - \vec{c}) $. Setting these equal, we have: $$\begin{array}{l} \alpha\vec{q} + (1-\alpha)\vec{b} = \beta\vec{p} + (1-\beta)\vec{c} \\ \alpha k_b \vec{c} + (1-\alpha)\vec{b} = \beta k_c \vec{b} + (1-\beta)\vec{c} \\ (1-\alpha - \beta k_b) \vec{b} = (1-\beta - \alpha k_c) \vec{c} \end{array} $$ and since $\vec{b}$ and $\vec{c}$ are not linearly dependent, this can only be true if $$\left\{ \begin{array}{l} 1-\alpha - \beta k_b = 0\\ 1-\beta - \alpha k_c =0 \end{array} \right. $$ then $$ \left\{ \begin{array}{l} \alpha = \frac{1-k_b}{1-k_c}\\ \beta = \frac{1-k_c}{1-k_b} \end{array} \right. $$ so the orthocenter is at $$ \vec{b} + \alpha (\vec q - \vec{b}) = \vec{b} + \frac{1-k_b}{1-k_c}(k_b\vec{c} - \vec{b}) $$ On the computer you calculate $k_b$ and $k_c$ and then combine in this way.

The perpendicular bisector points are trivial in this scheme: On $AB$ the point is $\vec{b}/2$ for example.

The three perpendicular bisectors meet at the circumcenter. At the circumcenter we are on the line from $\vec{c}$ to $\vec{b}/2$ and also on the line from $\vec{b}$ to $\vec{c}/2$ so $$ \begin{array}{l} \alpha \vec{c} + (1-\alpha)\vec{b}/2) = \beta \vec{b} + (1-\beta)\vec{c} \\ \left( \alpha - \frac{1-\beta}{2} \right) \vec{c}= \left( \beta - \frac{1-\alpha}{2} \right) \vec{b} \end{array} $$ and as before, each of those coefficients must be zero so $$ \begin{array}{l} \beta = \frac{1-\alpha}{2} \\ -\frac{1}{2} + \alpha + \frac{1-\alpha}{4} = 0 \\ \alpha = \frac{1}{3} \end{array} $$ and the circumcenter is at $$ \frac{\vec{b}+\vec{c}}{3} $$ The same sort of techniques work to find the other points. Probably your professor wanted you to do these calculations as part of your project, so I won't finish it all for you. The hardest one will be the incenter, which is the intersection of the angle bisectors.