The equation of the straight line cutting off | vedclass

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(D) The equation of a line in slope-intercept form is given by $y = mx + c$.Given that the line is inclined at $30^\circ$ to the positive direction of

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

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(D) The equation of a line in slope-intercept form is given by $y = mx + c$.Given that the line is inclined at $30^\circ$ to the positive direction of the $x$-axis,the slope $m = 	an(30^\circ) = \frac{1}{\sqrt{3}}$.The line cuts off an intercept of $2$ from the negative direction of the $y$-axis,so $c = -2$.Substituting these values into the equation,we get $y = \frac{1}{\sqrt{3}}x - 2$.Multiplying both sides by $\sqrt{3}$,we get $\sqrt{3}y = x - 2\sqrt{3}$.Rearranging the terms,we obtain $\sqrt{3}y - x + 2\sqrt{3} = 0$.

The equation of the straight line cutting off an intercept of $2$ from the negative direction of the $y$-axis and inclined at $30^\circ$ to the positive direction of the $x$-axis is:

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