The polar equation of the line perpendicular to the line $\sin heta-\cos heta=\frac{1}{r}$ and passing through the point $\left(2, \frac{\pi}{6} ight)$ is

Question

The polar equation of the line perpendicular to the line $\sin 	heta-\cos 	heta=\frac{1}{r}$ and passing through the point $\left(2, \frac{\pi}{6}ight)$ is

Aicharya · Accepted Answer

Given polar equation of line

and point $(2, \pi / 6)$ Let $\quad x=r \cos \theta=2 \cdot \cos \pi / 6=\sqrt{3}$ and $y=r \sin \theta=2 \cdot \sin \pi / 6=1$ The cartesian point is $(\sqrt{3}, 1)$. Now, we change the polar of line into cartesian form i.e., $r \sin \theta-r \cos \theta=1$

Equation of perpendicular line to Eq. (ii) is

which passes through $(\sqrt{3}, 1)$ $\lambda=\sqrt{3}+1$ From Eq. (iii), we get $x+y=\sqrt{3}+1$ Now, we convert this into polar form $\begin{aligned} r \cos \theta+r \sin \theta & =\sqrt{3}+1 \\ \Rightarrow \quad \sin \theta+\cos \theta & =\frac{\sqrt{3}+1}{r}\end{aligned}$

The polar equation of the line perpendicular to the line $\sin \theta-\cos \theta=\frac{1}{r}$ and passing…

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