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(B) The initial line passes through $A(2,0)$ and makes an angle of $30^{\circ}$ with the positive $x$-axis. When the line is rotated clockwise by $15^

Vedclass · Accepted Answer

$(2-\sqrt{3})x-y-4+2\sqrt{3}=0$

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(B) The initial line passes through $A(2,0)$ and makes an angle of $30^{\circ}$ with the positive $x$-axis. When the line is rotated clockwise by $15^{\circ}$, the new angle $	heta$ with the positive $x$-axis becomes $30^{\circ} - 15^{\circ} = 15^{\circ}$. The slope of the new line is $m = 	an 15^{\circ} = 	an(45^{\circ} - 30^{\circ}) = \frac{	an 45^{\circ} - 	an 30^{\circ}}{1 + 	an 45^{\circ} 	an 30^{\circ}} = \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$. Rationalizing the denominator: $m = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{3+1-2\sqrt{3}}{2} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$. The equation of the line passing through $(2,0)$ with slope $m = 2-\sqrt{3}$ is $y - 0 = (2-\sqrt{3})(x - 2)$. $y = (2-\sqrt{3})x - 4 + 2\sqrt{3}$. Rearranging the terms, we get $(2-\sqrt{3})x - y - 4 + 2\sqrt{3} = 0$.

$A$ line through the point $A(2,0)$ which makes an angle of $30^{\circ}$ with the positive direction of $x$-axis is rotated about $A$ in clockwise direction through an angle $15^{\circ}$. Then the equation of the straight line in the new position is

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