Normal form of (x - \sqrt{3} y + 6 = 0) is | Mathematics | SolveeIt

Normal form of $x - \sqrt{3} y + 6 = 0$ is

$x \cos \frac{\pi}{3} + y \sin \frac{\pi}{3} = 2$

$x \cos \frac{2\pi}{3} + y \sin \frac{2\pi}{3} = 3$

$x \cos \frac{\pi}{4} + y \sin \frac{\pi}{4} = 6$

$x \cos \frac{5 \pi}{6} + y \sin \frac{5\pi}{6} = 2$

Answer

$x \cos \frac{2\pi}{3} + y \sin \frac{2\pi}{3} = 3$

Explanation

Answer (b) $x \cos \frac{2\pi}{3} + y \sin \frac{2\pi}{3} = 3$

Mathematics Question on Straight lines