Question

Find the equations to the straight lines bisecting the angles between the following pairs of straight lines, placing first the bisector of the angle in which the origin lies.
$2x+y=4$ and $y+3x=5$

Answer 1

.$L_1:2x+y-4=0L_2:y+3x-5=0L_1(0,0)=2\left(0\right)+0-4=-4L_2(0,0)=0+3\left(0\right)-5=-5L_1(0,0)\times L_2(0,0)=-4\times -5=20\Rightarrow L_1(0,0)\times L_2(0,0)>0$

So the angle angle bisector in which origin lies is

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}$

$\frac{2x+y-4}{\sqrt{2^2+1^2}}=\frac{y+3x-5}{\sqrt{3^2+1^2}}\frac{2x+y-4}{\sqrt{5}}=\frac{y+3x-5}{\sqrt{10}}\sqrt{2}(2x+y-4)=y+3x-5(2\sqrt{2}-3)x+(\sqrt{2}-1)y+5-4\sqrt{2}=0$

The other angle bisector is

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=-\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}\frac{2x+y-4}{\sqrt{2^2+1^2}}=-\frac{y+3x-5}{\sqrt{3^2+1^2}}\frac{2x+y-4}{\sqrt{5}}=-\frac{y+3x-5}{\sqrt{10}}\sqrt{2}(2x+y-4)=-(y+3x-5)(2\sqrt{2}+3)x+(\sqrt{2}+1)y-5-4\sqrt{2}=0$