# Question 21 - Final question on higher paper

1. Apr 11, 2007

### thomas49th

I cant think of a way to "show that y=....". Do I compare to x values?

Thx

2. Apr 11, 2007

### neutrino

Given coordinates of two points on the plane, how would you find the distance between them? Write down mathematically what the statement says and solve for y.

3. Apr 11, 2007

### symbolipoint

Just use the Distance Formula. This is the typical way of helping new students of Intermediate Algebra to understand parabolas. Your textbook probably also shows an example like the question which you asked.

4. Apr 12, 2007

### HallsofIvy

Staff Emeritus
What IS the distance from a point (x,y) to (0, 2)?
What IS the distance from a point (x,y) to the x-axis?

Since the problem tells you that those two distances are the same write down those two formulas and set them equal.

5. Apr 12, 2007

### thomas49th

(x,y) - (x,0) = (x,y) -(0,2)

is that right?

6. Apr 12, 2007

### Hootenanny

Staff Emeritus
Does this ring any bells;

$$d = \sqrt{\left(x_1-x_0\right)^2+\left(y_1-y_0\right)^2}$$

7. Apr 12, 2007

### HallsofIvy

Staff Emeritus
I don't even know what that means!

I know how to subtract numbers.
I even know how to subtract vectors.

How do you subtract points?

8. Apr 12, 2007

### thomas49th

LHS

x - x = 0
y - 0 = y

RHS
x - 0 = x
y - 2 = y - 2

SO

y = x + y - 2

x = 2

that right?

9. Apr 12, 2007

### neutrino

Nooo!

Take a look at Hootenanny's post.

10. Apr 12, 2007

### thomas49th

ive never ever seen that formula in my life, and I think it is beyond GCSE level (that exams you do in UK when your 15 or 16). Is that formula the simplest way?

11. Apr 12, 2007

### neutrino

I thought most people would have seen that formula at least a couple of years before they turn 16!

Given any two points (x0,y0) and (x1,y1), the distance between them is given by that formula. It's just the Pythagorean theorem done on the Cartesian plane. In three dimensions, there would also be the (z1-z0)2 within the square root.

12. Apr 12, 2007

### cristo

Staff Emeritus
Is it not just Pythagoras' Theorem in a different guise?

13. Apr 12, 2007

### thomas49th

how would cristo go about solving it then?

14. Apr 12, 2007

### Rhythmer

It is. You're right.

15. Apr 12, 2007

### thomas49th

is it just me or are the quotes note displaying today?

16. Apr 12, 2007

### Data

Rhythmer, if your posts are being deleted it is because you have not read the forum guidelines! Answering questions for people is not the purpose in the homework forums; The idea is to get them to do at least a significant portion of the work on their own, so that they can actually learn to understand the material.

17. Apr 12, 2007

### cristo

Staff Emeritus
As data mentions above, any and all questions which are coursework/homework type questions (regardless of whether the homework is set to be handed in) must be answered in a tutorial type way. That means, waiting for the original poster to show some work before giving help, and providing hints or suggestions as opposed to complete solutions.

Well, cristo would follow Halls' hint, using Hootenanny's formula for the distance, and then set them equal.

18. Apr 12, 2007

### jing

The question show in from an examination paper for GCSE. Thomas49th will have been given past examination papers for practice. The examination will take place in May.

Thomas49th you will not have been shown the distance formula for GCSE but you will have covered pythagoras and should be able to apply it.

HallsOfIvy gave you the correct approach and at GCSE finding distance should make you think of Pythagoras if they cannot be found directly. (or perhaps trigonometry if angles are involved)

What you needed to do was to do some drawing on the diagram.

Draw a line from P perpendicular to the x axis. P has coordinates (x,y) so what is the height of P above the x axis?

Now join P to (0,2). From the point (0,2) draw a line parallel to the y axis. This will form a right angled triangle. You should be able to mark on the lengths of the two sides parallel to the x and y axes. Then use Pythagoras to find the length of the line from P to (0,2).

However looking at your reply to HallsofIvy it seems you do not fully comprehend the nature of coordinates and you will need to do some revision and extra practice to get a proper understanding. You will be back at school next week so I suggest you go and see your teacher.

Last edited: Apr 12, 2007
19. Apr 12, 2007

### thomas49th

What did I do wrong when I add/subtracted the co-ordinates? I know you can only add/subtract the x from x and y from y. I had a look here http://www.bbc.co.uk/schools/gcsebitesize/maths/shapeih/transformationshrev2.shtml [Broken]

Can someone point me in the right direction.

Is (x,y) - (x,0) = (x,y) -(0,2) right?

Last edited by a moderator: May 2, 2017
20. Apr 12, 2007

### jing

NO ABSOLUTELY NOT. Just as Hallsof Ivy said.
The webpage above is for vector displacement not distance between points.

Lots of people have been pointing you in the right direction including me. Sometimes to follow advice you have to let go of the idea of what you were attempting was right and study the advice.

Last edited by a moderator: May 2, 2017