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Edexcel FP1 Thread - 20th May, 2016

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Original post by iMacJack
Should be (12,12) not (27,12)


Thought it looked too easy :biggrin:

For some reason I used p=3 when I got the x coordinate, whoops :frown:
Original post by Music With Rocks
Just to check when rearranging matrices does the term always go to the front so if
AB=C

A=B-1C
and
B=A-1C

Or is this wrong? it looks wrong


It depends on what side your value is on, if we have

XA = B
and we have to isolate the 'X'
then:
XAA^-1 = BA^-1
XI = BA^-1
X = BA^-1

Basically you apply it to the same side of the term on either side
(Sorry for the bad explanation!)
Original post by TheMoon
I don't understand, isn't it like two right angled triangles which you can use 1/2 base times height with?

I tried drawing it, obviously I've drawn it very inaccurately (terrible paint skills) since it doesn't look like a normal but is it something like this? Might be wrong. Sorry if I'm completely wrong with how it looks.
This gets 225 anyway.


If p=2, sub that into (3p^2, 6p), you get (12, 12) not (27, 12)
I drew a diagram and it was pretty much irregular as far as I could tell :frown:
Would you get credit for using this in the exam? @TeeEm, @Zacken ?

Original post by WhatIsSleep
For that PQS question, I found an area formula which helped me work it out..

(I have attached it as an image but the source is http://www.mathopenref.com/coordtrianglearea.html)

So A is (3, 0)
B is (12, 12)
and C is (27, -18)

Ax=3, Ay=0
Bx=12, By=12
Cx=27, Cy=-18

Subbing those into the formula, I got 225 units^2

Honestly though, I tried drawing diagrams and everything...couldn't find a better way of doing it tbh
Original post by Brailey
If you look at the screenshot I have put examples in.

All you do is add on what r=0 is to the the summation for your soultion.

Hope this helps!


Where did the 2n come from in the end equation. If u add r=0 shouldnt it be just +2
Original post by WhatIsSleep
If p=2, sub that into (3p^2, 6p), you get (12, 12) not (27, 12)
I drew a diagram and it was pretty much irregular as far as I could tell :frown:


yeah sorry, I used p=3 for some reason when getting the x coordinate. not sure then :frown:

There has to be an easier way though.
Original post by TrueDAN
That is a beastly question haha!! Just done the first part now! How far have you got?


tbh I think that's fp3. They introduce Loci in FP2 so FP1 takers wouldn't know what a loci of points is
Original post by iMacJack
It depends on what side your value is on, if we have

XA = B
and we have to isolate the 'X'
then:
XAA^-1 = BA^-1
XI = BA^-1
X = BA^-1

Basically you apply it to the same side of the term on either side
(Sorry for the bad explanation!)

Oh I think I get what you mean so

AB=C
A=CB-1

AB=C
B=A-1C


Is this correct now?
Original post by WhatIsSleep
If p=2, sub that into (3p^2, 6p), you get (12, 12) not (27, 12)
I drew a diagram and it was pretty much irregular as far as I could tell :frown:


Just draw a diagram, find the gradients of SP and SQ and u will see that they are perpendicular, then its just half the area of the rectangle using length of a line formula to work out the lengths
Original post by TheMoon
yeah sorry, I used p=3 for some reason when getting the x coordinate. not sure then :frown:

There has to be an easier way though.


Maybe it's something that's not on the UK FP1 spec :/
Original post by Music With Rocks
Oh I think I get what you mean so

AB=C
A=CB-1

AB=C
B=A-1C

Is this correct now?


Correct :smile:
Original post by iMacJack
ii A is just finding the Det M, because the det is the scale factor, so ii (a) is 4, isn't it? Aren't you talking about part ii (b)?


Could you explain a bit more in depth please?
Original post by iMacJack
Correct :smile:


Thank you very much :smile: you explained it well


did you figure it out?
Nope.
Original post by AmarPatel98
did you figure it out?
Original post by connorbarr
Just draw a diagram, find the gradients of SP and SQ and u will see that they are perpendicular, then its just half the area of the rectangle using length of a line formula to work out the lengths


ahh seems you're right :P
I didn't think to find the gradients of the lines
Original post by TheRandomGenius
Nope.

I've done 6bi, i'm just doing ii. I'll post in a minute.
Original post by TheRandomGenius
Nope.



So you get the tangent at P from part a.

Similarly you get the tangent at Q.

The gradient of the tangent at P is 1/p, so similarly, the gradient at Q is 1/q. The two tangents are perpendicular, so you know their gradients must multiply to give -1. Form an eqn from this.

You are told the intersection point, so form another eqn from this.

Solve simultaneously to obtain p and q
Original post by TheRandomGenius
Nope.


I think you would make the gradient of p equal to the -1/gradient of q then you would substitute back into the equation to find p and q and substitute the R coordinates
Who's ready for tomorrow?! :eek:

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