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C1 - coordinate geometry Watch

1. (Original post by Galileo Galilei)
Mathematical logic
2. What exam board are you?
3. (Original post by zed963)
I am just kidding, I worked it out using the information given in the question
4. (Original post by Galileo Galilei)
I am just kidding, I worked it out using the information given in the question
5. (Original post by zed963)
He literally subbed x=p into the equation for l1.
(Original post by Super199)
How do I do part d? Any hints on how to start?
Thanks
I did this question, but I can't remember where, so if you could tell me the paper you're doing OP, that would be great.

Spoiler:
Show
6. (Original post by zed963)
Plug the value of P into the equation for L1 to find Y
7. If C has x-coordinate p, then it has y-coordinate -p/2 + 6.

Therefore: (A,C) = √((2 - p)^2 + (5 + p/2 - 6)^2) = 5,

√(4 - 4p + p^2 + (p/2 - 1)^2) = 5,

√(4 - 4p + p^2 + (p^2)/4 - p + 1) = 5,

√(16 - 16p + 4p^2 + p^2 - 4p + 4)/2 = 5,

√(5p^2 - 20p + 20) = 10,

5p^2 - 20p + 20 = 100,

p^2 - 4p + 4 = 20,

p^2 - 4p - 16 = 0.

Hope this helps
8. (Original post by H0PEL3SS)
He literally subbed x=p into the equation for l1.

I did this question, but I can't remember where, so if you could tell me the paper you're doing OP, that would be great.

Spoiler:
Show
9. (Original post by BuddingAchiever)
If C has x-coordinate p, then it has y-coordinate -p/2 + 6.

Therefore: (A,C) = √((2 - p)^2 + (5 + p/2 - 6)^2) = 5,

√(4 - 4p + p^2 + (p/2 - 1)^2) = 5,

√(4 - 4p + p^2 + (p^2)/4 - p + 1) = 5,

√(16 - 16p + 4p^2 + p^2 - 4p + 4)/2 = 5,

√(5p^2 - 20p + 20) = 10,

5p^2 - 20p + 20 = 100,

p^2 - 4p + 4 = 20,

p^2 - 4p - 16 = 0.

Hope this helps
Can you explain that?
10. (Original post by H0PEL3SS)
He literally subbed x=p into the equation for l1.

I did this question, but I can't remember where, so if you could tell me the paper you're doing OP, that would be great.

Spoiler:
Show
Jan 2009 . I've solved it now so all good. One more question I couldn't do was part c. Any hints on how to do this?
I can't seem to work it out.
Attached Images

11. My solution.
Attached Images

12. (Original post by Super199)
Jan 2009 . I've solved it now so all good. One more question I couldn't do was part c. Any hints on how to do this?
I can't seem to work it out.
Draw a triangle and think about it
13. (Original post by Galileo Galilei)
Draw a triangle and think about it
Using I can't seem to get it into that form.
14. (Original post by Galileo Galilei)
But that isnt Q10 C? Which question are you talking about?
This one haha
Attached Images

15. (Original post by Super199)
This one haha
Got ya, do you want the solution or hints mate?
16. (Original post by Galileo Galilei)
Got ya, do you want the solution or hints mate?
A solution I guess. I mean I know what to do but I can't seem to get it in that form.
17. (Original post by Super199)
Using I can't seem to get it into that form.
Just find an expression for n using the sum formula. You should double 2750 to get rid of the n/2. From there, expand and simplify.
18. For part C, take the equation of sum of n

n/2(2a+(n-1)d)
then substitute numbers

n/2(-17.5(2)+2.5n-2.5)=2750

times 2750 by 2

n(-35+2.5n-2.5)= 5500

expand
-35n+2.5n^2-2.5n=5500

times by 2

-75n+5n^2=11000
5n^2-75n-11000=0
n^2-15n-2200=0

Therefore

n^2-15n-55x40
19. (Original post by BuddingAchiever)
For part C, take the equation of sum of n

n/2(2a+(n-1)d)
then substitute numbers

n/2(-17.5(2)+2.5n-2.5)=2750

times 2750 by 2

n(-35+2.5n-2.5)= 5500

expand
-35n+2.5n^2-2.5n=5500

times by 2

-75n+5n^2=11000
5n^2-75n-11000=0
n^2-15n-2200=0

Therefore

n^2-15n-55x40

you can just check that 55x40=2200....if you can't do this question, at least you will be able to do part d
20. (Original post by Super199)
A solution I guess. I mean I know what to do but I can't seem to get it in that form.

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