Question

I : The length of the perpendicular from $(3,\frac{3\pi }{4})$ to the line $r(\cos \theta -\sin \theta )=6\sqrt{2}$ is $9$.
II: The points $(\sqrt{6},\frac{5\pi }{6}),(\sqrt{2},\frac{\pi }{3}),(\sqrt{2},\frac{4\pi }{3})$ forms an equilateral triangle.
Which of the above is true?

• A. Only I
• B. Only II
• C. both I and II
• D. neither I nor II

Answer 1

The correct option is C both I and II

Length of perpendicular from$(3,\frac{3\pi }{4})$to the line $r\cos \theta -r\sin \theta -6\sqrt{2}$ is
$\left|\frac{3\cos \frac{3\pi }{4}-3\sin \frac{3\pi }{4}-6\sqrt{2}}{\sqrt{1^2+1^2}}\right|=\left|\frac{-9\sqrt{2}}{\sqrt{2}}\right|=9$
Cartesian form of the given coordinates is $A(-\frac{3\sqrt{2}}{2},\frac{\sqrt{6}}{2}),B(\frac{\sqrt{2}}{2},\frac{\sqrt{6}}{2}),C(-\frac{\sqrt{2}}{2},-\frac{\sqrt{6}}{2})$
$AB=\sqrt{{\left(\frac{-4\sqrt{2}}{2}\right)}^2+0}=2\sqrt{2}$
Similarly, $BC=2\sqrt{2}$ and $AC=2\sqrt{2}$
Hence both statements are true.