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.
$12x+5y-4=0$ and $3x+4y+7=0$

Answer 1

$L_1:12x+4y-4=0L_2:3x+4y+7=0L_1(0,0)=12\left(0\right)+4\left(0\right)-4=-4L_2(0,0)=3\left(0\right)+4\left(0\right)+7=7L_1(0,0)\times L_2(0,0)=-4\times 7=-28L_1(0,0)\times L_2(0,0)<0$

So the angle bisector in which the origin lies is

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=-\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}\frac{12x+5y-4}{\sqrt{{\left(12\right)}^2+5^2}}=-\frac{3x+4y+7}{\sqrt{3^2+4^2}}\frac{12x+5y-4}{13}=-\frac{3x+4y+7}{5}5(12x+5y-4)=-13(3x+4y+7)60x+25y-20=-39x-52y-9199x+77y+71=0$

Other angle bisector is

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}\frac{12x+5y-4}{\sqrt{{\left(12\right)}^2+5^2}}=\frac{3x+4y+7}{\sqrt{3^2+4^2}}\frac{12x+5y-4}{13}=\frac{3x+4y+7}{5}5(12x+5y-4)=13(3x+4y+7)60x+25y-20=39x+52y+9121x-27y-111=07x-9y-37=0$