The line joining two points A(2, 0) and B(3, | vedclass

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(C) The slope of line $AB$ is $m = \frac{1 - 0}{3 - 2} = 1$.This implies $	an 	heta = 1$,so $	heta = 45^{\circ}$.The line is rotated anticlockwise

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

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(C) The slope of line $AB$ is $m = \frac{1 - 0}{3 - 2} = 1$.This implies $	an 	heta = 1$,so $	heta = 45^{\circ}$.The line is rotated anticlockwise by $15^{\circ}$,so the new angle of inclination is $	heta' = 45^{\circ} + 15^{\circ} = 60^{\circ}$.The slope of the new line is $m' = 	an 60^{\circ} = \sqrt{3}$.The equation of the line passing through $A(2, 0)$ with slope $\sqrt{3}$ is given by $y - 0 = \sqrt{3}(x - 2)$.Simplifying this,we get $y = \sqrt{3}x - 2\sqrt{3}$,which can be written as $\sqrt{3}x - y - 2\sqrt{3} = 0$.

The line joining two points $A(2, 0)$ and $B(3, 1)$ is rotated about $A$ in the anticlockwise direction through an angle of $15^{\circ}$. The equation of the line in the new position is:

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