Find the equation of the line which passes though

The slope of the line that inclines with the $x$-axis at an angle of $75^{\circ}$ is

$m=\tan 75^{\circ}$

$\Rightarrow m=\tan \left(45^{\circ}+30^{\circ}\right)=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \cdot \tan 30^{\circ}}=\frac{1+\frac{1}{\sqrt{3}}}{1-1 \cdot \frac{1}{\sqrt{3}}}=\frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

We know that the equation of the line passing through point $\left(x_{0}, y_{0}\right)$, whose slope is $m$, is $\left(y-y_{0}\right)=m\left(x-x_{0}\right)$.

Thus, if a line passes though $(2,2 \sqrt{3})$ and inclines with the $x$-axis at an angle of $75^{\circ}$, then the equation of the line is given as

$(y-2 \sqrt{3})=\frac{\sqrt{3}+1}{\sqrt{3}-1}(x-2)$

$(y-2 \sqrt{3})(\sqrt{3}-1)=(\sqrt{3}+1)(x-2)$

$y(\sqrt{3}-1)-2 \sqrt{3}(\sqrt{3}-1)=x(\sqrt{3}+1)-2(\sqrt{3}+1)$

$(\sqrt{3}+1) x-(\sqrt{3}-1) y=2 \sqrt{3}+2-6+2 \sqrt{3}$

$(\sqrt{3}+1) x-(\sqrt{3}-1) y=4 \sqrt{3}-4$

i.e., $(\sqrt{3}+1) x-(\sqrt{3}-1) y=4(\sqrt{3}-1)$