Question

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

Answer 1

Here $x_0=2,y_0=2\sqrt{3}$ and $\theta ={75}^{\circ }$

Now $m=tan{75}^{\circ }=tan({45}^{\circ }+{30}^{\circ })$

$=\frac{tan{45}^{\circ }+tan{30}^{\circ }}{1-tan{45}^{\circ }tan{30}^{\circ }}$

$=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

Putting these values in $y-y_0=m(x-x_0)$, we have

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

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

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

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