Question

Write the equation of the line passing through $(2,\frac{\pi }{2}),(3,\frac{\pi }{3})$ in polar form.

• A. $(3\sqrt{3}-4)\cos \theta -3\sin \theta$ + $\frac{6}{r}=0$
• B. $(3\sqrt{3}+4)\cos \theta -3\sin \theta +\frac{6}{r}=0$
• C. $(3\sqrt{3}-4)\cos \theta +3\sin \theta +\frac{6}{r}=0$
• D. $(3\sqrt{3}+4)\cos \theta +3\sin \theta +\frac{6}{r}=0$

Answer 1

The correct option is A $(3\sqrt{3}-4)\cos \theta -3\sin \theta$ + $\frac{6}{r}=0$

Equation of line joining $(r_1,{\theta }_1),(r_2,{\theta }_2)$ is given by
$\frac{\sin ({\theta }_2-{\theta }_1)}{r}=\frac{\sin (\theta -{\theta }_2)}{r_1}=\frac{\sin ({\theta }_1-\theta )}{r_2}$
So the equation of line passing through $(2,\frac{\pi }{2})$ and $(3,\frac{\pi }{3})$
$\Rightarrow \frac{\sin (-\frac{\pi }{6})}{r}=\frac{\sin (\theta -\frac{\pi }{3})}{2}=\frac{\sin (\frac{\pi }{2}-\theta )}{3}$
$\Rightarrow 3\sin (\theta -\frac{\pi }{3})+2\cos \theta =\frac{3}{r}$
$\Rightarrow \frac{3}{2}\sin \theta -\frac{3\sqrt{3}}{2}\cos \theta +2\cos \theta =\frac{3}{r}$
$\Rightarrow (3\sqrt{3}-4)cos\theta -3sin\theta$ + $\frac{6}{r}=0$