One vertex of the 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 ⊥ 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

⇒ a + 2p = 0 and b + 2q = 0 …(A)

⇒ a + 2p = b + 2q

⇒ 2p – 2q = b – a

⇒ 2(p – q) = b – a …(i)

It is given that BC = x + y – 2 = 0

Since, the above equation passes through (p, q)

⇒ p + q – 2 = 0 …(ii)

Now, we find the slope of line AP

and 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

Since, both the lines are perpendicular to each other.

∴ m_AP × m_BC = -1

⇒ b = a

Putting the value of b = a in eq. (i), we get

2(p – q) = b – b

⇒ 2(p – q) = 0

⇒ p = q

Now, putting the value p = q in eq. (ii), we get

p + p – 2 = 0

⇒ 2p = 2

⇒ p = 1

⇒ q = 1 [∵ p = q]

Putting the value of p and q in eq. (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)