A straight line (sqrt{3} - 1)x = (sqrt{3} + 1 | vedclass

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(A) The given line is $(\sqrt{3} - 1)x = (\sqrt{3} + 1)y$,which can be written as $y = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}x$.We know that $	an(15^{\cir

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$x = 0$

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(A) The given line is $(\sqrt{3} - 1)x = (\sqrt{3} + 1)y$,which can be written as $y = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}x$.We know that $	an(15^{\circ}) = 	an(45^{\circ} - 30^{\circ}) = \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$.Thus,the slope of the line is $m_1 = 	an(15^{\circ})$.Let the slope of the required line be $m_2$. The angle between the lines is $75^{\circ}$,so $	an(75^{\circ}) = |\frac{m_2 - m_1}{1 + m_1 m_2}|$.Since $	an(75^{\circ}) = \cot(15^{\circ}) = \frac{1}{	an(15^{\circ})}$,we have $\frac{1}{	an(15^{\circ})} = |\frac{m_2 - 	an(15^{\circ})}{1 + m_2 	an(15^{\circ})}|$.Solving this,we find $m_2$ corresponds to the slope of the $y$-axis,which is undefined,or $m_2 = 	an(15^{\circ} + 75^{\circ}) = 	an(90^{\circ}) = \infty$.The line passing through the origin with an undefined slope is the $y$-axis,given by $x = 0$.

$A$ straight line $(\sqrt{3} - 1)x = (\sqrt{3} + 1)y$ makes an angle of $75^{\circ}$ with another straight line which passes through the origin. Then the equation of the line is:

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