Question

One vertex of the equilateral triangle with centriod at origin and one side as $x+y-2=0$ is

• A. $(-2,-2)$
• B. $(2,-2)$
• C. $(-2,2)$
• D. $(2,2)$

Answer 1

The correct option is A $(-2,-2)$

Let the vertices of the triangle is $A,B,C$ and $D$ be the mid point of $BC$ and $BC$ lies in the line $x+y-2=0$
$\therefore A$ will lie in $3^{\text{rd}}$ quadrant.
$\because$ the triangle is equilateral the centroid $G(0,0)$ is also orthocenter and median $AD$ is also its altitude and $G$ divides it in the ratio $2:1$.

Let the coordinates of $A$ be $(a,b)$, then
$\frac{|a+b-2|}{\sqrt{2}}=3\times \frac{2}{\sqrt{2}}\Rightarrow -(a+b-2)=6\Rightarrow a+b+4=0\cdots \left(i\right)$

Equation of the line perpendicular to $BC$ and passing through origin is $x=y$ and $A$ lie on it.
$\therefore a=b$
Hence, coordinates of $A$ is $(-2,-2)$.