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.
$4x+3y-7=0$ and $24x+7y-31=0$

Answer 1

$L_1:4x+3y-7=0L_2:24x+7y-31=0L_1(0,0)=4\left(0\right)+3\left(0\right)-7=-7L_2(0,0)=24\left(0\right)+7\left(0\right)-31=-31L_1(0,0)\times L_2(0,0)=-7\times -31=217L_1(0,0)\times L_2(0,0)>0$

So, the angle bisector is

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}\frac{4x+3y-7}{\sqrt{4^2+3^2}}=\frac{24x+7y-31}{\sqrt{{\left(24\right)}^2+7^2}}\frac{4x+3y-7}{5}=\frac{24x+7y-31}{25}5(4x+3x-7)=24x+7y-3120x+15y-35=24x+7y-314x-8y+4=0x-2y+1=0$

The other angle bisector is

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=-\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}\frac{4x+3y-7}{\sqrt{4^2+3^2}}=-\frac{24x+7y-31}{\sqrt{{\left(24\right)}^2+7^2}}\frac{4x+3y-7}{5}=-\frac{24x+7y-31}{25}5(4x+3y-7)=-(24x+7y-31)20x+15y-35=-24x-7y+3144x+22y-66=02x+y-3=0$