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

Nov 3, 2017

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 $y = {m}_{1} x + {c}_{1}$ and $y = {m}_{2} x + {c}_{2}$

or ${m}_{1} x - y + {c}_{1} = 0$ and ${m}_{2} x - y + {c}_{2} = 0$

The distance of point $\left(x , y\right)$ from ${m}_{1} x - y + {c}_{1} = 0$ is

$| \frac{{m}_{1} x - y + {c}_{1}}{\sqrt{1 + {m}_{1}^{2}}} |$

and the distance of point $\left(x , y\right)$ from ${m}_{2} x - y + {c}_{2} = 0$ is

$| \frac{{m}_{2} x - y + {c}_{2}}{\sqrt{1 + {m}_{2}^{2}}} |$

and equation of angular bisectors would be

$\frac{{m}_{1} x - y + {c}_{1}}{\sqrt{1 + {m}_{1}^{2}}} = \pm \left[\frac{{m}_{2} x - y + {c}_{2}}{\sqrt{1 + {m}_{2}^{2}}}\right]$

For example , if two lines are $y = \sqrt{3} x$ and $y = \frac{1}{\sqrt{3}} x$,

then equations would be

$\frac{\sqrt{3} x - y}{\sqrt{1 + 3}} = \pm \left[\frac{\frac{1}{\sqrt{3}} x - y}{\sqrt{1 + \frac{1}{3}}}\right]$

or $\frac{\sqrt{3} x - y}{2} = \pm \left[\frac{x - \sqrt{3} y}{2}\right]$

i.e. $\frac{\sqrt{3} - 1}{2} x + \frac{\sqrt{3} - 1}{2} y = 0$ and $\frac{\sqrt{3} + 1}{2} x - \frac{\sqrt{3} - 1}{2} y = 0$

or $\left(\sqrt{3} - 1\right) x + \left(\sqrt{3} - 1\right) y = 0$ and $\left(\sqrt{3} + 1\right) x - \left(\sqrt{3} + 1\right) y = 0$

or $x + y = 0$ and $x - y = 0$ or $\left(x + y\right) \left(x - y\right) = 0$ i.e. ${x}^{2} - {y}^{2} = 0$

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