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.
$x+y\sqrt{3}=6+2\sqrt{3}$ and $x-y\sqrt{3}=6-2\sqrt{3}$.

Answer 1

$L_1:x+\sqrt{3}y-2\sqrt{3}-6=0L_2:x-\sqrt{3}y+2\sqrt{3}-6=0L_1(0,0)=0+\sqrt{3}\left(0\right)-2\sqrt{3}-6=-2\sqrt{3}-6L_2(0,0)=0-\sqrt{3}\left(0\right)+2\sqrt{3}-6=2\sqrt{3}-6L_1(0,0)\times L_2(0,0)=-(2\sqrt{3}+6)(2\sqrt{3}-6)=-(12-36)=24L_1(0,0)\times L_2(0,0)>0$

So the equation of angle bisector containing origin is

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}\frac{x+\sqrt{3}y-2\sqrt{3}-6}{\sqrt{1^2+{\left(\sqrt{3}\right)}^2}}=\frac{x-\sqrt{3}y+2\sqrt{3}-6}{\sqrt{{\left(\sqrt{3}\right)}^2{+1}^2}}x+\sqrt{3}y-2\sqrt{3}-6=x-\sqrt{3}y+2\sqrt{3}-62\sqrt{3}y-4\sqrt{3}=0y-2=0y=2$

Equation of other angle bisector is

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=-\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}\frac{x+\sqrt{3}y-2\sqrt{3}-6}{\sqrt{1^2+{\left(\sqrt{3}\right)}^2}}=-\frac{x-\sqrt{3}y+2\sqrt{3}-6}{\sqrt{{\left(\sqrt{3}\right)}^2{+1}^2}}x+\sqrt{3}y-2\sqrt{3}-6=-x+\sqrt{3}y-2\sqrt{3}+62x-12=0x-6=0x=6$