Question

Find the equation of a line for which

(i) p = 5, $\alpha ={60}^{\circ }$

(ii) p = 4, $\alpha ={150}^{\circ }$

(iii) p = 8, $\alpha ={225}^{\circ }$

(iv) p = 8, $\alpha ={300}^{\circ }$

Answer 1

$\left(i\right)p=5,\alpha ={60}^{\circ }$

$xcos\alpha +ysin\alpha =p$

$\Rightarrow xcos{60}^{\circ }+ysin{60}^{\circ }=5$

$\Rightarrow x\times \frac{1}{2}+y\times \frac{\sqrt{3}}{2}=5$

$\Rightarrow x+\sqrt{3}y=10$

$\left(ii\right)p=4,\alpha ={150}^{\circ }$

$xcos\alpha +ysin\alpha =p$

$\Rightarrow xcos{150}^{\circ }+ysin{150}^{\circ }=4$

$\Rightarrow -x\times \frac{\sqrt{3}}{2}+y\times \frac{1}{2}=4$

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

$\left(iii\right)p=8,\alpha ={225}^{\circ }$

$xcos\alpha +ysin\alpha =p$

$\Rightarrow xcos{225}^{\circ }+ysin{225}^{\circ }=8$

$\Rightarrow -x\times \frac{1}{\sqrt{2}}-y\times \frac{1}{\sqrt{2}}=8$

$\Rightarrow x+y+8\sqrt{2}=0$

$\left(iv\right)p=8,\alpha ={300}^{\circ }$

$xcos\alpha +ysin\alpha =p$

$\Rightarrow xcos{300}^{\circ }+ysin{300}^{\circ }=8$

$\Rightarrow xcos({360}^{\circ }-{60}^{\circ })+ysin({360}^{\circ }-{60}^{\circ })=8$

$\Rightarrow xcos{60}^{\circ }-ysin{60}^{\circ }=8$

$\Rightarrow \frac{x}{2}-\frac{\sqrt{3}y}{2}=8$

$\Rightarrow x-\sqrt{3}y=16$