Question

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

Answer 1

(i) The given equation is

$x-\sqrt{3}y+8=0$

which can be written as
$x-\sqrt{3}y=-8$
$\Rightarrow -x+\sqrt{3}y=8$
On dividing both sides by $\sqrt{(-1{)}^2+(\sqrt{3}{)}^2}=\sqrt{4}=2$, we get
$-\frac{x}{2}+\frac{\sqrt{3}}{2}y=\frac{8}{2}$
$\Rightarrow (-\frac{1}{2})x+\left(\frac{\sqrt{3}}{2}\right)y=4$
$\Rightarrow x\cos {120}^{\circ }+y\sin {120}^{\circ }=4$
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

$\alpha ={120}^{\circ }$ and $p=4$
So, the perpendicular distance of the line from the origin is $4$ and the angle between the perpendicular and the positive $x$-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 $0.x+1.y=2$
$\Rightarrow x\cos {90}^{\circ }+y\sin {90}^{\circ }=2$
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

$\alpha ={90}^{\circ }$ and $p=2$
So, the perpendicular distance of the line from the origin is $2$ and the angle between the perpendicular and the positive $x$--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

$\frac{1}{\sqrt{2}}x+(-\frac{1}{\sqrt{2}})y=\frac{4}{\sqrt{2}}$
$\Rightarrow x\cos (2\pi -\frac{\pi }{4})+y\sin (2\pi -\frac{\pi }{4})=2\sqrt{2}$ (Since, cosine is positive and sine is negative in fourth quadrant )
$\Rightarrow x\cos {315}^{\circ }+y\sin {315}^{\circ }=2\sqrt{2}$
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

$\alpha ={315}^{\circ }$ 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 $x$-axis is ${315}^{\circ }$