Reduce the equation


Reduce the equation $x+y-\sqrt{2}=0$ to the normal form $x \cos \alpha+y \sin \alpha=p$, and hence find the values of $\propto$ and $\mathrm{p}$.



Given equation is $x+y-\sqrt{2}=0$

If the equation is in the form of $a x+b y=c$, to get into the normal form, we should divide it by $\sqrt{a^{2}+b^{2}}$, so now

Divide by $\sqrt{1+1}=\sqrt{2}$

$\Rightarrow \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=1$

This is in the form of $x \cos \alpha+y \sin \alpha=p$

$\cos \alpha=\frac{1}{\sqrt{2}} \Rightarrow \alpha=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$\Rightarrow \alpha=\frac{\pi}{4}$ And

$\Rightarrow \mathrm{p}=1$

Conclusion: $\alpha=\frac{\pi}{4}$ and $p=1$


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