A straight line through the point (3, -2) is | vedclass

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(B) Given line: $\sqrt{3}x + y = 1$, which can be written as $y = -\sqrt{3}x + 1$. The slope $m_2 = -\sqrt{3}$.Let the slope of the required line be $

Vedclass · Accepted Answer

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

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(B) Given line: $\sqrt{3}x + y = 1$, which can be written as $y = -\sqrt{3}x + 1$. The slope $m_2 = -\sqrt{3}$.Let the slope of the required line be $m_1$. The angle between the lines is $	heta = 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{m_1 - (-\sqrt{3})}{1 + m_1(-\sqrt{3})} ight|$$\sqrt{3} = \left| \frac{m_1 + \sqrt{3}}{1 - \sqrt{3}m_1} ight|$This gives two cases:Case $1$: $\sqrt{3} = \frac{m_1 + \sqrt{3}}{1 - \sqrt{3}m_1} \implies \sqrt{3} - 3m_1 = m_1 + \sqrt{3} \implies 4m_1 = 0 \implies m_1 = 0$.The equation of the line passing through $(3, -2)$ with slope $0$ is $y - (-2) = 0(x - 3) \implies y + 2 = 0$.Case $2$: $-\sqrt{3} = \frac{m_1 + \sqrt{3}}{1 - \sqrt{3}m_1} \implies -\sqrt{3} + 3m_1 = m_1 + \sqrt{3} \implies 2m_1 = 2\sqrt{3} \implies m_1 = \sqrt{3}$.The equation of the line passing through $(3, -2)$ with slope $\sqrt{3}$ is $y - (-2) = \sqrt{3}(x - 3) \implies y + 2 = x\sqrt{3} - 3\sqrt{3} \implies y - x\sqrt{3} + 2 + 3\sqrt{3} = 0$.Comparing with the given options, the correct equation is $y - x\sqrt{3} + 2 + 3\sqrt{3} = 0$.

$A$ straight line through the point $(3, -2)$ is inclined at an angle $60^{\circ}$ to the line $\sqrt{3}x + y = 1$. If it intersects the $X$-axis, then its equation will be

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