Reduce the following equations into normal form.
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) $x-y=4$

Solution:

(i) The given equation is $x-\sqrt{3} y+8=0$.

It can be reduced 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 obtain

$-\frac{x}{2}+\frac{\sqrt{3}}{2} y=\frac{8}{2}$

$\Rightarrow\left(-\frac{1}{2}\right) x+\left(\frac{\sqrt{3}}{2}\right) y=4$

$\Rightarrow x \cos 120^{\circ}+y \sin 120^{\circ}=4$ ..(1)

Equation (1) is in the normal form.

On comparing equation (1) with the normal form of equation of line

$x \cos \omega+y \sin \omega=p$, we obtain $\omega=120^{\circ}$ and $p=4$

Thus, the perpendicular distance of the line from the origin is 4, while the angle between the perpendicular and the positive xaxis is 120°.

(ii) The given equation is – 2 = 0.

It can be reduced as 0.x + 1.= 2

On dividing both sides by $\sqrt{0^{2}+1^{2}}=1$, we obtain $0 \cdot x+1 \cdot y=2$

$\Rightarrow x \cos 90^{\circ}+y \sin 90^{\circ}=2 \ldots$ (1)

Equation (1) is in the normal form.

On comparing equation (1) with the normal form of equation of line

$x \cos \omega+y \sin \omega=p$, we obtain $\omega=90^{\circ}$ and $p=2$.

Thus, the perpendicular distance of the line from the origin is 2, while the angle between the perpendicular and the positive xaxis is 90°.

(iii) The given equation is $x-y=4$.

It can be reduced as $1 \cdot x+(-1) y=4$

On dividing both sides by $\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$, we obtain

$\frac{1}{\sqrt{2}} x+\left(-\frac{1}{\sqrt{2}}\right) y=\frac{4}{\sqrt{2}}$

$\Rightarrow x \cos \left(2 \pi-\frac{\pi}{4}\right)+y \sin \left(2 \pi-\frac{\pi}{4}\right)=2 \sqrt{2}$

$\Rightarrow x \cos 315^{\circ}+y \sin 315^{\circ}=2 \sqrt{2}$ $\ldots(1)$

Equation (1) is in the normal form.

On comparing equation (1) with the normal form of equation of line

$x \cos \omega+y \sin \omega=p$, we obtain $\omega=315^{\circ}$ and $\mathrm{p}=2 \sqrt{2}$.

Thus, the perpendicular distance of the line from the origin is $2 \sqrt{2}$, while the angle between the perpendicular and the positive $x$-axis is $315^{\circ}$.