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(A) The slope $m$ of the line making an angle $	heta = \frac{2 \pi}{3}$ with the positive $x$-axis is $m = 	an\left(\frac{2 \pi}{3}ight) = -\sqrt{

Vedclass · Accepted Answer

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

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(A) The slope $m$ of the line making an angle $	heta = \frac{2 \pi}{3}$ with the positive $x$-axis is $m = 	an\left(\frac{2 \pi}{3}ight) = -\sqrt{3}$.The equation of the line passing through $(0, 2)$ with slope $m = -\sqrt{3}$ is given by the point-slope form $(y - y_1) = m(x - x_1)$:$(y - 2) = -\sqrt{3}(x - 0)$$y - 2 = -\sqrt{3}x$$\sqrt{3}x + y - 2 = 0$.For the second line,it is parallel to the first line,so its slope is also $m = -\sqrt{3}$.It crosses the $y$-axis $2$ units below the origin,meaning it passes through the point $(0, -2)$.Using the point-slope form $(y - (-2)) = -\sqrt{3}(x - 0)$:$y + 2 = -\sqrt{3}x$$\sqrt{3}x + y + 2 = 0$.

Find the equation of the line passing through the point $(0, 2)$ making an angle of $\frac{2 \pi}{3}$ with the positive $x$-axis. Also,find the equation of the line parallel to it and crossing the $y$-axis at a distance of $2$ units below the origin.

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