Question

Reduce the equation $\sqrt{3}x+y-8=0$ into normal form. Find the values at $p$ and $\omega$.

Answer 1

$\sqrt{3}x+y-8=0$

$\sqrt{3}x+y=8$

$\frac{\sqrt{3}x}{2}+\frac{y}{2}=4$

$x\left(\cos \left(\pi /6\right)\right)+y\left(\sin \left(\pi /6\right)\right)=4$ ...... $1$

$x\left(\cos \theta \right)+y\left(\sin \theta \right)=p$ ....... $2$

On comparing

$\theta =\pi /6$

and $p=4$

$w=\frac{2\pi }{\theta }=\frac{2\pi }{\pi /6}=6\times 2=12$

$\therefore w=12$ and $p=4$