Question

Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular x-axis.

(i) $x-\sqrt{3}y+8=0$, (ii) y - 2 = 0, (iii) x - y = 4

Answer 1

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

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

Dividing both sides by $\sqrt{(-1{)}^2+(\sqrt{3}{)}^2}=2$, we have

$\frac{-x}{2}+\frac{\sqrt{3}}{2}y=\frac{8}{2}\Rightarrow \frac{-1}{2}x+\frac{\sqrt{3}}{2}y=4$

Put $cos\alpha =\frac{-1}{2}$ and $sin\alpha =\frac{\sqrt{3}}{2}$

$\Rightarrow$ $\alpha$ lies in IInd quadrant

$\therefore cos\alpha =\frac{-1}{2}$

$=-cos60=cos(180-60)$

$\Rightarrow \alpha ={120}^{\circ }$

$\therefore$ Equation of line in normal form is

$xcos\frac{2\pi }{3}+ysin\frac{2\pi }{3}=4$

Comparing it with $xcos\alpha +ysin\alpha =p$, we have

$\alpha =\frac{2\pi }{3}$ and p = 4

(ii) Here y - 2 = 0

$\Rightarrow$ y = 2 $\Rightarrow$ 0x + y = 2

Dividing both sides by $\sqrt{(0{)}^2+(1{)}^2}=1$, we have

0x + y = 2

Put $cos\alpha =0$ and $sin\alpha =1$

$\Rightarrow \alpha =\frac{\pi }{2}$

$\therefore$ Equation of line in normal form is

$xcos\frac{\pi }{2}+ysin\frac{\pi }{2}=2$

Comparing it with $xcos\alpha +ysin\alpha =p$,we have

$\alpha =\frac{\pi }{2}$ and p = 2

(iii) Here x - y = 4

Dividing both sides by $\sqrt{(1{)}^2+(-1{)}^2}=\sqrt{2},$ we have

$\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}=\frac{4}{\sqrt{2}}$

$\Rightarrow \frac{1}{\sqrt{2}}x-\frac{1}{\sqrt{2}}y=2\sqrt{2}$

Put $cos\alpha =\frac{1}{\sqrt{2}}$ and $sin\alpha =\frac{-1}{\sqrt{2}}$

$\Rightarrow$ $\alpha$ lies in $I{V}^{th}$ quadrant.

$\therefore cos\alpha =\frac{1}{\sqrt{2}}=cos(2\pi -\frac{\pi }{4})$

$\Rightarrow \alpha =\frac{7\pi }{4}$

$\therefore$ Equation of line in normal form is

$xcos\frac{7\pi }{4}+ysin\frac{7\pi }{4}=2\sqrt{2}$

Comparing it with $xcos\alpha +ysin\alpha =p$, we have

$\alpha =\frac{7\pi }{4}$ and $p=2\sqrt{2}$