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FP2: How to do this argand intersection question?

Question (c). I cannot do it. Do you convert them into cartesian equations and then find the point of intersection? I don't think this is the efficient method as it is only worth 2 marks.

Any idea?

Thank you.
Reply 1
Avatar for Dingo749
Dingo749
9
Ahhh I've got this question now :smile: Two seconds and I'll give you help :smile:
(edited 9 years ago)
Reply 2
Avatar for Dingo749
Dingo749
9
If you just look at the diagram it should look obvious. I know that's not much help but it is the easiest way of doing it, by sight.

Once you've got it, to express it in terms of z1 think of what the first part of the question asked you to do :smile:

Something you could do is if you can't see it straight away, imagine the point of intersection is z3, then imagine z1 and z2 are vector lines, how can you use z1 and z2 to get to z3?
(edited 9 years ago)
Original post by Sayonara
...


To get the z2 you need iz1 (from part (a)). Then to get to the point of intersection you need z1.
(edited 9 years ago)
Reply 4
Avatar for Sayonara
Sayonara
OP
11
Original post by Dingo749
If you just look at the diagram it should look obvious. I know that's not much help but it is the easiest way of doing it, by sight.

Once you've got it, to express it in terms of z1 think of what the first part of the question asked you to do :smile:

Something you could do is if you can't see it straight away, imagine the point of intersection is z3, then imagine z1 and z2 are vector lines, how can you use z1 and z2 to get to z3?


Let the vector from O to Z3 be OC.

How can one assume OB + OA = OC?

Is there a tutorial that proves this or states this?

Sorry, I have had to learn some of the maths content myself.
Reply 5
Avatar for Dingo749
Dingo749
9
Original post by Sayonara
Let the vector from O to Z3 be OC.

How can one assume OB + OA = OC?

Is there a tutorial that proves this or states this?

Sorry, I have had to learn some of the maths content myself.


OB + BC = OC
OA + AC = OC

So

OB + BC = OA + AC

OB ×=× OA (×=× means not equal to :') )
AC ×=× BC

But

The shape has rotational symmetrical around the line OC
because OA is parallel to BC and OB is parallel to AC.
So

OB = AC
OA = BC

So

OA + OB = OC

I think thats it? Personally I'd just look and state, thats kinda what the mark scheme does :P
(edited 9 years ago)

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