Question 512 Marks
Find the equation of the line whose perpendicular distance from the origin is $4$ units and the angle which the normal makes with positive direction of $x-$axis is $15^\circ .$
Answer
View full question & answer→We are given that, $p = 4$ and $\omega = 15^\circ$
Now, $\cos 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
and $\sin 15^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
The equation of the line is $x \cos \omega + y \sin \omega = p$
$x \cos 15^{\circ}+y \sin 15^{\circ}$
or $\frac{\sqrt{3}+1}{2 \sqrt{2}} x+\frac{\sqrt{3}-1}{2 \sqrt{2}} y=4$
or $(\sqrt{3}+1) x+(\sqrt{3}-1) y=8 \sqrt{2}$
This is the required equation.
Now, $\cos 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
and $\sin 15^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
The equation of the line is $x \cos \omega + y \sin \omega = p$
$x \cos 15^{\circ}+y \sin 15^{\circ}$
or $\frac{\sqrt{3}+1}{2 \sqrt{2}} x+\frac{\sqrt{3}-1}{2 \sqrt{2}} y=4$
or $(\sqrt{3}+1) x+(\sqrt{3}-1) y=8 \sqrt{2}$
This is the required equation.