Find the equation of a line


Find the equation of a line

which makes an angle of $\frac{2 \pi}{3}$ with the positive direction of the $x-$ axis and passes through the point $(0,2)$



We have given angle so we have to find slope first given by m = tanθ.

$\mathrm{m}=\tan \theta \Rightarrow \tan \left(\frac{2 \pi}{3}\right)=\tan \left(\pi-\frac{\pi}{3}\right)$

$\mathrm{m} \Rightarrow-\tan \left(\frac{\pi}{3}\right)=-(\sqrt{3})(\tan x$ is negative in $\|$ quadrant $)$


Now the line is passing through the point $(0,2)$. Using the slope - intercept form of the equation of the line, we will find intercept

$y=m x+c$ ..................(1)

$2=-(\sqrt{3})(0)+c \Rightarrow c=2$

Putting the value of c in equation(1),we have

$y=-(\sqrt{3}) x+2$

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

So, required equation of line is $-(\sqrt{3}) x-y+2=0$.


Leave a comment