All points on lines bisecting are equidistant from the two given lines.
Let the equation of lines be y=m_1x+c_1y=m1x+c1 and y=m_2x+c_2y=m2x+c2
or m_1x-y+c_1=0m1x−y+c1=0 and m_2x-y+c_2=0m2x−y+c2=0
The distance of point (x,y)(x,y) from m_1x-y+c_1=0m1x−y+c1=0 is
|(m_1x-y+c_1)/sqrt(1+m_1^2)|∣∣
∣
∣∣m1x−y+c1√1+m21∣∣
∣
∣∣
and the distance of point (x,y)(x,y) from m_2x-y+c_2=0m2x−y+c2=0 is
|(m_2x-y+c_2)/sqrt(1+m_2^2)|∣∣
∣
∣∣m2x−y+c2√1+m22∣∣
∣
∣∣
and equation of angular bisectors would be
(m_1x-y+c_1)/sqrt(1+m_1^2)=+-[(m_2x-y+c_2)/sqrt(1+m_2^2)]m1x−y+c1√1+m21=±⎡⎢
⎢⎣m2x−y+c2√1+m22⎤⎥
⎥⎦
For example , if two lines are y=sqrt3xy=√3x and y=1/sqrt3xy=1√3x,
then equations would be
(sqrt3x-y)/sqrt(1+3)=+-[(1/sqrt3x-y)/sqrt(1+1/3)]√3x−y√1+3=±⎡⎢
⎢⎣1√3x−y√1+13⎤⎥
⎥⎦
or (sqrt3x-y)/2=+-[(x-sqrt3y)/2]√3x−y2=±[x−√3y2]
i.e. (sqrt3-1)/2x+(sqrt3-1)/2y=0√3−12x+√3−12y=0 and (sqrt3+1)/2x-(sqrt3-1)/2y=0√3+12x−√3−12y=0
or (sqrt3-1)x+(sqrt3-1)y=0(√3−1)x+(√3−1)y=0 and (sqrt3+1)x-(sqrt3+1)y=0(√3+1)x−(√3+1)y=0
or x+y=0x+y=0 and x-y=0x−y=0 or (x+y)(x-y)=0(x+y)(x−y)=0 i.e. x^2-y^2=0x2−y2=0
graph{(x+y)(x-y)(sqrt3x-y)(x-sqrt3y)=0 [-10, 10, -5, 5]}
