Question

If the centroid and a vertex of an equilateral triangle are $(2,3)$ and $(4,3)$ respectively, then the other two vertices of the triangle are

• A. $(1,3+\sqrt{3})\text{and}(1,3-\sqrt{3})$
• B. $(2,3+\sqrt{3})\text{and}(2,3-\sqrt{3})$
• C. $(-1,3+\sqrt{3})\text{and}(-1,3-\sqrt{3})$
• D. $(1,3+2\sqrt{3})\text{and}(1,3-2\sqrt{3})$

Answer 1

The correct option is A $(1,3+\sqrt{3})\text{and}(1,3-\sqrt{3})$

Assuming the triangle to be $ABC$ and $G$ be the centroid.
Let $AD$ be the median through $A$.
We know that the centroid divides the median in ratio of $2:1$.
So $\frac{AG}{GD}=\frac{2}{1}$


Let the coordinates of $D$ be $(x,y)$.
Then $G=(\frac{2x+4}{3},\frac{2y+3}{3})$
$\Rightarrow (2,3)=(\frac{2x+4}{3},\frac{2y+3}{3})\Rightarrow x=1,y=3\Rightarrow D=(1,3)$

As $△ABC$ is equilateral triangle, so median is same as altitude of the triangle. Then
$\tan {60}^{\circ }=\frac{AD}{BD}\Rightarrow \sqrt{3}=\frac{3}{BD}\Rightarrow BD=\sqrt{3}=CD$

Now, $A(4,3),G(2,3)$ and $D(1,3)$ lie on $y=3$
and $BC$ is perpendicular to $AD.$
$\Rightarrow B,C,D$ have same $x-$coordinate i.e. $1.$
Let $y$-coordinate of $B$ be $(1,y)$
Then $\sqrt{(1-1{)}^2+(y-3{)}^2}=\sqrt{3}$
$\Rightarrow (y-3{)}^2=3$
$\Rightarrow y=3\pm \sqrt{3}$

Therefore, the other coordinates of the triangle are $(1,3+\sqrt{3})$ and $(1,3-\sqrt{3})$