Question

The centroid of an equilateral triangle is $(0,0)$. If two vertices of the triangle lie on $x+y=2\sqrt{2}$, then one of them will have its coordinates

• A. $(\sqrt{2}+\sqrt{6},\sqrt{2}-\sqrt{6})$
• B. $(\sqrt{2}+\sqrt{3},\sqrt{2}-\sqrt{3})$
• C. $(\sqrt{2}+\sqrt{5},\sqrt{2}-\sqrt{5})$
• D. $(\sqrt{2},\sqrt{2})$

Answer 1

Answer: (A)

$(\sqrt{2}+\sqrt{6},\sqrt{2}-\sqrt{6})$

Given line

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

Centroid $O(0,0)$

Let B and C lie on given line $x+y=2\sqrt{2}$

$OD=\frac{2\sqrt{2}}{\sqrt{2}}=2$ (O is centroid and D is mid point of BC)

The eq of OD is

$y=x=\sqrt{2}$

Also $BD=OD$

$\tan 60=2\sqrt{3}$ for coordinates of B and C

Using parametric eq ,we get

$\frac{x-\sqrt{2}}{\frac{-1}{\sqrt{2}}}=\frac{y-\sqrt{2}}{\frac{1}{\sqrt{2}}}=\pm 2\sqrt{3}$

$C(\sqrt{2}+\sqrt{6},\sqrt{2}-\sqrt{6})$