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Reduce the following equations into normal form. Find their perpendicular distance from the origin and angle between perpendicular and the positive $x$-axis.
(i) $x-\sqrt{3} y+8=0$
(ii) $y-2=0$
(iii) $x-y=4$
(i) $x-\sqrt{3} y+8=0$
(ii) $y-2=0$
(iii) $x-y=4$
Solution:
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Verified Answer
(i) $x-\sqrt{3} y+8=0$
$-x+\sqrt{3} y=8$
Now $\quad a=1, b=\sqrt{3}$
Here $\sqrt{a^2+b^2}=\sqrt{1+3}=\sqrt{4}=2$
Thus, $-\frac{x}{2}+\frac{\sqrt{3}}{2} y=\frac{8}{2}$
$-\frac{x}{2}+\frac{\sqrt{3}}{2} y=4$
or $x \cos 120^{\circ}+y \sin 120^{\circ}=4$ (normal form)
Where $p=4$ and $\omega=120^{\circ}$
(ii) $y-2=0$
$y=2$
Now $a=0, b=1$
Here, $\sqrt{a^2+b^2}=\sqrt{0+1}=1$
Thus, $0 x+1 y=2$
or $x \cos 90^{\circ}+y \sin 90^{\circ}=2$ (normal form)
Where $\omega=90^{\circ}$ and $p=2$
(iii) $x-y=4$
Now $a=1, b=-1$
Here, $\sqrt{a^2+b^2}=\sqrt{1+1}=\sqrt{2}$
Thus, $\frac{1}{\sqrt{2}} x-\frac{1}{\sqrt{2}} y=4$ or $x \cos 315^{\circ}+y \sin \left(315^{\circ}\right)=4$ (normal form)
Where $\omega=315^{\circ}$ and $p=4$
$-x+\sqrt{3} y=8$
Now $\quad a=1, b=\sqrt{3}$
Here $\sqrt{a^2+b^2}=\sqrt{1+3}=\sqrt{4}=2$
Thus, $-\frac{x}{2}+\frac{\sqrt{3}}{2} y=\frac{8}{2}$
$-\frac{x}{2}+\frac{\sqrt{3}}{2} y=4$
or $x \cos 120^{\circ}+y \sin 120^{\circ}=4$ (normal form)
Where $p=4$ and $\omega=120^{\circ}$
(ii) $y-2=0$
$y=2$
Now $a=0, b=1$
Here, $\sqrt{a^2+b^2}=\sqrt{0+1}=1$
Thus, $0 x+1 y=2$
or $x \cos 90^{\circ}+y \sin 90^{\circ}=2$ (normal form)
Where $\omega=90^{\circ}$ and $p=2$
(iii) $x-y=4$
Now $a=1, b=-1$
Here, $\sqrt{a^2+b^2}=\sqrt{1+1}=\sqrt{2}$
Thus, $\frac{1}{\sqrt{2}} x-\frac{1}{\sqrt{2}} y=4$ or $x \cos 315^{\circ}+y \sin \left(315^{\circ}\right)=4$ (normal form)
Where $\omega=315^{\circ}$ and $p=4$
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