Question

One vertex of the equilateral triangle with centroid at the origin and one side as $\mathrm{x}+\mathrm{y}-2=0$ is

• A. $(–1,–1)$
• B. $(2,2)$
• C. $(–2,–2)$
• D. None of these

Answer 1

The correct option is C

$(–2,–2)$

Explanation for the correct option:

Finding one vertex of the equilateral triangle:

Let $ABC$ be the equilateral triangle.

Given centroid is at the origin $(0,0)$

Let the equation of side $BC$ is $\mathrm{x}+\mathrm{y}–2=0$

$AD$ is the altitude.

We know that the centroid of an equilateral triangle divides $AD$ in the ratio $2:1$.

Let $(a,b)$ is foot of median on $BC$

$$\begin{gathered} \mathrm{a}+\mathrm{b}–2=0 \\ \Rightarrow \mathrm{a}+\mathrm{b}=2\dots \left(\mathrm{i}\right) \end{gathered}$$

Let $A$ is the point $(h,k)$.

$$\begin{gathered} \frac{(2a+h)}{3}=0 \\ \mathrm{and} \\ \frac{(2b+k)}{3}=0 \end{gathered}$$

$\Rightarrow \mathrm{h}=-2\mathrm{a}\mathrm{and}\mathrm{k}=-2\mathrm{b}\dots \left(\mathrm{ii}\right)$

Since $AD$ is perpendicular to $BC$

$\mathrm{The}\mathrm{slope}\mathrm{of}AD\times \mathrm{slope}\mathrm{of}\mathrm{BC}=-1$

$$\begin{gathered} \frac{(\mathrm{k}-\mathrm{b})}{(\mathrm{h}-\mathrm{a})}\times (-1)=(-1) \\ \Rightarrow (\mathrm{k}–\mathrm{b})=(\mathrm{h}–\mathrm{a}) \\ \Rightarrow -3\mathrm{b}=-3\mathrm{a}(\mathrm{from}(\mathrm{ii}\left)\right) \\ \Rightarrow \mathrm{b}=\mathrm{a} \\ \mathrm{Substitute}\mathrm{b}=\mathrm{a}\mathrm{in}\left(\mathrm{i}\right) \\ 2\mathrm{a}=2 \\ \Rightarrow \mathrm{a}=1 \\ \Rightarrow \mathrm{b}=1 \end{gathered}$$

So $\mathrm{h}=-2,\mathrm{k}=-2$ (from (ii))

The vertex is $(-2,-2).$

Hence, the correct answer is option (C).