Additional Mathematics 0606 / Straight-line graphs / 3.883.88: Perpendicular bisector of OA0606/21/M/J/21 — Question 5 · 8 marksMark as done · Save for later · Show all solutionsThe curves y=x2y = x^{2}y=x2 and y=27x2y = \dfrac{27}{x^{2}}y=x227 intersect at O(0,0)O(0, 0)O(0,0) and at the point AAA. Find the equation of the perpendicular bisector of the line OAOAOA.[8]▸ Answer▸ Official mark scheme← 3.14: Linear law: lny\ln ylny against lnx\ln xlnx3.1: Perpendicular bisector of ABABAB; point DDD →