A straight line parallel to the line y=sqrt{3 | vedclass

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(C) The slope of the line $y=\sqrt{3}x$ is $m=\sqrt{3}$. Since the required line is parallel to this line,its slope is also $\sqrt{3}$.Thus,$	an 	he

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$2\sqrt{3}-1$

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(C) The slope of the line $y=\sqrt{3}x$ is $m=\sqrt{3}$. Since the required line is parallel to this line,its slope is also $\sqrt{3}$.Thus,$	an 	heta = \sqrt{3}$,which implies $	heta = 60^{\circ}$.Let the distance $PQ = r$. The coordinates of any point $P$ on the line passing through $Q(2,3)$ at a distance $r$ are given by $(2+r \cos 60^{\circ}, 3+r \sin 60^{\circ}) = (2+\frac{r}{2}, 3+\frac{r\sqrt{3}}{2})$.Since $P$ lies on the line $2x+4y-27=0$,we substitute these coordinates into the equation:$2(2+\frac{r}{2}) + 4(3+\frac{r\sqrt{3}}{2}) - 27 = 0$$4 + r + 12 + 2\sqrt{3}r - 27 = 0$$r(1+2\sqrt{3}) - 11 = 0$$r = \frac{11}{2\sqrt{3}+1}$Rationalizing the denominator:$r = \frac{11(2\sqrt{3}-1)}{(2\sqrt{3}+1)(2\sqrt{3}-1)} = \frac{11(2\sqrt{3}-1)}{12-1} = \frac{11(2\sqrt{3}-1)}{11} = 2\sqrt{3}-1$.Therefore,the length of the line segment $PQ$ is $2\sqrt{3}-1$.

$A$ straight line parallel to the line $y=\sqrt{3}x$ passes through $Q(2,3)$ and cuts the line $2x+4y-27=0$ at $P$. Then the length of the line segment $PQ$ is

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