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# Coordinate Geometry Core 1 help (2 questions) watch

1. Hi guys,

I'm doing some maths revision for core 1 AQA maths and I've got stuck on these two questions.

5. The straight line joining the point P(5,6) to the point Q(q,2) is perpendicular to the straight line joining point Q to point R (9,-1). Calculate the possible values of q

and...

8. Show that points A(2,3), B(4,8), C(8,9) and D(4,-1) form a trapezium.

Thanks in advance for any help,
Blake
2. The product of perpendicular gradients is -1.
3. For 8, remember the properties of a trapezium, it has a pair of parallel sides, what does this mean for the gradients?
4. (Original post by NotNotBatman)
For 8, remember the properties of a trapezium, it has a pair of parallel sides, what does this mean for the gradients?
This means they must have the same gradients

(Original post by NotNotBatman)
The product of perpendicular gradients is -1.
Yep I know this and have used this but still don't get the answer which is in the back of the book, I'm beginning to wonder whether the answer might be wrong? (it wouldn't be the first time)

5. (Original post by Blake Jones)
This means they must have the same gradients
Yes.

(Original post by Blake Jones)
Yep I know this and have used this but still don't get the answer which is in the back of the book, I'm beginning to wonder whether the answer might be wrong? (it wouldn't be the first time)

Could you show your working out?
6. (Original post by Blake Jones)
This means they must have the same gradients

Yep I know this and have used this but still don't get the answer which is in the back of the book, I'm beginning to wonder whether the answer might be wrong? (it wouldn't be the first time)

Well I've just done this and got q = 11 or q = 3? Could you let me know if this is correct and then I can guide you.
7. (Original post by iMacJack)
Well I've just done this and got q = 11 or q = 3? Could you let me know if this is correct and then I can guide you.
Yep, this is what the answer in the books says
8. For 5 I did:

m = change in y / change in x
2-6 / q-5
and 2+1 / q-9
so then because they're perpendicular
4/q-5 * 3/q-9 = -1
so 12/q^2-14q+45 = -1
meaning q^2-14q+57 = 0???
9. (Original post by Blake Jones)
Yep, this is what the answer in the books says
Okay - well have you tried working out the gradients of PQ and then QR?

For PQ I got the gradient as 4/(5-q), did you get this part?

After this, work out the gradient of QR,. (y2-y1)/(x2-x1) in any order as long as it's consistent (you start from the same bracket when using y2 and x2).

I got, for the gradient of 'QR': 3/(q-9)

And we know, since they are perpendicular to one another, by multiplying them together the product of them must = -1

Multiplying them together and then taking these terms over to form a quadratic will leave you with a quadratic that factorizes.

In the spoiler below will be my answer but don't look until you've tried it from this knowledge
Spoiler:
Show
(4/5-q) * (3/q-9) = -1
12/(5-q)(q-9) = -1
12/(-q^2+14q-45) = -1
12 = q^2 - 14q + 45
q^2 - 14q + 33 = 0
(q-11)(q-3) = 0
q = 11 or q = 3
10. (Original post by Blake Jones)
For 5 I did:

m = change in y / change in x
2-6 / q-5
and 2+1 / q-9
so then because they're perpendicular
4/q-5 * 3/q-9 = -1
so 12/q^2-14q+45 = -1
meaning q^2-14q+57 = 0???
You forgot to multiply it by -1

11. (Original post by Blake Jones)
For 5 I did:

m = change in y / change in x
2-6 / q-5
and 2+1 / q-9
so then because they're perpendicular
4/q-5 * 3/q-9 = -1
so 12/q^2-14q+45 = -1
meaning q^2-14q+57 = 0???
You're missing out the negative - this should be a negative. Read my above comment and check the spoiler!
12. Oh! I see! Thank you both! Did either of you manage to get the other one to work? I think I need a break haha!! Been doing maths for 3 hours straight!! XD
13. (Original post by Blake Jones)
Oh! I see! Thank you both! Did either of you manage to get the other one to work? I think I need a break haha!! Been doing maths for 3 hours straight!! XD
Pleasure

If you draw out those points you'll notice something about the gradients between them and then as a result you can show these gradients have a certain relationship

Posted from TSR Mobile
14. (Original post by iMacJack)
Pleasure

If you draw out those points you'll notice something about the gradients between them and then as a result you can show these gradients have a certain relationship

Posted from TSR Mobile
Thanks, will do!

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