It is given that the slope of the first line, \(m_1 = 2. \)
Let the slope of the other line be \(m_2\).
The angle between the two lines is \(60°.\)
\(∴tan60º=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\)
\(⇒\sqrt3=\left|\frac{2-m_2}{1+2m_2}\right|\)
\(⇒\sqrt3=±\left(\frac{2-m_2}{1+2m_2}\right)\)
\(⇒\sqrt3=\left(\frac{2-m_2}{1+2m_2}\right) or \sqrt3=-\left(\frac{2-m_2}{1+2m_2}\right)\)
\(⇒\sqrt3(1+2m_2)=2-m_2 \space or \sqrt3(1+2m_2)=-(2-m_2)\)
\(⇒\sqrt3+2\sqrt3m_2+m_2=2 \space or \sqrt3+2\sqrt3m_2-m_2=-2\)
\(⇒m_2=\frac{2-\sqrt3}{\left(2\sqrt3+1\right)}\space or\space m_2=-\frac{\left(2+\sqrt3\right)}{\left(2\sqrt3-1\right)}\)
Case I: \(m_2=\frac{2-\sqrt3}{\left(2\sqrt3+1\right)}\)
The equation of the line passing through point (2, 3) and having a slope of\(\frac{\left(2-\sqrt3\right)}{\left(2\sqrt3+1\right)}\) is
\((y-3)=\frac{2-\sqrt3}{2\sqrt3+1}(x-2)\)
\((2\sqrt3+1)(y-3)=(2-\sqrt3)x-2(2-\sqrt3)\)
\((\sqrt3-2)x+(2\sqrt3+1)y=-4+2\sqrt3+6\sqrt3+3\)
\((\sqrt3-2)x+(2\sqrt3+1)y=-1+8\sqrt3\)
In this case, the equation of the other line is \((\sqrt3-2)x+(2\sqrt3+1)y=-1+8\sqrt3\)
Case II: \(m_2=\frac{-\left(2+\sqrt3\right)}{\left(2\sqrt3-1\right)}\)
The equation of the line passing through point (2, 3) and having a slope of\( \frac{-\left(2+\sqrt3\right)}{\left(2\sqrt3-1\right)}\) is
\((y-3)=\frac{-\left(2+\sqrt3\right)}{\left(2\sqrt3-1\right)}\left(x-2\right)\)
\((2\sqrt3-1)y-3(2\sqrt3-1)=-(2+\sqrt3)x+2(2+\sqrt3)\)
\((2\sqrt3-1)y+(2+\sqrt3)x=4+2\sqrt3+6\sqrt3-3\)
\((2+\sqrt3)x+(2\sqrt3-1)y=1+8\sqrt3\)
In this case, the equation of the other line is\( (2+\sqrt3)x+(2\sqrt3-1)y=1+8\sqrt3\)
Thus, the required equation of the other line is\( (\sqrt3-2)x+(2\sqrt3+1)y\)\(=-1+8\sqrt3\) or \( (2+\sqrt3)x+(2\sqrt3-1)y=1+8\sqrt3\)