Question

An equilateral triangle has its centroid at origin and one side is $x+y-1=0$ then the vertex of the triangle not on the given side is

• A. $(1,1)$
• B. $(-1,-1)$
• C. $(2,2)$
• D. $(-2,-2)$

Answer 1

Answer: (B)

$(-1,-1)$

Let the coordinates of vertex $A$ be $(h,k)$.

Then, $AD$ is perpendicular to $BC$,

$\therefore OA\perp BC\Rightarrow \frac{k-0}{h-0}\times \frac{-1}{1}=-1\Rightarrow k=h$ ...(1)

Let the coordinates of $D$ be $(\alpha ,\beta )$

Then the coordinates of $O$ are $(\frac{2\alpha +h}{2+1},\frac{2\beta +k}{2+1})$, therefore

$\frac{2\alpha +h}{3}=0$ and $\frac{2\beta +h}{3}=0$

$\Rightarrow \alpha =-\frac{h}{2},\beta =-\frac{k}{2}$

Since $(\alpha ,\beta )$ lies on $x+y-1=0,$ therefore

$\alpha +\beta -1=0\Rightarrow -\frac{h}{2}-\frac{k}{2}-1=0\Rightarrow h+k+2=0\Rightarrow 2h+2=0\Rightarrow h=k=-1$

Hence the coordinates of vertex $A$ are $(-1,-1)$