The equation of the lines which pass through | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

The equation of the lines which pass through the point $(3, -2)$ and are inclined at $60^o$ to the line $\sqrt{3}x + y = 1$ is:

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