What is the slope between the lines $y-sqrt{3} x-5=0$ and $sqrt{3} y-x+6=0$

We know that,

Angle between two lines is

$an heta=left|frac{m_{1}-m_{2}}{1+m_{1} m_{2}} ight|$

Here,

$m_{1}=sqrt{3}$ and

$m_{2}=frac{1}{sqrt{3}}$

$an heta=left|frac{sqrt{3}-frac{1}{sqrt{3}}}{1+sqrt{3} frac{1}{sqrt{3}}} ight|=left|frac{3-1}{2 sqrt{3}} ight|$

$an heta=frac{1}{sqrt{3}}$

$herefore$ Slope $=frac{1}{sqrt{3}}$