The circumcentre of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points ( a 2 + 1 , a 2 + 1 ) and 2 a , - 2 a , a ≠0 . Then for any a , the orthocentre of this triangle lies on the line

Step-by-step solution

The mid-point of a line segment joining the points x 1 ,   y 1 and x 2 ,   y 2 is x 1 + x 2 2 ,   y 1 + y 2 2 Given, the centroid G is the mid-point of the line segment joining the points a 2 + 1 ,   a 2 + 1 and 2 a ,   - 2 a , thus G ≡ a 2 + 1 + 2 a 2 ,   a 2 + 1 - 2 a 2 ⇒ G ≡ a + 1 2 2 ,   a - 1 2 2 . Also, given circumcentre C is origin ⇒ C ≡ 0 ,   0 . We know that, for a triangle circumcentre, orthocentre and centroid are collinear. Thus, C ,