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

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(B) The slope of the line $AB$ is given by $m = \frac{1-0}{3-2} = 1$.Since $m = 	an 	heta = 1$,the angle of inclination is $	heta = 45^{\circ}$.Whe

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

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(B) The slope of the line $AB$ is given by $m = \frac{1-0}{3-2} = 1$.Since $m = 	an 	heta = 1$,the angle of inclination is $	heta = 45^{\circ}$.When the line is rotated about point $A$ in the anticlockwise direction by $15^{\circ}$,the new angle of inclination becomes $	heta' = 45^{\circ} + 15^{\circ} = 60^{\circ}$.The slope of the new line is $m' = 	an 60^{\circ} = \sqrt{3}$.Since the line passes through $A(2,0)$,the equation of the line in point-slope form is $(y - 0) = \sqrt{3}(x - 2)$.This simplifies to $y = \sqrt{3}x - 2\sqrt{3}$.

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

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