# Reduce the equation

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 $x \cos \alpha+y \sin \alpha=p$

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