Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive xaxis.(i) $x-\sqrt{3}y+8=0$, (ii) $y-2=0$, (iii) $x-y=4$.

(i) The given equation is
$x-\sqrt{3}y+8=0$
which can be written as
$\begin{array}{rl}& x-\sqrt{3}y=-8\\ & \Rightarrow -x+\sqrt{3}y=8\end{array}$
On dividing both sides by $\sqrt{(-1{)}^2+(\sqrt{3}{)}^2}=\sqrt{4}=2$, we get
$\begin{array}{rl}& -\frac{x}{2}+\frac{\sqrt{3}}{2}y=\frac{8}{2}\\ & \Rightarrow \left(-\frac{1}{2}\right)x+\left(\frac{\sqrt{3}}{2}\right)\mathrm{y}=4\\ & \Rightarrow x\cos 120^{\circ }+\mathrm{y}\sin 120^{\circ }=4\end{array}$
which is the normal form.
On comparing it with the normal form of equation of line $\mathrm{x}\cos \alpha +\mathrm{y}\sin \alpha =\mathrm{p}$, we get
$\alpha =120^{\circ }\text{ and }\mathrm{p}=4$
So, the perpendicular distance of the line from the origin is 4 and the angle between the perpendicular and the positive $\mathrm{x}$-axis is $120^{\circ }$

(ii) The given equation is
$y-2=0$
which can be written as $0.x+1.y=2$
On dividing both sides by $\sqrt{0^2+1^2}=1$, we get $0.x+1.y=2$
$\Rightarrow \mathrm{x}\cos 90^{\circ }+\mathrm{y}\sin 90^{\circ }=2$
which is the normal form.
On comparing it with the normal form of equation of line $x\cos \alpha +y\sin \alpha =p$ ,we get
$\alpha =90^{\circ }\text{ and }p=2$
So, the perpendicular distance of the line from the origin is 2 and the angle between the perpendicular and the positive $\mathrm{x}$--axis is $90^{\circ }$


(iii) The given equation is $x-y=4$
which can be written as
$1.x+(-1)y=4$
On dividing both sides by $\sqrt{1^2+(-1{)}^2}=\sqrt{2}$, we get
$\frac{1}{\sqrt{2}}x+\left(-\frac{1}{\sqrt{2}}\right)y=\frac{4}{\sqrt{2}}$
$\Rightarrow \mathrm{x}\cos \left(2\pi -\frac{\pi }{4}\right)+\mathrm{y}\sin \left(2\pi -\frac{\pi }{4}\right)=2\sqrt{2}$ (Since, cosine is positive and sine is negative in fourth quadrant )
$\Rightarrow \mathrm{x}\cos 315^{\circ }+\mathrm{y}\sin 315^{\circ }=2\sqrt{2}$
which is the normal form.
On comparing it with the normal form of equation of line $x\cos \alpha +y\sin \alpha =p$ ,we get
$\alpha =315^{\circ }\text{ and }p=2\sqrt{2}$
So, the perpendicular distance of the line from the origin is $2\sqrt{2}$, and the angle between the perpendicular and the positive $\mathrm{x}$-axis is $315^{\circ }$