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(B) The given line is $\sqrt{3} x + y = 1$,which can be written as $y = -\sqrt{3} x + 1$. The slope $m_1 = -\sqrt{3}$.Let the slope of the required li

Vedclass · Accepted Answer

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

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(B) The given line is $\sqrt{3} x + y = 1$,which can be written as $y = -\sqrt{3} x + 1$. The slope $m_1 = -\sqrt{3}$.Let the slope of the required line be $m_2$. The angle between the two lines is $60^{\circ}$.Using the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$,we have:$	an 60^{\circ} = \left| \frac{-\sqrt{3} - m_2}{1 + (-\sqrt{3})m_2} ight|$$\sqrt{3} = \left| \frac{-\sqrt{3} - m_2}{1 - \sqrt{3} m_2} ight|$Squaring both sides:$3 = \frac{(-\sqrt{3} - m_2)^2}{(1 - \sqrt{3} m_2)^2}$$3(1 - \sqrt{3} m_2)^2 = (\sqrt{3} + m_2)^2$$3(1 + 3m_2^2 - 2\sqrt{3} m_2) = 3 + m_2^2 + 2\sqrt{3} m_2$$3 + 9m_2^2 - 6\sqrt{3} m_2 = 3 + m_2^2 + 2\sqrt{3} m_2$$8m_2^2 - 8\sqrt{3} m_2 = 0$$8m_2(m_2 - \sqrt{3}) = 0$So,$m_2 = 0$ or $m_2 = \sqrt{3}$.For $m_2 = \sqrt{3}$,the equation of the line passing through $(3, 2)$ is:$y - 2 = \sqrt{3}(x - 3)$$y - 2 = \sqrt{3} x - 3\sqrt{3}$$\sqrt{3} x - y + (2 - 3\sqrt{3}) = 0$.

The equation of the straight line passing through the point $(3, 2)$ and inclined at an angle of $60^{\circ}$ with the line $\sqrt{3} x + y = 1$ is

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