Reduce the equation x-sqrt{3} y+8=0 into norm | vedclass

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(A) The given equation is $x-\sqrt{3} y+8=0$.It can be written as $x-\sqrt{3} y=-8$.Multiplying by $-1$,we get $-x+\sqrt{3} y=8$.Dividing both sides b

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$p=4, \omega=120^{\circ}$

Vedclass · Answer

(A) The given equation is $x-\sqrt{3} y+8=0$.It can be written as $x-\sqrt{3} y=-8$.Multiplying by $-1$,we get $-x+\sqrt{3} y=8$.Dividing both sides by $\sqrt{(-1)^{2}+(\sqrt{3})^{2}}=\sqrt{1+3}=2$,we obtain:$-\frac{1}{2} x+\frac{\sqrt{3}}{2} y=4$.This is in the form $x \cos \omega+y \sin \omega=p$,where $\cos \omega = -\frac{1}{2}$ and $\sin \omega = \frac{\sqrt{3}}{2}$.Since $\cos \omega$ is negative and $\sin \omega$ is positive,$\omega$ lies in the second quadrant.Thus,$\omega = 180^{\circ}-60^{\circ} = 120^{\circ}$.The perpendicular distance $p = 4$ and the angle $\omega = 120^{\circ}$.

Reduce the equation $x-\sqrt{3} y+8=0$ into normal form. Find the perpendicular distance from the origin and the angle between the perpendicular and the positive $x$-axis.

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