Question

Reduce the following equations to the normal form and find p and α in each case:
(i) $x+\sqrt{3}y-4=0$
(ii) $x+y+\sqrt{2}=0$
(iii) $x-y+2\sqrt{2}=0$
(iv) x − 3 = 0
(v) y − 2 = 0.

Answer 1

(i) $x+\sqrt{3}y-4=0$

$$\begin{gathered} \Rightarrow x+\sqrt{3}y=4 \\ \Rightarrow \frac{x}{\sqrt{1^2+{\left(\sqrt{3}\right)}^2}}+\frac{\sqrt{3}y}{\sqrt{1^2+{\left(\sqrt{3}\right)}^2}}=\frac{4}{\sqrt{1^2+{\left(\sqrt{3}\right)}^2}}\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\sqrt{{\left(\mathrm{coefficient}\mathrm{of}x\right)}^2+{\left(\mathrm{coefficient}\mathrm{of}y\right)}^2}\right] \\ \Rightarrow \frac{x}{2}+\frac{\sqrt{3}y}{2}=2 \end{gathered}$$

This is the normal form of the given line, where p = 2, $\mathrm{cos\alpha }=\frac{1}{2}$ and $\mathrm{sin\alpha }=\frac{\sqrt{3}}{2}\Rightarrow \mathrm{\alpha }=\frac{\mathrm{\pi }}{3}$.

(ii) $x+y+\sqrt{2}=0$

$$\begin{gathered} \Rightarrow -x-y=\sqrt{2} \\ \Rightarrow -\frac{x}{\sqrt{{\left(-1\right)}^2+{\left(-1\right)}^2}}-\frac{y}{\sqrt{{\left(-1\right)}^2+{\left(-1\right)}^2}}=\frac{\sqrt{2}}{\sqrt{{\left(-1\right)}^2+{\left(-1\right)}^2}}\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\sqrt{{\left(\mathrm{coefficient}\mathrm{of}x\right)}^2+{\left(\mathrm{coefficient}\mathrm{of}y\right)}^2}\right] \\ \Rightarrow -\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}=1 \end{gathered}$$

This is the normal form of the given line, where p = 1, $\mathrm{cos\alpha }=-\frac{1}{\sqrt{2}}$ and
$$\begin{gathered} \sin \alpha =-\frac{1}{\sqrt{2}} \\ \Rightarrow \alpha ={225}^{\circ }\left[\because \mathrm{The}\mathrm{coefficent}\mathrm{of}x\mathrm{and}y\mathrm{are}\mathrm{negative}.\mathrm{So},\alpha \mathrm{lies}\mathrm{in}\mathrm{third}\mathrm{quadrant}\right] \end{gathered}$$

(iii) $x-y+2\sqrt{2}=0$

$$\begin{gathered} \Rightarrow -x+y=2\sqrt{2} \\ \Rightarrow -\frac{x}{\sqrt{{\left(-1\right)}^2+{\left(1\right)}^2}}+\frac{y}{\sqrt{{\left(-1\right)}^2+{\left(1\right)}^2}}=\frac{2\sqrt{2}}{\sqrt{{\left(-1\right)}^2+{\left(1\right)}^2}}\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\sqrt{{\left(\mathrm{coefficient}\mathrm{of}x\right)}^2+{\left(\mathrm{coefficient}\mathrm{of}y\right)}^2}\right] \\ \Rightarrow -\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=2 \end{gathered}$$

This is the normal form of the given line, where p = 2, $\mathrm{cos\alpha }=-\frac{1}{\sqrt{2}}$ and
$$\begin{gathered} \sin \alpha =\frac{1}{\sqrt{2}} \\ \Rightarrow \alpha ={135}^{\circ }\left[\because \mathrm{The}\mathrm{coefficent}\mathrm{of}x\mathrm{and}y\mathrm{are}\mathrm{negative}\mathrm{and}\mathrm{positive}\mathrm{respectively}.\mathrm{So},\alpha \mathrm{lies}\mathrm{in}\mathrm{second}\mathrm{quadrant}\right] \end{gathered}$$.

(iv) x − 3 = 0

$$\begin{gathered} \Rightarrow x=3 \\ \Rightarrow x+0\times y=3 \\ \Rightarrow \frac{x}{\sqrt{1^2+0^2}}+0\times \frac{y}{\sqrt{1^2+0^2}}=\frac{3}{\sqrt{1^2+0^2}}\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\sqrt{{\left(\mathrm{coefficient}\mathrm{of}x\right)}^2+{\left(\mathrm{coefficient}\mathrm{of}y\right)}^2}\right] \\ \Rightarrow x+0\times y=3 \end{gathered}$$

This is the normal form of the given line, where p = 3, $\mathrm{cos\alpha }=1$ and $\mathrm{sin\alpha }=0\Rightarrow \mathrm{\alpha }=0$.

(v) y − 2 = 0

$$\begin{gathered} \Rightarrow y=2 \\ \Rightarrow 0\times x+y=2 \\ \Rightarrow \frac{0\times x}{\sqrt{0^2+1^2}}+\frac{y}{\sqrt{0^2+1^2}}=\frac{2}{\sqrt{0^2+1^2}}\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\sqrt{{\left(\mathrm{coefficient}\mathrm{of}x\right)}^2+{\left(\mathrm{coefficient}\mathrm{of}y\right)}^2}\right] \\ \Rightarrow 0\times x+y=2 \end{gathered}$$

This is the normal form of the given line, where p = 2, $\mathrm{cos\alpha }=0$ and $\mathrm{sin\alpha }=1\Rightarrow \mathrm{\alpha }={90}^{\circ }$.