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AQA Core 1 jan 2011 Q3 help!

Hi all,
Does anyone have or could possible work through the solution to the following question for me:
Not sure if the links working so I'll type it out:

"point E has coordinates (5,k). Given that CE has length 5 find the two possible values of the constant k"

Part c is the part I'm stuck on and I just feel like im missing something, the mark scheme really isn't helping either, I have my exam tomorrow so any help would be very much appreciated thank you :smile:

Jan 2011 Q3.docx222.7KB

Reply 1

ghostwalker

Original post by Science94

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Just use the formula for the length of a line based on the co-ordinates of the end points. You'll get a quadratic in k, and solve.

Reply 2

Science94

what do you do with the 5 though? do you square it to 25 or leave it as 5? and then do you bring it to the other side with k?
could someone do a worked solution for me please?

Reply 3

Amber May

well you know that the point c is (2,-7), to get to the point (5,k) you have to go along 3

If you go along 3 you have to go up or down 4, to make the total distance 5 (3,4,5 triangle)

so k either = -7 + 4, or -7-4, so k = -3 or -11

let me know if that makes sense :smile:

dont know about quadratic method, seems over complicated

(edited 10 years ago)

Reply 4

ghostwalker

Original post by Science94

what do you do with the 5 though? do you square it to 25 or leave it as 5? and then do you bring it to the other side with k?
could someone do a worked solution for me please?

5^2= (5-2)^2+(k-(-7))^2

using the co-ordinates of C and E.

We don't generally do fully worked solutions on here - see the forum guidelines if you want to know why.

Reply 5

Science94

Original post by ghostwalker

5^2= (5-2)^2+(k-(-7))^2

using the co-ordinates of C and E.

We don't generally do fully worked solutions on here - see the forum guidelines if you want to know why.

oh okay where does the square root sign go though? as I think this was why my answer was coming out dodgy :smile:

Reply 6

Science94

Original post by Amber May

dont know about quadratic method, seems over complicated

yeah that makes perfect sense and is much simpler that the method we have been taught, thank you :smile:

Reply 7

ghostwalker

Original post by Science94

oh okay where does the square root sign go though? as I think this was why my answer was coming out dodgy :smile:

You don't need the square root sign, as we've used Pythagoras, and rather than square rooting, we have squared the length CE on the lefthand side.

Edit: Amber May's method works nicely because you have a nice 3,4,5 triangle, but you do need to know the quadratic method, or equivalent, as you're not guaranteed to get such a nice example.

(edited 10 years ago)

Reply 8

Science94

Original post by ghostwalker

You don't need the square root sign, as we've used Pythagoras, and rather than square rooting, we have squared the length CE on the lefthand side.

oh right, I see. I must be merging the two different formula in my head, Thank you both ever so much I always struggle with these and I knew if I didn't sort it and one came up I would be really annoyed at myself.
Right gonna go do some similar questions, thanx again :smile:

Reply 9

Amber May

i haven't been taught this quadratic method as such, but basically what you are doing is writing the equation of a circle, with center the point you are given and radius the distance, then substituting in x

makes sense i suppose, especially when you draw it