One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is
A. (–1, –1)
B. (2, 2)
C. (–2, –2)
D. (2, –2)
Let ABC be an equilateral triangle with vertex A (a,b)
Let AD be perpendicular to BC and let (p,q) be the coordinates of D
Given that the centroid P lies at the origin (0,0)
We know that, the centroid of a triangle divides the median in the ratio 1:2
Now, using the section formula, we get
and
a+2p=0 and b+2q=0…A…
a+2p=b+2q
2p-2q=b-a …i
It is given that BC=x+y-2=0
Since, the above equation passes through p,q
p+q-2=0
Now we find the slope of line AP
Equation of line BC is x+y-2=0
y= -x+2
y=-(1)x+2
Since the above equation is in y=mx+b form
So, slope of line BC is mBC=-1
Since both lines rae perpendicular
b=a
Putting this value in equation (i )we get 2p-q=b-b=0
p=q
Now putting this value in equation (i) we get p+q-2=0
2p=2
p=1
q=1
Putting the value of p and q in equation A , we get a+2*1=0 and b+2*1=0
a=-2 and b=-2
So, the coordinates of vertex A (a,b) is (-2,-2)
Hence, the correct option is c