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DAY 1: 25 May

Type a 100-word reflection for today's lesson.

Today I have learnt how to find the distance of a line and the mid point of the line. To find the distance of a line, we just have se the line as the hypotenuse of a right angled tringle and apply the pythagoras theorem. To find the mid point of a line, we just have to find the average of the coordinates of the 2 points. However, can we find the mid point by first finding the length then diving it by 2? Is there any way to find out the coordinates of the mid point like that?

Can is can but it's not ideal. Longwinded and troublesome. If you take the length/2, you will be finding half the length of the line. Then you have to make use of the length to figure out the coordinates. It's not easy if the length is not a "nice" number.

So it's A LOT faster to use the formula for finding mid-point.


DAY 2: 26 May


Type a 100-word reflection for today's lesson and comment on your work.

It is a lot harder to find the equation of a circle. I had to spend quite a bit of time doing the GC calculator logo and had to ask around. There was also an added difficulty as we had to convert the equation into that of y=... Even though it was quite difficult today, i feel that i have at least learnt something but i will need to do more on this topic.


Submission of Designs:
Design 1
(LONDON 2010)

Design 2
("RU4")

Design 3
(SYRINGE)

Design 3 (Optional)
(BOAT)
2P210_YiJie_OlympicsSign.JPG
y1=2 - squareroot(2^2-(X-0)^2)
y2=2 + squareroot(2^2-(X-0)^2)
y3=0 + squareroot(2^2-(X+2)^2
y4=0 - squareroot(2^2-(X+2)^2
y5=2 + squareroot(2^2-(X+4)^2
y6=2 - squareroot(2^2-(X+4)^2
y7=0 + squareroot(2^2-(X-2)^2
y8=0 - squareroot(2^2-(X-2)^2
y9=2 + squareroot(2^2-(X-4)^2
y10=2 - squareroot(2^2-(X-4)^2