Question

Reduce the following equations to the normal form and find p and $\alpha$ in each case :

$\left(i\right)x+\sqrt{3}y-4=0$ $\left(ii\right)x+y+\sqrt{2}=0$ $\left(iii\right)x-y+2\sqrt{2}=0$ $\left(iv\right)x-3=0$ $\left(v\right)y-2=0$

Answer 1

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

$\text{Divide the equation by 2, we get}$

$\frac{1}{2}X+\frac{\sqrt{3}}{2}y=2$

$xcos60+ysin60=2$

$So,p=2and\alpha ={60}^{\circ }=\frac{\pi }{3}=60$

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

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

$\text{Dividing each term by}\sqrt{(1{)}^2+(1{)}^2}=\sqrt{2}$

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

$\text{Comparing with x}cos\alpha +ysin\alpha =p$

$cos\alpha =\frac{-1}{\sqrt{2}},sin\alpha =\frac{-1}{\sqrt{2}},p=1$

$\text{Both are negative}$

$\alpha \text{is in III quadrant}$

$\Rightarrow \alpha =\pi \frac{\pi }{4}=\frac{5\pi }{4}={225}^{\circ }$

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

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

$\text{Dividing each term by}\sqrt{(1{)}^2+(1{)}^2}=\sqrt{2}$

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

$\text{Comparing with}xcos\alpha +ysin\alpha =p$

$cos\alpha =\frac{-1}{\sqrt{2}},sin\alpha =\frac{-1}{\sqrt{2}},p=2$

$\alpha \text{is in II quadrant}$

$\Rightarrow a=\frac{\pi }{4}+\frac{\pi }{2}=\frac{3\pi }{4}={135}^{\circ },p=2$

$\left(iv\right)x-3=0$

$x=3$

$\text{Comparing with}xcos\alpha +ysin\alpha =p$

$cos\alpha =1$

$=cos0$

$\Rightarrow \alpha =0$

$p=3$

$\left(v\right)y-2=0$

$y=2$

$\text{Comparing with}xcos\alpha +ysin\alpha =p$

$sin\alpha =1$

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