Question

Six points (x_i, y_i), i = 1, 2, …, 6 are taken on the circle x² + y² = 4 such that $\sum _ { i = 1 } ^ { 6 } x _ { i } = 8$ and $\sum _ { i = 1 } ^ { 6 } y _ { i } = 4$. The line segment joining orthocentre of a triangle made by any three points and the centroid of the triangle made by other three points passes through a fixed points (h, k), then h + k is –

• A. 3
• B. 4
• C. 5
• D. 2

Answer 1

Answer: (A)

3

Answer 2

Let $\sum _ { i = 1 } ^ { 6 } x _ { i } = \alpha$and $\sum _ { i = 1 } ^ { 6 } y _ { i } = \beta$

Let O be the orthocentre of the triangle made by (x₁, y₁),

(x₂, y₂) and (x₃, y₃)

̃ O is (x₁ + x₂ + x₃, y₁ + y₂ + y₃) º (a₁ , b₁)

Similarly let G be the centroid of the triangle made by other three points.

= G is

= G is $\left( \frac { \alpha - \alpha _ { 1 } } { 3 } , \frac { \beta - \beta _ { 1 } } { 3 } \right)$

The point dividing OG is the ratio 3 : 1 is $\left( \frac { \alpha } { 4 } , \frac { \beta } { 4 } \right)$ º (2, 1)

̃ h + k = 3