Question

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

Answer 1

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

Explanation:

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]}