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A straight line parallel to the line $y=\sqrt{3} x$ passes through $Q(2,3)$ and cuts the line $2 x+4 y-27=0$ at $P$. Then the length of the line segment $\mathrm{PQ}$ is
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Verified Answer
The correct answer is:
$\sqrt{3}-1$
Slope of parallel lines are equal
$\therefore$ slope of line parallel to $P Q=\sqrt{3}$
$$
\therefore \theta=60^{\circ}
$$
$\mathrm{P}$ is a point at $\mathrm{r}$ distance from $\mathrm{Q}$
$\therefore$ Co-ordinates of $\mathrm{Q}$
$$
\begin{aligned}
& \left(2+\mathrm{r} \cos 60^{\circ}, 3+\mathrm{r} \sin 60^{\circ}\right) \\
& \left(2+\frac{r}{2}, 3+\frac{r \sqrt{3}}{2}\right) \\
& \because \text { P lies on } 2 x+4 y-27=0 \\
& \therefore 2\left(2+\frac{r}{2}\right)+4\left(3+\frac{r \sqrt{3}}{2}\right)-27=0 \\
& 4+r+12+2 \sqrt{3} r-27=0 ;(2 \sqrt{3}+1) r=11
\end{aligned}
$$
$\begin{aligned} & r=\frac{11}{2 \sqrt{3}+1}=\frac{11}{2 \sqrt{3}+1} \times \frac{(2 \sqrt{3}-1)}{(2 \sqrt{3}-1)} \\ & \therefore \mathrm{PQ}=r=2 \sqrt{3}-1\end{aligned}$
$\therefore$ slope of line parallel to $P Q=\sqrt{3}$
$$
\therefore \theta=60^{\circ}
$$
$\mathrm{P}$ is a point at $\mathrm{r}$ distance from $\mathrm{Q}$
$\therefore$ Co-ordinates of $\mathrm{Q}$
$$
\begin{aligned}
& \left(2+\mathrm{r} \cos 60^{\circ}, 3+\mathrm{r} \sin 60^{\circ}\right) \\
& \left(2+\frac{r}{2}, 3+\frac{r \sqrt{3}}{2}\right) \\
& \because \text { P lies on } 2 x+4 y-27=0 \\
& \therefore 2\left(2+\frac{r}{2}\right)+4\left(3+\frac{r \sqrt{3}}{2}\right)-27=0 \\
& 4+r+12+2 \sqrt{3} r-27=0 ;(2 \sqrt{3}+1) r=11
\end{aligned}
$$
$\begin{aligned} & r=\frac{11}{2 \sqrt{3}+1}=\frac{11}{2 \sqrt{3}+1} \times \frac{(2 \sqrt{3}-1)}{(2 \sqrt{3}-1)} \\ & \therefore \mathrm{PQ}=r=2 \sqrt{3}-1\end{aligned}$
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