What is the perpendicular bisector of the point (0,2) and (1,0)?

First things first it will definately be useful to make a quick sketch of the x-y axes, the two points, and roughtly what the perpendicular bisector will look like.

This kind of question can be made much more painless if you make the following observation: for any point (x,y) on the perpendicular bisector (lets call this P) the distance from (x,y) to (0,2) and the distance from (x,y) to (1,0) are equal. 

Let (x,y) lie on P. Using Pythagoras' Theorem the distance (x,y) to (0,2) is  sqrt[(x - 0)2 +  (y - 2)2]. Similairly the distance from (x,y) to (1,0) is sqrt[(x - 1)2 + (y - 0)2]. These two distances must then be equal so:

 sqrt[(x - 0)2 +  (y - 2)2] = sqrt[(x - 1)2 + (y - 0)2] , now squaring gives:

(x - 0)2 +  (y - 2)2 = (x - 1)2 + (y - 0)2 , then simplifying gives:

x2 + (y - 2)2 = (x - 1)2 + y2 , expanding the brackets gives:

x2 + y2 - 4y +4 = x- 2x +1 + y2 , then cancelling the x2 and y2 terms:

-4y + 4 = -2x + 1 , which rearranges to give the line:

y = x/2 + 3/4 

There was nothing special here about finding the perpendicular bisector of (0,2) and (1,0), the exact some method can be used for any two points in the plane. The only thing that must be altered is our two expressions for the distances from (x,y) to the first point and the distance form (x,y) to the second. 

JR
Answered by Josh R. Maths tutor

5650 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

The cost of a ticket increases by 10% to £19.25. What is the original cost?


Please factorise and solve x^2 -1 = 0


How does differentiation work?


What is the difference between unconditional and conditional probability?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences