Reduce the equation


Reduce the equation $x+\sqrt{3} y-4=0$ to the normal form x cos ∝ + y sin ∝ = p, and hence find the values of ∝ and p



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


Now we get

$\Rightarrow \frac{x}{2}+\frac{\sqrt{3} y}{2}=1$

This is in the form of

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


$\cos \alpha=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{3}$

And p = 1

Conclusion: $\alpha=\frac{\pi}{3}$ and $p=1$




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