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1. http://www.thomas-reddington.com/upl...p_jan_2009.pdf

For question 5a)

In the mark scheme they are finding the gradeints for PQ and QR, then multiplying them, I have no idea why?
2. Find the gradient of PQ, find the perpendicular for the gradient of QR, find the equation of QR, substitute the y value for a.

You should be familiar with m1m2 = -1, another way of finding the perpendicular gradient.
3. theyre perpandicular (because the triangle inside a semicircle is right angled) and hence the product of their gradients is -1.
So from this it should be easy to work out your value for a
4. (Original post by Joshmeid)
Find the gradient of PQ, find the perpendicular for the gradient of QR, find the equation of QR, substitute the y value for a.

You should be familiar with m1m2 = -1, another way of finding the perpendicular gradient.
This is the first time i've seen of m1m2 = -1.
5. (Original post by zoxe)
This is the first time i've seen of m1m2 = -1.
m1 is the gradient of PQ.

Lets say PQ = 5 for example.

we have m1m2 = -1

We now substitute m1 = 5

5m2 = -1

m2 = -1/5
6. ive solved it with Pythagoras, however i'm still baffled on how theyve done it
7. (Original post by Joshmeid)
m1 is the gradient of PQ.

Lets say PQ = 5 for example.

we have m1m2 = -1

We now substitute m1 = 5

5m2 = -1

m2 = -1/5

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