Question

The equation of line passing through $\left( - 1,\frac{\pi}{2} \right)$ and perpendicular to $\sqrt{3}\sin\theta + 2\cos\theta = \frac{4}{r}$ is

• A. $2 = \sqrt{3}r\cos\theta - 2r\sin\theta$
• B. $5 = - 2\sqrt{3}r\sin\theta + 4r\cos\theta$
• C. $2 = \sqrt{3}r\cos\theta + 2r\sin\theta$
• D. $5 = 2\sqrt{3}r\sin\theta + 4r\cos\theta$

Answer 1

Answer: (A)

$2 = \sqrt{3}r\cos\theta - 2r\sin\theta$

Answer 2

Equation of a line, perpendicular to $\sqrt{3}\sin\theta + 2\cos\theta = \frac{4}{r}$ is $\sqrt{3}\sin\left( \frac{\pi}{2} + \theta \right) + 2\cos\left( \frac{\pi}{2} + \theta \right) = \frac{k}{r}$

It is passing through $\left( - 1,\frac{\pi}{2} \right)$. Hence,

$\sqrt{3}\sin\pi + 2\cos\pi = k/ - 1$ ⇒ $k = 2$

$\therefore$ $\sqrt{3}\cos\theta - 2\sin\theta = \frac{2}{r}$ ⇒ $2 = \sqrt{3}r\cos\theta - 2r\sin\theta$.