Question:
Reduce the equation $x+\sqrt{3} y-4=0$ to the normal form $x \cos \alpha+y \sin \alpha=$ $p$, and hence find the values of $\propto$ and $p$.
Solution:
Given equation is
$x+\sqrt{3} y-4=0$
If the equation is in the form of ax + by = c, to get into the normal form, we should divide
it by $\sqrt{a^{2}+b^{2}}$ so now
Divide by $\sqrt{\sqrt{3}^{2}+1^{2}}=2$
Now we get $\Rightarrow \frac{\mathrm{x}}{2}+\frac{\sqrt{3} \mathrm{y}}{2}=1$
This is in the form of $x \cos \alpha+y \sin \alpha=p$