[zoxe]

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?

[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.

[Coulson72]

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

[zoxe]

(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.

[Joshmeid]

(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

[zoxe]

ive solved it with Pythagoras, however i'm still baffled on how theyve done it

[zoxe]

(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

Why does PQ = gradient?