Suppose that the sides passing through the vertex (α,β) of a triangle are bisected at right angles by the lines y2−8xy−ax2=0. Then the centroid of the triangle is
y2−8xy−9x2=0 y2−9xy+xy−9x2=0 (y−9x)(y+x)=0 The two given lines are y=9x and y=−x Slope of line y=9x is 9 and slope of line y=−x is −1. Let AB and BC be the line perpendicular to y=9x and y=−x respectively. Slope of AB=‌−1∕9 and slope of line BC is 1. Equation of AB is y−β=‌−1∕9(x−α) 9y−9β‌=−x+α ⇒‌‌x+9y−α−9β‌=0 M is Mid-point of the AB and y=9x. ‌ So, ‌x+9(9x)+α+9β So,x+9(9x)+α+9β 82x=α+9β ˙x=‌α+9β∕82 Coordinate of M is (‌α+9β∕82,‌9α+81β∕82). M is Mid-point of AB, Let coordinate of A be h,k ‌
α+9β
82
=‌
α+h
2
and ‌
9α+81β
82
=‌
β+k
2
⇒2α+18β=82α+82h ⇒82h=−80α+18β ⇒h=‌
−80α+18β
82
‌‌ and ‌18α+162β‌=82β+82k ⇒‌82k‌=18α−80β ⇒k‌=‌
18α−80β
82
Equation of BC is y−β=1(x−α) x−y=α−β N is intersection point of BC and y=−x ∴‌x+x=α−β ⇒‌2x=α−β ⇒x=‌
α−β
2
‌y=‌
β−α
2
Coordinate of N is (‌
α−β
2
,‌
β−α
2
). N is Mid-point of BC. Let coordinate of C be (a,b) ‌‌
α−β
2
=‌
α+a
2
‌ and ‌‌
β−α
2
=‌
β+b
2
‌⇒a=−β‌ and ‌b=−α ∴ Coordinate of C is (−β,−α). Centroid of ABC is ‌(‌