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 as

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let the vertices 'B' and 'C' lie on the given line.
The distance from the origin (O) to the line x + y = 2$\sqrt{2}$ is $OD=\frac{2\sqrt{2}}{\sqrt{1^2+1^2}}$
$=\frac{2\sqrt{2}}{\sqrt{2}}$
$\therefore OD=2$
Equation of OD is
$y=x\Rightarrow x=y$
Solving given equation and above equation, we get D$(\sqrt{2},\sqrt{2})$

Also, from Right angled triangle $\Delta BDO$
$\Rightarrow \tan {30}^{\circ }=\frac{OD}{BD}$
$BD=\frac{OD}{\tan {30}^{\circ }}=2\sqrt{3}$ for the coordinates of B and C, using parametric equation of line, we get
$\frac{x-\sqrt{2}}{\frac{1}{\sqrt{2}}}=\frac{y-\sqrt{2}}{\frac{1}{\sqrt{2}}}=\pm 2\sqrt{3}\Rightarrow C\equiv (\sqrt{2}+\sqrt{6},\sqrt{2}-\sqrt{6})andD\equiv (\sqrt{2}-\sqrt{6},\sqrt{2}+\sqrt{6})$