Question 21 - Final question on higher paper

[thomas49th]

I can't think of a way to "show that y=...". Do I compare to x values?

Thx

 

[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.

 

[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.

 

[HallsofIvy]

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.

 

[thomas49th]

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

is that right?

 

[Hootenanny]

Does this ring any bells;

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

 

[HallsofIvy]

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

is that right?

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?

 

[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?

 

[neutrino]

LHS

x - x = 0
y - 0 = y

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

SO

y = x + y - 2

x = 2

that right?

Nooo!

Take a look at Hootenanny's post.

 

[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?

 

[neutrino]

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?

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.

 

[cristo]

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?

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

 

[thomas49th]

how would cristo go about solving it then?

 

[Rhythmer]

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

It is. You're right.

 

[thomas49th]

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

 

[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.

Guidelines: https://www.physicsforums.com/showthread.php?t=94379

 

[cristo]

Well, I've already experienced some moderators deleting my posts in this section on which I answer homework questions, but I think I'm allowed to answer this one simply because it's not homework.

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.

how would cristo go about solving it then?

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

 

[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.

 

[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

Can someone point me in the right direction.

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

 

[jing]

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

Can someone point me in the right direction.

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

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.

 

[thomas49th]

The webpage above is for vector displacement not distance between points.

So how do I find the distance between points?

 

[neutrino]

So how do I find the distance between points?

Have you read any of the posts at all? The formula has been stated at least three times in this thread.

 

[symbolipoint]

So how do I find the distance between points?

thomas49th, if you are serious about this, then this suggests that you are either misplaced in your course, or you are studying on your own and beyond your level of achievement. What course are you enrolled in? IF you are enrolled in "Algebra 2" or Intermediate Algebra, then you have been taught the distance formula. In case you are just now in the course being introduced to the distance formula, then study your book very carefully. Again, as reminder, you probably will find an example in your book much like the question which you have here been asking.

 

[jing]

So how do I find the distance between points?

Follow the instructions in my post #18 and do the drawing.

You do know Pythagoras' Theorem don't you?

 

[jing]

thomas49th, if you are serious about this, then this suggests that you are either misplaced in your course, or you are studying on your own and beyond your level of achievement. What course are you enrolled in? IF you are enrolled in "Algebra 2" or Intermediate Algebra, then you have been taught the distance formula. In case you are just now in the course being introduced to the distance formula, then study your book very carefully. Again, as reminder, you probably will find an example in your book much like the question which you have here been asking.

thomas49th is taking what is called General Secondary Certficate Examination (GCSE) in Maths in England. It includes number, algebra, statistics and geometry and is taken at the age of 16. The examination is graded A,B,C,D,E,F, U (unclassified). Students achieving grades A or B will be allowed to go on to a pre-university maths course.

This question is the last on the paper and is meant to differentiate between those who can do formulas in maths and those who understand maths. The distance formula is not part of the course. Those who can apply the maths they have been taught through their knowledge of coordinates and pythagoras get the marks.

 

[Data]

So how do I find the distance between points?

Okay, let's give this a try:

Suppose you have two points, p = (a, b), and q = (x, y). What's the distance between p and q?

Well, we could try something like what you did in order to define distance: Subtract the components of the two to get a new point (a - x, b - y), and then add the components of the new point and call that the distance, so our distance is d(p, q) = (a - x) + (b - y). But, there's a problem!

What if, say, a = y and x = b, so our two points are really p = (y, x) and q = (x, y). If $x \neq y$, then those two points don't have the same coordinates. However, according to our distance formula, d(p, q) = (y - x) + (x - y) = 0. Does it make sense to have a distance of zero between two different points? No!So that doesn't work. What can we do to fix it? Well, we could try taking the positive version of each of the components of (a - x, b - y); in other words, we could set d(p, q) = |a - x| + |b - y|.

The earlier problem disappears now: as you can check for yourself, d(p, q) is never 0 unless a = x and b = y; in other words, it's not zero unless p = q. Good!

However, there's still a problem!

Suppose I draw a right-angled triangle. Its vertices are p = (0, 0), q = (1, 1), and r = (1, 0).

Now, our distance formula tells us that: d(p, q) = |0 - 1| + |0 - 1| = 1 + 1 = 2, d(p, r) = |0 - 1| + |0 - 0| = 1, and d(r, q) = |1 - 1| + |0 - 1| = 1.

But, now remember the Pythagorean theorem! The Pythagorean theorem says that, if u is the length of the longest side of a right-angled triangle, and t and v are the lengths of the other sides, then

$$u^2 = t^2 + v^2.$$

Let's try it on our triangle, using our distance formula: The longest side has length 2, and each of the others have length 1. Does the Pythagorean theorem work out? No: $4 = 2^2 \neq 1^2 + 1^2 = 2$.

What does this tell us? Well, the longest side of a right triangle with those vertices, according to the two smaller lengths (both 1) and the Pythagorean theorem, is $$\sqrt{1^2 + 1^2} = \sqrt{2}$$.

So, according to the Pythagorean theorem, there's a line between p and q shorter than the distance between p and q!

That definitely seems off! After all, the distance should represent the length of the shortest path between points, shouldn't it (in technical terms, we want to make the plane into a geodesic metric space - but you definitely don't need to worry about what that means! Edit: whoops! I've mixed up my definitions a bit. This little comment wasn't right anyways, so you should definitely ignore it!)?

So we need to fix our definition for distance again. How can we do that? Well, right now if p = (a, b) and q = (x, y), we set d(p, q) = |a - x| + |b - y|. But what's that really doing? It's taking the vertical distance between the points and just adding it to the horizontal distance between the points.

The problem (illustrated above with the Pythagorean theorem) is that we can put a sloped line between the points that is shorter than our "distance!"

How can we fix that? Well, remember the Pythagorean theorem again. If u is the length of the longest side of a right angled triangle and t and v are the lengths of the shorter sides, then

$$u^2 = t^2 + v^2.$$

In other words, the length of the longest side is

$$u = \sqrt{t^2 + v^2}.$$

What does this really say, though? Well, take points p = (a, b) and q = (x, y). Well, we can form a right triangle by using a vertex r = (x, b).

If we do that, then the shorter distances $t$ and $v$ in the Pythagorean theorem correspond to the lengths $t = |a - x|$ and $v = |b - y|$. So, the length of the longest side, which we also want to be the distance between p and q, is

$$d(p, q) = u = \sqrt{t^2 + v^2} = \sqrt{|a - x|^2 + |b - y|^2}.$$

This is the general formula for distance between points. In other words, if $p_1 = (x_1, y_1)$ and $p_2 = (x_2, y_2)$ are points on the Cartesian plane, then the distance between them is

$$d(p_1, p_2) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$$

(You should figure out for yourself why I was able to drop the | |'s).

At this point, you should definitely be able to figure out your problem .[Yes, I know that my argument using the Pythagorean theorem is circular, but I'm just trying to give some sort of natural motivation without too much hassle ]

 

[neutrino]

Nicely done, Data. Great post!

geodesic metric space? :rofl:

 

[Data]

The thread was getting a bit tedious!