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Question

(A) The slope of the line that inclines with the $x-$axis at an angle of $75^{\circ}$ is $m = 	an 75^{\circ}$.$m = 	an(45^{\circ} + 30^{\circ}) = \f

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The slope of the line that inclines with the $x-$axis at an angle of $75^{\circ}$ is $m = 	an 75^{\circ}$.$m = 	an(45^{\circ} + 30^{\circ}) = \frac{	an 45^{\circ} + 	an 30^{\circ}}{1 - 	an 45^{\circ} \cdot 	an 30^{\circ}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.The equation of a line passing through $(x_0, y_0)$ with slope $m$ is $(y - y_0) = m(x - x_0)$.Substituting $(x_0, y_0) = (2, 2\sqrt{3})$ and $m = \frac{\sqrt{3}+1}{\sqrt{3}-1}$:$(y - 2\sqrt{3}) = \frac{\sqrt{3}+1}{\sqrt{3}-1}(x - 2)$.$(y - 2\sqrt{3})(\sqrt{3}-1) = (\sqrt{3}+1)(x - 2)$.$y(\sqrt{3}-1) - 2\sqrt{3}(\sqrt{3}-1) = x(\sqrt{3}+1) - 2(\sqrt{3}+1)$.$(\sqrt{3}+1)x - (\sqrt{3}-1)y = 2\sqrt{3} + 2 - (6 - 2\sqrt{3}) = 4\sqrt{3} - 4$.$(\sqrt{3}+1)x - (\sqrt{3}-1)y = 4(\sqrt{3}-1)$.

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

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