Question

Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of x-axis is ${15}^{\circ }.$

Answer 1

$\text{Here,}$

$p=4and\alpha ={15}^{\circ }$

$\text{The equation of line is}$

$xcos\alpha +ysin\alpha =p...\left(i\right)$

$xcos{15}^{\circ }+ysin{15}^{\circ }=4$

$cos{15}^{\circ }=cos(45-30)$

$=cos45cos30+sin45sin30$

$(\because cos(\theta -\Phi )=cos\theta cos\Phi +sin\Phi sin\Phi )$

$=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}$

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

$=\frac{1}{2\sqrt{2}}(\sqrt{3}+1)$

$sin15=sin(45-30)$

$sin45cos30cos45sin30$

$=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{1}{2\sqrt{2}}(\sqrt{3}-1)$

$\text{Putting in (1)}$

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

$x(\sqrt{3}+1)+y(\sqrt{3}-1)=8\sqrt{2}$