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If the normal drawn from the origin to the straight line $2 x+7 y+6=0$ makes an angle 0 with the positive $X$-axis, then $0=$
Options:
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Verified Answer
The correct answer is:
$\pi+\tan ^{-1} \frac{7}{2}$
Given equation of straight line,
$2 x+7 y+6=0 \Rightarrow \frac{x}{-3}+\frac{y}{-6 / 7}=1$
So, angle $\theta=\pi+\angle A O M$
$\begin{aligned} & =\pi+\alpha[\text { Let } \angle A O M=\alpha], \\ & \because(\tan \alpha) \times \text { Slope of line }=-1 \\ & \Rightarrow(\tan \alpha) \times\left(-\frac{2}{7}\right)=-1 \\ & \Rightarrow \quad \tan \alpha=7 / 2\end{aligned}$
So, required angle $\theta=\pi+\tan ^{-1} \frac{7}{2}$ Hence, option (c) is correct.
$2 x+7 y+6=0 \Rightarrow \frac{x}{-3}+\frac{y}{-6 / 7}=1$
So, angle $\theta=\pi+\angle A O M$
$\begin{aligned} & =\pi+\alpha[\text { Let } \angle A O M=\alpha], \\ & \because(\tan \alpha) \times \text { Slope of line }=-1 \\ & \Rightarrow(\tan \alpha) \times\left(-\frac{2}{7}\right)=-1 \\ & \Rightarrow \quad \tan \alpha=7 / 2\end{aligned}$
So, required angle $\theta=\pi+\tan ^{-1} \frac{7}{2}$ Hence, option (c) is correct.
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