A line passing through the point A(9,0) makes | vedclass

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Question

(A) The initial line passes through $A(9,0)$ and makes an angle of $30^{\circ}$ with the positive $x$-axis.When the line is rotated clockwise by $15^{

Vedclass · Accepted Answer

$\frac{y}{\sqrt{3}-2}+x=9$

Vedclass · Answer

(A) The initial line passes through $A(9,0)$ and makes an angle of $30^{\circ}$ with the positive $x$-axis.When the line is rotated clockwise by $15^{\circ}$ about $A$,the new angle it makes with the positive $x$-axis is $	heta = 30^{\circ} - 15^{\circ} = 15^{\circ}$.The slope of the new line is $m = 	an 15^{\circ}$.Using the value $	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} = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$.The equation of the line passing through $(x_1, y_1) = (9,0)$ with slope $m$ is $y - y_1 = m(x - x_1)$.Substituting the values: $y - 0 = (2-\sqrt{3})(x - 9)$.Rearranging the equation: $y = (2-\sqrt{3})(x - 9)$ $\Rightarrow \frac{y}{2-\sqrt{3}} = x - 9$ $\Rightarrow x - \frac{y}{2-\sqrt{3}} = 9$.Since $\frac{1}{2-\sqrt{3}} = \frac{2+\sqrt{3}}{4-3} = 2+\sqrt{3}$,this does not match the options directly.Let's re-examine the options. The equation $y = (2-\sqrt{3})(x-9)$ can be written as $x-9 = \frac{y}{2-\sqrt{3}}$.Alternatively,$y = (2-\sqrt{3})(x-9)$ $\Rightarrow \frac{y}{\sqrt{3}-2} = -(x-9) = 9-x$ $\Rightarrow x + \frac{y}{\sqrt{3}-2} = 9$.

$A$ line passing through the point $A(9,0)$ makes an angle of $30^{\circ}$ with the positive direction of $x$-axis. If this line is rotated about $A$ through an angle of $15^{\circ}$ in the clockwise direction,then its equation in the new position is

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