The distance of the point (2, 3) from the lin | vedclass

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(D) Let the point be $A(2, 3)$ and the line be $L: 2x - 3y + 28 = 0$. The distance is measured parallel to the line $\sqrt{3}x - y + 1 = 0$,which has

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$4 + 6\sqrt{3}$

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(D) Let the point be $A(2, 3)$ and the line be $L: 2x - 3y + 28 = 0$. The distance is measured parallel to the line $\sqrt{3}x - y + 1 = 0$,which has a slope $m = \sqrt{3}$.Thus,$	an 	heta = \sqrt{3}$,which implies $	heta = 60^\circ$. So,$\cos 	heta = \frac{1}{2}$ and $\sin 	heta = \frac{\sqrt{3}}{2}$.The coordinates of any point $P$ on the line passing through $A(2, 3)$ at a distance $r$ are given by $(2 + r \cos 	heta, 3 + r \sin 	heta) = (2 + \frac{r}{2}, 3 + \frac{\sqrt{3}r}{2})$.Since $P$ lies on the line $2x - 3y + 28 = 0$,we substitute these coordinates into the equation:$2(2 + \frac{r}{2}) - 3(3 + \frac{\sqrt{3}r}{2}) + 28 = 0$$4 + r - 9 - \frac{3\sqrt{3}r}{2} + 28 = 0$$23 + r(1 - \frac{3\sqrt{3}}{2}) = 0$$r(\frac{2 - 3\sqrt{3}}{2}) = -23$$r = \frac{46}{3\sqrt{3} - 2}$.Rationalizing the denominator:$r = \frac{46(3\sqrt{3} + 2)}{(3\sqrt{3})^2 - 2^2} = \frac{46(3\sqrt{3} + 2)}{27 - 4} = \frac{46(3\sqrt{3} + 2)}{23} = 2(3\sqrt{3} + 2) = 6\sqrt{3} + 4$.

The distance of the point $(2, 3)$ from the line $2x - 3y + 28 = 0$,measured parallel to the line $\sqrt{3}x - y + 1 = 0$,is equal to

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