Sehan7
Badges: 7
Rep:
?
#1
Report Thread starter 3 years ago
#1
Is there any other way rather than this wait I attach http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
0
reply
Sehan7
Badges: 7
Rep:
?
#2
Report Thread starter 3 years ago
#2
Name:  20180730_175035.jpg
Views: 75
Size:  100.6 KB
0
reply
MiladA
Badges: 14
Rep:
?
#3
Report 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,
1
reply
username3535256
Badges: 17
Rep:
?
#4
Report 3 years ago
#4
This is a screenshot: Name:  C75D8008-D9B4-492C-AFF2-0DBF70B15785.jpg.jpeg
Views: 80
Size:  26.0 KB
1
reply
Sehan7
Badges: 7
Rep:
?
#5
Report Thread starter 3 years ago
#5
So I have to find y in terms of p first
0
reply
Sehan7
Badges: 7
Rep:
?
#6
Report Thread starter 3 years ago
#6
Oh so it's the same eq.y = -1/2p+6
0
reply
Sehan7
Badges: 7
Rep:
?
#7
Report Thread starter 3 years ago
#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
0
reply
Sehan7
Badges: 7
Rep:
?
#8
Report Thread starter 3 years ago
#8
O ty all! probably its a error when simplifying
1
reply
MiladA
Badges: 14
Rep:
?
#9
Report 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.
0
reply
Sehan7
Badges: 7
Rep:
?
#10
Report Thread starter 3 years ago
#10
👍✌
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

What support do you need with your UCAS application?

I need help researching unis (7)
11.67%
I need help researching courses (5)
8.33%
I need help with filling out the application form (3)
5%
I need help with my personal statement (26)
43.33%
I need help with understanding how to make my application stand out (14)
23.33%
I need help with something else (let us know in the thread!) (2)
3.33%
I'm feeling confident about my application and don't need any help at the moment (3)
5%

Watched Threads

View All