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(A) The equation of a line passing through $(3, -2)$ is given by $y + 2 = m(x - 3)$.The slope of the given line $\sqrt{3}x + y = 1$ is $m_1 = -\sqrt{3

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of a line passing through $(3, -2)$ is given by $y + 2 = m(x - 3)$.The slope of the given line $\sqrt{3}x + y = 1$ is $m_1 = -\sqrt{3}$.The angle $	heta = 60^{\circ}$ between the lines is given by $	an(60^{\circ}) = \left| \frac{m - m_1}{1 + m \cdot m_1} ight|$.$\sqrt{3} = \left| \frac{m - (-\sqrt{3})}{1 + m(-\sqrt{3})} ight| = \left| \frac{m + \sqrt{3}}{1 - m\sqrt{3}} ight|$.Case $1$: $\frac{m + \sqrt{3}}{1 - m\sqrt{3}} = \sqrt{3} \implies m + \sqrt{3} = \sqrt{3} - 3m \implies 4m = 0 \implies m = 0$.Equation: $y + 2 = 0(x - 3) \implies y + 2 = 0$.Case $2$: $\frac{m + \sqrt{3}}{1 - m\sqrt{3}} = -\sqrt{3} \implies m + \sqrt{3} = -\sqrt{3} + 3m \implies 2m = 2\sqrt{3} \implies m = \sqrt{3}$.Equation: $y + 2 = \sqrt{3}(x - 3) \implies y + 2 = \sqrt{3}x - 3\sqrt{3} \implies \sqrt{3}x - y - 2 - 3\sqrt{3} = 0$.

Find the equation of the line passing through the point $(3, -2)$ and making an angle of $60^{\circ}$ with the line $\sqrt{3}x + y = 1$.

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