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 $\alpha .$

Answer 1

$\text{(i) Slope intercept form}(y=mx+c)$

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

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

$\Rightarrow m=-\sqrt{3},c=-\sqrt{2}$

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

$\text{(ii) Intercept form}(\frac{x}{a}+\frac{y}{b}=1)$

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

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

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

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

$\Rightarrow \text{x intercept =}\frac{-2}{\sqrt{3}},\text{y intercept = -2}$

$\text{(iii) Normal form}(xcos\alpha +ysin\alpha =p)$

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

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

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

$\Rightarrow cos\alpha =\frac{-\sqrt{3}}{2}=cos{210}^{\circ }andsin\alpha =\frac{-1}{2}$

$=sin{210}^{\circ }$

$\Rightarrow p=1,\alpha ={210}^{\circ }$