Question

Find the lines through the point (0, 2) making angles $\frac{\mathrm{\pi }}{3}\mathrm{and}\frac{2\mathrm{\pi }}{3}$ with the x-axis. Also, find the lines parallel to them cutting the y-axis at a distance of 2 units below the origin.

Answer 1

The inclinations of the two lines with the positive x-axis are $\frac{\mathrm{\pi }}{3}\mathrm{and}\frac{2\mathrm{\pi }}{3}$.
So, their slopes are $m_1=\tan \left(\frac{\mathrm{\pi }}{3}\right)=\sqrt{3}\mathrm{and}{m}_2=\tan \left(\frac{2\mathrm{\pi }}{3}\right)=-\tan \left(\frac{\mathrm{\pi }}{3}\right)=-\sqrt{3}$.

Now, the equations of the lines that pass through (0, 2) and have slopes $m_1\mathrm{and}{m}_2$ are

$$\begin{gathered} y-2=\sqrt{3}\left(x-0\right)\mathrm{and}y-2=-\sqrt{3}\left(x-0\right) \\ \Rightarrow y-\sqrt{3}x-2=0\mathrm{and}y+\sqrt{3}x-2=0 \\ \mathrm{or}\sqrt{3}x-y+2=0\mathrm{and}\sqrt{3}x+y-2=0 \end{gathered}$$

Now, the equation of the line parallel to the line having slope m₁ and intercept c = $-2$ is

$$\begin{gathered} y=m_{1}x+c \\ \Rightarrow y=\sqrt{3}x-2 \\ \Rightarrow \sqrt{3}x-y-2=0 \end{gathered}$$

Similarly, the equation of line parallel to the line having slope m₂ and intercept c = $-2$ is

$$\begin{gathered} y=m_{2}x+c \\ \Rightarrow y=-\sqrt{3}x-2 \\ \Rightarrow \sqrt{3}x+y+2=0 \end{gathered}$$