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If $\pi / 3$ is the angle between the straight lines $p x+q y+r=0$ and $x \sin \alpha+y \cos \alpha=r(r \neq 0)$ which meet at a point $A$ and the straight line $x \cos \alpha-y \sin \alpha=0$ also passes through the point $A$, then
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Verified Answer
The correct answer is:
$p^2+q^2=4$
Lines
$\begin{array}{ll}
& p x+q y+r=0 \\
& x \sin \alpha+y \cos \alpha=r \\
\text { and } \quad & x \cos \alpha-y \sin \alpha=0
\end{array}$
and
$x \cos \alpha-y \sin \alpha=0
$are intersecting at $A$
$\therefore$ lines are concurrent
$\begin{aligned}
& \Rightarrow \quad\left|\begin{array}{ccc}
p & q & r \\
\sin \alpha & \cos \alpha & -r \\
\cos \alpha & -\sin \alpha & 0
\end{array}\right|=0 \\
& \quad p \sin \alpha+q \cos \alpha+1=0
\end{aligned}$
Angle between Eqs. (i) and (ii) is $\frac{\pi}{3}$
$\therefore \tan \frac{\pi}{3}=\frac{\frac{-p}{q}+\frac{\sin \alpha}{\cos \alpha}}{1+\frac{p \sin \alpha}{q \cos \alpha}} \Rightarrow \sqrt{3}=\frac{-p \cos \alpha+q \sin \alpha}{q \cos \alpha+p \sin \alpha}$
$\begin{aligned}
\Rightarrow & \sqrt{3}(q \cos \alpha+p \sin \alpha)=-p \cos \alpha+q \sin \alpha \\
& -\sqrt{3}=-p \cos \alpha+q \sin \alpha \quad[\because q \cos \alpha+p \sin \alpha=-1] \\
\Rightarrow & p \cos \alpha-q \sin \alpha=\sqrt{3} \text { and } p \sin \alpha+q \cos \alpha=-1
\end{aligned}$
Adding and squaring, we get
$p^2+q^2=3+1 \Rightarrow p^2+q^2=4$
$\begin{array}{ll}
& p x+q y+r=0 \\
& x \sin \alpha+y \cos \alpha=r \\
\text { and } \quad & x \cos \alpha-y \sin \alpha=0
\end{array}$
and
$x \cos \alpha-y \sin \alpha=0
$are intersecting at $A$
$\therefore$ lines are concurrent
$\begin{aligned}
& \Rightarrow \quad\left|\begin{array}{ccc}
p & q & r \\
\sin \alpha & \cos \alpha & -r \\
\cos \alpha & -\sin \alpha & 0
\end{array}\right|=0 \\
& \quad p \sin \alpha+q \cos \alpha+1=0
\end{aligned}$
Angle between Eqs. (i) and (ii) is $\frac{\pi}{3}$
$\therefore \tan \frac{\pi}{3}=\frac{\frac{-p}{q}+\frac{\sin \alpha}{\cos \alpha}}{1+\frac{p \sin \alpha}{q \cos \alpha}} \Rightarrow \sqrt{3}=\frac{-p \cos \alpha+q \sin \alpha}{q \cos \alpha+p \sin \alpha}$
$\begin{aligned}
\Rightarrow & \sqrt{3}(q \cos \alpha+p \sin \alpha)=-p \cos \alpha+q \sin \alpha \\
& -\sqrt{3}=-p \cos \alpha+q \sin \alpha \quad[\because q \cos \alpha+p \sin \alpha=-1] \\
\Rightarrow & p \cos \alpha-q \sin \alpha=\sqrt{3} \text { and } p \sin \alpha+q \cos \alpha=-1
\end{aligned}$
Adding and squaring, we get
$p^2+q^2=3+1 \Rightarrow p^2+q^2=4$
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