Find the coordinates of one vertex of an equilateral triangle with centroid at the origin and the opposite side `x +y-2 =0`.

Options

A. `(-1,-1)`
B. `(2,2)`
C. `(-2,-2)`
D. `(2,-2)`

Answer

C

Explanation

Let ABC be the equilateral triangle iwth vertex A(h,k).
Let the coordinates of D are `(alpha,beta)`.

We know that `2:1` from the vertexA.
`therefore 0=(2alpha+h)/(3)` and `0=(2beta+k)/(3)`
`rArr2alpha=-h` ......(i)
and `2beta=-k`
Also, D `(alpha,beta)`lies on the line `x+Y-2=0`......(ii) .
`therefore alpha+beta-2=0`
`ADbotBC`
Since, the slope of line BC i.e., `m_(BC)=-1`
and slope of the line AG i.e., `m_(AG)=(k-0)/(h-0)=(k)/(h)`
`rArr (-1).((k)/(h))=-1`
`rArrh=k` ......(iii)
From Eqs. (i) and (iii),
`2alpha=-h` and `2beta=-h`
`therefore alpha=beta`
From Eq. (ii), `2alpha-2=0rArralpha=1`
If `alpha=1` then `beta=1`
So, the vertex A is `(-2,-2)`.