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Question
Find the lines through the point (0, 2) making angles \[\frac{\pi}{3} \text { and } \frac{2\pi}{3}\] with the x-axis. Also, find the lines parallel to them cutting the y-axis at a distance of 2 units below the origin.
Solution
The inclinations of the two lines with the positive x-axis are \[\frac{\pi}{3} \text { and }\frac{2\pi}{3}\].
So, their slopes are \[m_1 = \tan\left( \frac{\pi}{3} \right) = \sqrt{3} \text { and } m_2 = \tan\left( \frac{2\pi}{3} \right) = - \tan\left( \frac{\pi}{3} \right) = - \sqrt{3}\].
Now, the equations of the lines that pass through (0, 2) and have slopes \[m_1\text { and } m_2\] are
\[y - 2 = \sqrt{3}\left( x - 0 \right) \text { and } y - 2 = - \sqrt{3}\left( x - 0 \right)\]
\[ \Rightarrow y - \sqrt{3}x - 2 = 0\text { and } y + \sqrt{3}x - 2 = 0\]
\[\text { or }\sqrt{3}x - y + 2 = 0 \text { and } \sqrt{3}x + y - 2 = 0\]
Now, the equation of the line parallel to the line having slope m1 and intercept c = \[- 2\] is
\[y = m_1 x + c\]
\[ \Rightarrow y = \sqrt{3}x - 2\]
\[ \Rightarrow \sqrt{3}x - y - 2 = 0\]
\[y = m_2 x + c\]
\[ \Rightarrow y = - \sqrt{3}x - 2\]
\[ \Rightarrow \sqrt{3}x + y + 2 = 0\]
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