Find the locus of a point equidistant from two lines #y=sqrt3x# and #y=1/sqrt3x#?

Question

5728 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-03T16:05:01

Locus is given by pair of lines given by #x^2-y^2=0# i.e. #x+y=0# and #x-y=0#

All points on lines bisecting are equidistant from the two given lines.

Let the equation of lines be #y=m_1x+c_1# and #y=m_2x+c_2#

or #m_1x-y+c_1=0# and #m_2x-y+c_2=0#

The distance of point #(x,y)# from #m_1x-y+c_1=0# is

#|(m_1x-y+c_1)/sqrt(1+m_1^2)|#

and the distance of point #(x,y)# from #m_2x-y+c_2=0# is

#|(m_2x-y+c_2)/sqrt(1+m_2^2)|#

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)]#

For example , if two lines are #y=sqrt3x# and #y=1/sqrt3x#,

then equations would be

#(sqrt3x-y)/sqrt(1+3)=+-[(1/sqrt3x-y)/sqrt(1+1/3)]#

or #(sqrt3x-y)/2=+-[(x-sqrt3y)/2]#

i.e. #(sqrt3-1)/2x+(sqrt3-1)/2y=0# and #(sqrt3+1)/2x-(sqrt3-1)/2y=0#

or #(sqrt3-1)x+(sqrt3-1)y=0# and #(sqrt3+1)x-(sqrt3+1)y=0#

or #x+y=0# and #x-y=0# or #(x+y)(x-y)=0# i.e. #x^2-y^2=0#

graph{(x+y)(x-y)(sqrt3x-y)(x-sqrt3y)=0 [-10, 10, -5, 5]}