Question

Find the equations to the bisectors of the internal angles of the triangles the equations of whose sides are respectively
$3x+4y=6,12x-5y=3$, and $4x-3y+12=0$.

Answer 1

$3x+4y-6=012x-5y-3=0a_1{a}_2+b_1{b}_2=3\left(12\right)+4(-5)=26-20=6a_1{a}_2+b_1{b}_2>0$

So the equation of internal angle bisector is

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=-\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}\frac{3x+4y-6}{\sqrt{3^2+4^2}}=-\frac{12x-5y-3}{\sqrt{{\left(12\right)}^2+5^2}}13(3x+4y-6)=-5(12x-5y-3)39x+52y-78=-60x+25y+1599x+27y-93=033x+9y-31=0$

Taking the other two sides

$12x-5y-3=04x-3y+12=0a_1{a}_2+b_1{b}_2=12\left(4\right)+(-5)(-3)=63a_1{a}_2+b_1{b}_2>0$

So the equation of internal angle bisector is

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=-\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}\frac{12x-5y-3}{\sqrt{{\left(12\right)}^2+5^2}}=-\frac{4x-3y+12}{\sqrt{4^2+3^2}}5(12x-5y-3)=-13(4x-3y+12)60x-25y-15=-52x+39y-156112x-64y+141=0$

Taking the remaining two sides

$4x-3y+12=03x+4y-6=0a_1{a}_2+b_1{b}_2=4\left(3\right)+(-3)4=0a_1{a}_2+b_1{b}_2=0$

So the equation of internal angle bisector is
$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}\frac{4x-3y+12}{\sqrt{4^2+3^2}}=\frac{3x+4y-6}{\sqrt{3^2+4^2}}4x-3y+12=3x+4y-67y-x-18=07y-x=18$