Find the equation of a straight line which passes through the point $(-1,-1)$ and makes an angle…

Find the equation of a straight line which passes through the point $(-1,-1)$ and makes an angle $150^{\circ}$ with positive direction of $X$-axis.
  1. $\sqrt{3} x+y=1$
  2. $\sqrt{3} y+x+(1+\sqrt{3})=0$
  3. $x+\sqrt{3} y+(\sqrt{3}-1)=0$
  4. $x+y=0$

Solution


Now, $\tan \theta=\tan 150^{\circ}=\tan \left(\pi-30^{\circ}\right)$ $ =-\tan 30^{\circ}=-\frac{1}{\sqrt{3}} $ $\therefore$ Slope of given line $=-\frac{1}{\sqrt{3}}$ Since, line passes through $(-1,-1)$. $\therefore$ Equation of line will be $ \begin{array}{rlrl} y-y_1 & =m\left(x-x_1\right) \\ \Rightarrow & & \\ y+1 & =-\frac{1}{\sqrt{3}}(x+1) \\ & & \sqrt{3} y+\sqrt{3} & =-x-1 \\ & \text { or } \sqrt{3} y+x+(\sqrt{3}+1) & =0 \end{array} $

Asked in: AP EAMCET 2021 (23 Aug Shift 1)

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