Let the vertices of a triangle are A = - 3 + 2 sin ⁡ θ , 4 + 2 cos ⁡ θ , B = - 3 + 2 cos ⁡ θ , 4 - 2 sin ⁡ θ and C = - 3 - 2 sin ⁡ θ , 4 - 2 cos ⁡ θ , then the distance between the centroid and the circumcentre of Δ A B C is

Question

Let the vertices of a triangle are A = - 3 + 2 sin ⁡ θ , 4 + 2 cos ⁡ θ , B = - 3 + 2 cos ⁡ θ , 4 - 2 sin ⁡ θ and C = - 3 - 2 sin ⁡ θ , 4 - 2 cos ⁡ θ , then the distance between the centroid and the circumcentre of Δ A B C is, 2 3 units, 3 2 units, 1 2 units, 1 3 units

MARKS App · Accepted Answer

A , B , C lie on the circle x + 3 2 + y - 4 2 = 2 2 ⇒ circumcentre of Δ A B C is - 3,4 = S & the centroid of Δ A B C is - 3 + 2 3 cos ⁡ θ , 4 - 2 3 sin ⁡ θ = G Distance between S & G is 2 3 2 c o s 2 θ + 2 3 2 s i n 2 θ = 2 3 units

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