Question

Find the equations to the bisectors of the internal angles of the triangles the equations of whose sides are respectively
$3x+5y=15,x+y=4$, and $2x+y=6$.

Answer 1

$3x+5y=\mathrm{15......}\left(i\right)x+y=\mathrm{4.....}\left(ii\right)a_1{a}_2+b_1{b}_2=3\left(1\right)+5\left(1\right)=8>0$

So the first internal angle bisector is

$\frac{3x+5y-15}{\sqrt{3^3+5^2}}=\pm \frac{x+y-4}{\sqrt{1^2+1^2}}\frac{3x+5y-15}{\sqrt{34}}=-\frac{x+y-4}{\sqrt{2}}3x+5y-15=-\sqrt{17}(x+y-4)(3+\sqrt{17})x+(5+\sqrt{17})y=4\sqrt{17}+15$

$--------------------------------------$

$x+y=\mathrm{4....}\left(ii\right)2x+y=\mathrm{6......}\left(iii\right)a_1{a}_2+b_1{b}_2=1\left(2\right)+1\left(1\right)=3>0$

So, the second internal angle bisector is

$\frac{x+y-4}{\sqrt{1^2+1^2}}=-\frac{2x+y-6}{\sqrt{2^2+1^2}}\sqrt{5}(x+y-4)=-\sqrt{2}(2x+y-6)(\sqrt{5}+2\sqrt{2})x+(\sqrt{5}+\sqrt{2})y=6\sqrt{2}+4\sqrt{5}$

$--------------------------------------$

$2x+y=\mathrm{6....}\left(iii\right)3x+5y=\mathrm{15.....}\left(i\right)a_1{a}_2+b_1{b}_2=2\left(3\right)+1\left(5\right)=11>0$

So the third internal angle bisector is

$\frac{2x+y-6}{\sqrt{2^2+1^2}}=-\frac{3x+5y-15}{\sqrt{3^2+5^2}}\sqrt{34}(2x+y-6)=-\sqrt{5}(3x+5y-15)(2\sqrt{34}+3\sqrt{5})x+(\sqrt{34}+5\sqrt{5})y=15\sqrt{5}+6\sqrt{34}$