Discutimos
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Can someone help with this one, not sure which triangles i should be forming:

http://filestore.aqa.org.uk/subjects...W-QP-JAN09.PDF

question 7

I'm thinking that i am wrong with c/d = b-r / r-a
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lizard54142
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(Original post by Discutimos)
Can someone help with this one, not sure which triangles i should be forming:

http://filestore.aqa.org.uk/subjects...W-QP-JAN09.PDF

question 7

I'm thinking that i am wrong with c/d = b-r / r-a
PQ is a straight line, so has a constant gradient. Hence the gradient from (a,c) to (r,0) is the same as from (a,c) to (b,d). You can form two equations for the gradient, and then equate them.
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Discutimos
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(Original post by lizard54142)
PQ is a straight line, so has a constant gradient. Hence the gradient from (a,c) to (r,0) is the same as from (a,c) to (b,d). You can form two equations for the gradient, and then equate them.
Ah.

so, c/(r-a) = d/(b-r) ?

EDIT:

Nope, (c-d)/(b-a) = c/(r-a) rearrange and boom
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lizard54142
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(Original post by Discutimos)
Ah.

so, c/(r-a) = d/(b-r) ?
Looks fine to me. It might be easier if you find an expression for the gradient not in terms of r, i.e. from (a,c) to (b,d), so you don't have to do a lot of rearranging, but it doesn't really matter.

EDIT: You were one step ahead
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Discutimos
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(Original post by lizard54142)
Looks fine to me. It might be easier if you find an expression for the gradient not in terms of r, i.e. from (a,c) to (b,d), so you don't have to do a lot of rearranging, but it doesn't really matter.
Yeah i suppose XD, finished the question now. Cheers.

EDIT: Cheeky
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