# C1 2009 Jan Q10 (d)

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#1
Is there any other way rather than this wait I attach http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
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#2
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3 years ago
#3
Well, the point A is known. Write the equation of line which I assume you have done.

It says the point C also lies on the line with an X - coordinate equal to p, so when X = p, Y = whatever value you get in terms of p.

The length AC can be found using the distance between two points formula. Since you know the length is 5, you can rearrange to get that equation shown on question paper,
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3 years ago
#4
This is a screenshot:
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#5
So I have to find y in terms of p first
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#6
Oh so it's the same eq.y = -1/2p+6
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#7
As y= -1/2p+6
A (2,5)
C (p,-1/2p+6)
5= root (p-2)^2 +[(-1/p+6)-5]^2
I'm getting 5p^2 +28p+400 something
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#8
O ty all! probably its a error when simplifying
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3 years ago
#9
Y = (-1/2)X + 6

When X = P, Y = 6 - (P/2) ==> C(P, 6 - P/2)

AC^2 = (P - 2)^2 + (6 -P/2 - 5)^2 = 5. This is (X2 - X1)^2 + (Y2 - Y1)^2

Expand and you should get (5/4)P^2 - 5P - 20 = 0

Then multiply by 4/5 and you get what's required.
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#10
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