Find the equation of a straight line which pa | vedclass

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(B) The slope $m$ of the line is given by $m = 	an 	heta$,where $	heta = 150^{\circ}$.$m = 	an 150^{\circ} = 	an(180^{\circ} - 30^{\circ}) = -	a

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$\sqrt{3} y + x + (1 + \sqrt{3}) = 0$

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(B) The slope $m$ of the line is given by $m = 	an 	heta$,where $	heta = 150^{\circ}$.$m = 	an 150^{\circ} = 	an(180^{\circ} - 30^{\circ}) = -	an 30^{\circ} = -\frac{1}{\sqrt{3}}$.The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.Substituting the point $(-1, -1)$ and $m = -\frac{1}{\sqrt{3}}$:$y - (-1) = -\frac{1}{\sqrt{3}}(x - (-1))$$y + 1 = -\frac{1}{\sqrt{3}}(x + 1)$$\sqrt{3}(y + 1) = -(x + 1)$$\sqrt{3}y + \sqrt{3} = -x - 1$$x + \sqrt{3}y + \sqrt{3} + 1 = 0$Thus,the equation is $x + \sqrt{3}y + (\sqrt{3} + 1) = 0$.

Find the equation of a straight line which passes through the point $(-1, -1)$ and makes an angle $150^{\circ}$ with the positive direction of the $X$-axis.

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