Coordinates of incentre without finding side lengths

Question

If I am given the equations of sides of a triangle and I need to find incentre what is the shortest method ? Is it possible without having to find lengths of sides of triangle?

Answer 1

Suppose you are given an equation of a line $l$ (in the plane) \begin{equation*} ax+by+c=0~, \quad (a,b)\neq(0,0)\,. \end{equation*} For the task you have ahead of you it is best to divide the equation by $\pm d$, where $d=\sqrt{a^2+b^2}$, obtaining one of the two of its normalized forms \begin{equation*} \pm\,\frac{a}{d}x \,\pm\, \frac{b}{d}y \,\pm\, \frac{c}{d} \,=\, 0~. \end{equation*} Here $u:=\pm(a,b)/d$ is the unit vector orthogonal to the line $l$. Writing $e:=\pm c/d$ and $X:=(x,y)$, we rewrite the normalized equation of the line as \begin{equation*} u\cdot X +\, e \,=\, 0~, \end{equation*} where $u\cdot X$ is the usual scalar product (a.k.a. the dot product) of pairs of real numbers: $(a_1,a_2)\cdot(b_1,b_2):=a_1b_1+a_2b_2$. For any point $X$ in the plane the value of $u\!\cdot\!X+e$ is the signed distance of the point $X$ from the line $l$, positive on the side of the line to which the vector $u$ points, and negative on the opposite side.

Now to your problem. You are given the equations of the sides of a triangle. First thing you do, you determine the coordinates of the triangle's vertices $A$, $B$, $C$ by solving three systems of two linear equations in two unknowns (the coordinates of the vertex). Next, you obtain the normalized equations of the sides: \begin{align*} u\cdot X +\, d &\,=\, 0 \qquad \text{$\,$(side $B\lor C\,$)}~, \\ v\cdot X +\, e &\,=\, 0 \qquad \text{$\,$(side $C\lor A\,$)}~, \\ w\cdot X +\, f &\,=\, 0 \qquad \text{$\,$(side $A\lor B\,$)}~, \end{align*} where the unit normals $u$, $v$, $w$ point towards the triangle; you achive this by multiplying each linear form measuring the distance from a tringle's side by a factor $\pm1$ chosen so that the vertex opposite to the side will have a positive distance from it. As next to the last step you write down the equations of two of the three internal angle bisectors, say of the bisectors $l_A$ and $l_B$:

The bisector $l_A$ is the line that passes through the vertex $A$ and has the direction vector $v+w$
(here it is important that the vectors $v$ and $w$ are of the same length); similarly the bisector $l_B$ is the line that passes through the point $B$ and has the direction vector $w+u$. This is easier to do than it seems: the equations of the bisectors are \begin{align*} l_A\,&:\quad~\mspace{1mu} v\cdot X +\, e \,=\, w\cdot X +\, f ~, \\ l_B\,&:\quad w\cdot X +\, f \,=\, u\cdot X +\, e ~. \end{align*} Indeed, the first equation says that $l_A$ is the set of all points $X$ that are at the same distance from the side $C\lor A$ as from the side $A\lor B$, and the equation for $l_B$ is similarly intepreted. (Here it is important that the signed distances have correct signs).

Finally you solve the system of the two equations for the bisectors $l_A$ and $l_B$, obtaining the coordinates of the incenter $I$.

Remark. $~$The line passing through the vertex $A$ with the direction vector $w-v$ is the bisector $l'_{\!A}$ of the external angle at $A$, whose equation is \begin{equation*} l'_{\!A}\,:\quad~\mspace{1mu} v\cdot X +\, e \,=\, -(w\cdot X +\, f) ~; \end{equation*} the equation of the bisector $l'_B$ of the external angle at $B$ is obtained analogously. With the bisectors of both the internal and the external angles at the vertices $A$ and $B$ in hand you can compute the coordinates of the three triangle's excenters.

Remark. $~$According to the recipe above we have to calculate the coordinates of the triangle's vertices just because we need them to choose correct signs of the linear forms measuring the distances of a point from the triangle's sides. Actually this is unnecessary: we can choose properly aligned signs of the vectors $u$, $v$, $w$ straightaway, without involving the triangle's vertices.

Here is how. For any two pairs of real numbers $x=(x_1,x_2)$, $y=(\smash{y}_1,\smash{y}_2)$ we define $[x,y] := x_1\smash{y}_2-x_2\smash{y}_1$. Initially we determine the unit normals $u$, $v$, $w$ of the triangle's sides without caring about the signs. Then we look at the numbers $[u,v]$, $[v,w]$, and $[w,u]$. If they are all of the same sign, then all three vectors $u$, $v$, $w$ point towards the triangle or all three point away from the triangle (which is equally good). If the three numbers are not of the same sign, then we flip the sign of one of the three vectors: if, say, $[u,v]$ and $[v,w]$ are negative and $[w,u]$ is positive, we replace $v$ by $v_1=-v$ and then have all three numbers $[u,v_1]$, $[v_1,w]$, and $[w,u]$ positive.