Competition - post a photo you've taken of nature >>

Fp2 complex transformation

FA296E9D-7FFA-4723-B187-9CB3F4478F26.jpg.jpeg

Can someone please explain part c . I don’t get what to do . What’s the easiest way to go about this question. we get a gradient of 1 according to exam solutions . I don’t get this though . How does the arg give us a negative number and how does this help us . If we are trying to find the intersection between the circle and the half line theta =-3/4pi are we supposed to get the half line in Cartesian form and sub into the equation of the circle . How can we change its form ?

Scroll to see replies

Reply 1

Angels1234

https://postimg.cc/image/kpn3d8tev/

Reply 2

Notnek

Original post by Angels1234

Firstly you need a sketch - I assume you have one? It would have been useful if you posted it.

You do need the cartesian equation of the line. The argument is

-\frac{3\pi}{4}

which is halfway between

-\frac{\pi}{2}

and

-\pi

. So this half-line goes "south-west" if you like. Can you see why the gradient of the line is 1?

Reply 3

Angels1234

Original post by Notnek

Firstly you need a sketch - I assume you have one? It would have been useful if you posted it.

You do need the cartesian equation of the line. The argument is

-\frac{3\pi}{4}

which is halfway between

-\frac{\pi}{2}

and

-\pi

. So this half-line goes "south-west" if you like. Can you see why the gradient of the line is 1?

My drawings a bit wrong . I think my line should be more towards the left and I should of drawn a dash line parallel to the x axis
https://postimg.cc/image/futeem04n/

Reply 4

Notnek

Original post by Angels1234

My drawings a bit wrong . I think my line should be more towards the left and I should of drawn a dash line parallel to the x axis
https://postimg.cc/image/futeem04n/

You need a better diagram. The radius of the circle is

\sqrt{20}

which is between 4 and 5. And the horizontal distance between the centre (4,-2) and the point (6,0) is only 2 units. But look where you've marked the point (6,0) on your diagram.

Try to draw something better and post it again.

Reply 5

AspiringUnderdog

Original post by Notnek

Firstly you need a sketch - I assume you have one? It would have been useful if you posted it.

You do need the cartesian equation of the line. The argument is

-\frac{3\pi}{4}

which is halfway between

-\frac{\pi}{2}

and

-\pi

. So this half-line goes "south-west" if you like. Can you see why the gradient of the line is 1?

isn't the gradient -1?

Reply 6

Notnek

Original post by AspiringUnderdog

isn't the gradient -1?

No, if you've drawn a correct diagram it should clearly be 1 and not -1.

Reply 7

AspiringUnderdog

Original post by Notnek

No, if you've drawn a correct diagram it should clearly be 1 and not -1.

-3pi/4 goes downwards though so it must be a negative gradient

Reply 8

Notnek

Original post by AspiringUnderdog

-3pi/4 goes downwards though so it must be a negative gradient

You're getting confused by what gradient means. Look at your line and ignore everything else on the diagram. Isn't it clear that the line has a positive gradient?

Reply 9

AspiringUnderdog

Original post by Notnek

You're getting confused by what gradient means. Look at your line and ignore everything else on the diagram. Isn't it clear that the line has a positive gradient?

Okay so the gradient is 1 fine I've realised. What can you do with that though?

Reply 10

Notnek

@Angels1234 Looking at this question again, there's a simple geometric approach that I'd recommend instead of using algebra. But you need a correct diagram first.

Reply 11

Notnek

Original post by AspiringUnderdog

Okay so the gradient is 1 fine I've realised. What can you do with that though?

You should be able to then form a cartesian equation of the line since you know if goes through (6,0). Then you can solve the equations simultaneously to find the point of intersection.

But there's a quicker way by considering the geometry as I mentioned above.

Reply 12

AspiringUnderdog

Original post by Notnek

If the half line crosses the centre of the circle is that correct?

Reply 13

Angels1234

Original post by Notnek

https://postimg.cc/image/ooow9aruv/ is this okay ?

Reply 14

Notnek

Original post by AspiringUnderdog

If the half line crosses the centre of the circle is that correct?

Yes that's right.

Reply 15

AspiringUnderdog

Original post by Notnek

Yes that's right.

tbh I'm still a bit stuck after that though

Reply 16

Notnek

Original post by Angels1234

https://postimg.cc/image/ooow9aruv/ is this okay ?

It's much better but needs to be improved to get all the marks in b). Firstly if you consider x=0 or y=0 you'll see the circle must pass through (0,0). Then your line should be 45 degrees from the vertical - your line is too steep.

Also (not sure if you need this to get all the marks), it's a line with gradient 1 so if you go 2 left and 2 down from (6,0) you get to (4,-2) so the line must pass through the centre of the circle.

Try one more time :smile:

Reply 17

Notnek

Original post by AspiringUnderdog

tbh I'm still a bit stuck after that though

Where? Do you have the equation of the line?

Reply 18

AspiringUnderdog

Original post by Notnek

Where? Do you have the equation of the line?

Can you explain it geometrically without changing to cartesian?

In AQA it generally comes up and tbh I'm not feeling another method. Sorry.

Reply 19

Notnek

Original post by AspiringUnderdog

Can you explain it geometrically without changing to cartesian?

In AQA it generally comes up and tbh I'm not feeling another method. Sorry.

Even though there's often a faster geometric approach, you should really be able to use the algebra method. Once you've got the equations it's just GCSE algebra.

Geometric approach:

Spoiler

Let me know if you need more help.