C1 2009 Jan Q10 (d)Watch
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,
5= root (p-2)^2 +[(-1/p+6)-5]^2
I'm getting 5p^2 +28p+400 something
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.