94. Equations of angle bisectors (interior and exterior).

by Virgil Nicula, Aug 29, 2010, 2:39 AM

$\blacktriangleright$ Let $\left\|\begin{array}{c} \Delta_1 (x,y)\equiv a_1x+b_1y+c_1=0\\\ \Delta_2 (x,y)\equiv a_2x+b_2y+c_2=0\end{array}\right\|$ be the equations of two lines which aren't perpendiculary, i.e. $a_1a_2+b_1b_2\ne 0$ .
Then the equation of the acute angle between these lines is $\frac {\Delta_1 (x,y)}{\sqrt {a_1^2+b_1^2}}+\frac {\Delta_2 (x,y)}{\sqrt {a_2^2+b_2^2}}\cdot \mathrm{sgn}\ (a_1a_2+b_1b_2)=0$ .

$\blacktriangleright$ Let $A_1A_2A_3$ be an equilateral triangle, where $A_k\left(x_k,y_k\right)$ , $k\in\overline{1,3}$ . Then the equation of interior $A_1$-bisector is $\left|\begin{array}{ccc} x & y & 1\\\\ x_1 & y_1 & 1\\\\ x_2 & y_2 & 1\end{array}\right|\ +\ \left|\begin{array}{ccc} x & y & 1\\\\ x_1 & y_1 & 1\\\\ x_3 & y_3 & 1\end{array}\right|\ =\ 0$ .

$\blacktriangleright$ Let $A_1A_2A_3$ be a triangle, where equations of the its sidelines are $\left\|\begin{array}{cc} A_2A_3\ : & D_1(x,y)\equiv a_1x+b_1y+c_1=0\\\\ A_3A_1\ : & D_2(x,y)\equiv a_2x+b_2y+c_2=0\\\\ A_1A_2\ : & D_3(x,y)\equiv a_3x+b_3y+c_3=0\end{array}\right\|$ .

Then the equation of interior $A_1$-bisector is $\boxed {\ \mathrm{sign}\left|\begin{array}{cc} a_1 & b_1\\\ a_2 & b_2\end{array}\right|\cdot\frac {D_2(x,y)}{\sqrt {a_2^2+b_2^2}}+\mathrm{sign}\left|\begin{array}{cc} a_1 & b_1\\\ a_3 & b_3\end{array}\right|\cdot\frac {D_3(x,y)}{\sqrt {a_3^2+b_3^2}}=0\ }$ .

See the article "Distinctia analitica intre bisectoarele interioara si exterioara a unui unghi intr-un triunghi" from romanian review G.M.B., nr.3/1976.

Example. Let $ABC$ be a triangle for which the equations of its sidelines are $\left\|\begin{array}{cc} AB\ : & 4x-3y-1=0\\\\ AC\ : & 3x+4y-7=0\\\\ BC\ : & 7x+y-33=0\end{array}\right\|$ . Then the equation

of interior $A$-bisector is $\mathrm{sign}\left|\begin{array}{cc} 7 & 1\\\ 4 & -3\end{array}\right|\cdot\frac {4x-3y-1}{5}+\mathrm{sign}\left|\begin{array}{cc} 7 & 1\\\ 3 & 4\end{array}\right|\cdot\frac {3x+4y-7}{5}=0$ , i.e. $x-7y+6=0$ .
This post has been edited 28 times. Last edited by Virgil Nicula, Nov 23, 2015, 1:52 PM

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