Question

Reduce $x+\sqrt{3}y+4=0$ to the :
Slope intercepts form and find its slope and $y-$
Intercept.Intercepts form and find its intercepts on the axes
Normal form and find values of $P$ and $\alpha$.

Answer 1

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

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

Dividing by $\sqrt{{\left(\sqrt{3}\right)+1}^2}=2$

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

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

Normal form of any line is

$x\cos \alpha +y\sin \alpha =p$ .........$\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$

we get $p=4$

and $\cos \alpha =\frac{\sqrt{3}}{2}=\cos {30}^{\circ }$ and $\sin \alpha =\frac{1}{2}=\sin {30}^{\circ }$

Thus,$\alpha ={30}^{\circ }$

So $\alpha ={30}^{\circ }$ and $p=4$