Thank you for reporting, we will resolve it shortly
Q.
The normal form of the equation $x-\sqrt{3}y+8=0$ is
Straight Lines
Solution:
Given equation of line is $x-\sqrt{3}y+8=0$
$\Rightarrow -x+\sqrt{3}y=8\quad\ldots\left(i\right)$
Now, divide $\left(i\right)$ by $\sqrt{(coefficient\, of \,x)^{2}+(coefficient \,of \,y)}^{2}$
$=\sqrt{\left(-1\right)^{2}+\left(\sqrt{3}\right)^{2}}=\sqrt{1+3}=2$, we get normal form as
$-\frac{1}{2}x+\frac{\sqrt{3}}{2}y=\frac{8}{2}$
$\Rightarrow \left(-cos \,60^{\circ}\right)x + \left(sin \,60^{\circ}\right)y = 4$
$\Rightarrow x\,cos \left(180^{\circ} - 60^{\circ}\right) +y\, sin \left(180^{\circ} - 60^{\circ}\right) = 4$
$\Rightarrow x\,cos\, 120^{\circ} +y\, sin \,120^{\circ} = 4$