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Degree: Mathematics (Masters) - Warwick University

#### Subjects offered

SubjectQualificationPrices
Further Mathematics A Level £36 /hr
Maths A Level £36 /hr
Maths GCSE £36 /hr
Maths IB £36 /hr
Maths 13 Plus £36 /hr

#### Qualifications

MathematicsA-levelA2A*
Further MathematicsA-levelA2A*
PhysicsA-levelA2A*
EconomicsA-levelA2A*
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Please forgive the very late reply to this. Had some family crisis and totally forgot. First time I've logged in since June 2016, it's now January 2017. Callum was very pleased with Josh, he helped tremendously and was able to boost Callums expected outcomes and he managed to get an A in his AS level paper. He just missed out on an A* by one mark. Josh is a very friendly approachable and knowledgeable tutor, who we rate very highly. Please wish him every success in his future career. regards Stephen Brown Callums father.

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### 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 perpendic...

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.

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2 years ago

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