Question

Find the equation of the line passing through $(2,2\sqrt{3})$ and inclined with the $x$-axis at an angle of 75${}^{\circ }$

Answer 1

We have, $m=\tan {75}^{\circ }$
$\Rightarrow m=\tan ({45}^{\circ }+{30}^{\circ })$$=\frac{\tan {45}^{\circ }+{30}^{\circ }}{1-\tan {45}^{\circ }.\tan {30}^{\circ }}$
$=\frac{1+\frac{1}{\sqrt{3}}}{1-1\frac{1}{\sqrt{3}}}=\frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
Thus equation of line passing through $(2,2\sqrt{3})$ and inclined with the $x$-axis at an angle of ${75}^{\circ }$ is given by,
$(y-2\sqrt{3})=\frac{\sqrt{3}+1}{\sqrt{3}-1}(x-2)$
$(y-2\sqrt{3})(\sqrt{3}-1)=(\sqrt{3}+1)(x-2)$
$y(\sqrt{3}-1)-2\sqrt{3}(\sqrt{3}-1)=x(\sqrt{3}+1)-2(\sqrt{3}+1)$
$(\sqrt{3}+1)x-(\sqrt{3}-1)y=2\sqrt{3}+2-6+2\sqrt{3}$
$(\sqrt{3}+1)x-(\sqrt{3}-1)y=4\sqrt{3}-4$
i.e;$(\sqrt{3}+1)x-(\sqrt{3}-1)y=4(\sqrt{3}-1)$