I am a third year undergraduate at Warwick studying maths. In my first and second years I achieved a first (with an average score of 86% and 88% respectively). In my Maths and Further Maths A levels I scored an average of 95%. I have always wanted to study Maths from a young age. I have had a year of experience tutoring students on mytutor and in person. I am currently also working for the Warwick Mathematics Institute teaching first year undergraduate maths students.
I hope you be able to instil an appreciation of the beauty of the Maths whilst simultaneously preparing your child to effectively deal with exam questions.
Every child will learn best in a slightly different way. We will work to find the style of learning that suits your child best by adjusting the pace and depth in which we cover topics, the number and type of examples, the use of visual aids, and so on.
Whilst I hope to teach your child to love Maths, the primary aim of tutoring will probably be to achieve the highest grades in exams. My A level scores show that I know how to execute an exam perfectly.
According to your child’s needs we can:
-supplement school teaching through more and varied examples,
- approach a topic from fresh,
- revise a topic by covering the main ideas and then by doing questions,
- work on the nitty gritty aspects of exam technique and how to approach problems,
- approach a tricky topic from a different perspective to help with understanding.
I will make sure that parents are fully informed about their child’s progress and which areas we will be working on.
|Further Mathematics||A Level||£24 /hr|
|Maths||A Level||£24 /hr|
|Maths||13 Plus||£22 /hr|
|STEP I||Uni Admissions Test||1|
Amanda (Parent) October 19 2016
Callum (Student) June 1 2016
Callum (Student) February 17 2016
Callum (Student) October 18 2015
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 = x2 - 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.see more