Question

Reduce the equation $\sqrt{3}$ x + y + 2 = 0 to:
(i) slope-intercept form and find slope and y-intercept;
(ii) intercept form and find intercept on the axes;
(iii) the normal form and find p and α.

Answer 1

(i) $\sqrt{3}$x + y + 2 = 0

$\Rightarrow y=-\sqrt{3}x-2$

This is the slope intercept form of the given line.

Here, slope = $-\sqrt{3}$ and y-intercept = $-$2

(ii) $\sqrt{3}$x + y + 2 = 0

$$\begin{gathered} \Rightarrow \sqrt{3}x+y=-2 \\ \Rightarrow \frac{\sqrt{3}x}{-2}+\frac{y}{-2}=1\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}-2\right] \\ \Rightarrow \frac{x}{-{\displaystyle \frac{2}{\sqrt{3}}}}+\frac{y}{-2}=1 \end{gathered}$$

This is the intercept form of the given line.

Here, x-intercept = $-\frac{2}{\sqrt{3}}$ and y-intercept = $-$2

(iii) $\sqrt{3}$x + y + 2 = 0

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

This is the normal form of the given line.
Here, p = 1, $\mathrm{cos\alpha }=-\frac{\sqrt{3}}{2}$ and $\mathrm{sin\alpha }=-\frac{1}{2}$
$\Rightarrow \mathrm{\alpha }={210}^{\circ }$