FP2 weird loci question help

From the FP2 textbook, it's asking for a sketch of the locus and it's Cartesian equation for

z

\frac{\left | z+3i \right |}{\left | z-6i \right |} = 1

The textbook reckons I'm wrong, and the Cartesian equation is:

y=-0.5x+9/4

Can anyone point out where I've gone wrong? I couldn't find any examples of a division question, so maybe it's wrong for me to multiply by

|z-6i|

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Reply 1

ValerieKR

Original post by alexhazmat

From the FP2 textbook, it's asking for a sketch of the locus and it's Cartesian equation for

z

\frac{\left | z+3i \right |}{\left | z-6i \right |} = 1

The textbook reckons I'm wrong, and the Cartesian equation is:

y=-0.5x+9/4

Can anyone point out where I've gone wrong? I couldn't find any examples of a division question, so maybe it's wrong for me to multiply by

|z-6i|

are you sure the numerator is z+3i and not z+3, or the equivalent for the denominator?

based off the top of my head and on the actual answer I'd guess that's where the problem is

(edited 7 years ago)

Reply 2

alexhazmat

Original post by ValerieKR

are you sure the numerator is z+3i and not z+3, or the equivalent for the denominator?

based off the top of my head and on the actual answer I'd guess that's where the problem is

I was thinking that, but its definitely z+3i :

Reply 3

ValerieKR

Original post by alexhazmat

I was thinking that, but its definitely z+3i :

Then it's a typing mistake in the textbook/worksheet/whatever

That provided answer is an answer to a question where that i in the numerator isn't real (yay imaginary number jokes)
Your solution is correct for the given question

(edited 7 years ago)

Reply 4

alexhazmat

Original post by ValerieKR

Okay thanks, I knew I wasn't losing my mind. Sadly it seems these textbooks are riddled with errors :frown:

Reply 5

ValerieKR

Original post by alexhazmat

Okay thanks, I knew I wasn't losing my mind. Sadly it seems these textbooks are riddled with errors :frown:

btw there is a much quicker way to do those ones

once you have mod(z-a)=mod(z-b)
The equation is the equation of the perpendicular bisector of AB (because the distance from A and B is equal)
(and if A=B it's just everywhere)

Reply 6

alexhazmat

Original post by ValerieKR

I knew it was a bisector of AB, I just like doing it the long way to show a bit of reasoning.

However I don't understand why, for example:

|z-6| = 2|z+6-9i|

ends up being a circle, instead of a bisector?

Algebraically I see that it ends up being a circle, but I don't see why (the textbook fails at explaining why doubling the 2nd part completely changes it from a bisector to a circle) :smile:

Reply 7

ValerieKR

Original post by alexhazmat

I knew it was a bisector of AB, I just like doing it the long way to show a bit of reasoning.

However I don't understand why, for example:

|z-6| = 2|z+6-9i|

Because of the '2' - it's always twice as far away from one point as another, so you end up with a solution on one side of one complex number, and on the other

think
-----a------b------
if we mark on every point equidistant from a and b we get
-----a---x---b------

and every one twice as far from a as b
-----a----x--b----x--

it's an extension of that extra solution appearing to 2D

the factor not being 1 destroys the symmetry in a sense and allows solutions to appear on either side of either a or b on that 1D line

(edited 7 years ago)

Reply 8

alexhazmat

Original post by ValerieKR

Ahhhh that makes sense now. I never thought of it that way, initially I thought the coordinates of B would be doubled. Thanks for the insight!

Reply 9

IYGB

Original post by alexhazmat

Okay thanks, I knew I wasn't losing my mind. Sadly it seems these textbooks are riddled with errors :frown:

That is why they are free.
I had similar issues

Reply 10

alexhazmat

Original post by IYGB

That is why they are free.
I had similar issues

They're free? Each book is £16 per book on amazon. I'm only complaining because these books are my only source of practice questions (apart from past papers)

Reply 11

ValerieKR

Original post by alexhazmat

They're free? Each book is £16 per book on amazon. I'm only complaining because these books are my only source of practice questions (apart from past papers)

Bloody hell - I know for certain you can get resources for this type of question of the AQA website for free
and I suspect a lot of other places too

Reply 12

IYGB

Original post by alexhazmat

They're free? Each book is £16 per book on amazon. I'm only complaining because these books are my only source of practice questions (apart from past papers)