Coordinate Geometry

[Vastor]

hey guys,

 

from the (math)textbook, The distance between the points [math]P(x_1, y_1)[/math] and point [math]Q(x_2, y_2)[/math]

 

is [math] \sqrt{(x_2 - x_1)^2 + (y_2- y_1)^2}[/math]

 

what does the intuition behind the statement? it look like much Phytaghoras theorem, does anything related here?

 

 

another thing is the intercept form [math] \frac{x}{a} + \frac{y}{b} = 1 [/math], where a = x-intercept, and b = y-intercept.

 

anyone can give how does this formula formed from? or atleast any intuitive behind this equation?

 

 

hope anyonce can help, btw, if anyone got good website/tutorial about coordinate geometry, please send me links.. thnx

[ajb]

It is essentially Pythagoras theorem. You use this theorem to define the notion of distance on a plane.

[Appolinaria]

 

Compare that to when you're given two points and need to find the distance;

Using the distance formula is the same as using the Pythagorean theorem to find the hypotenuse of a triangle;

[Vastor]

thnx for both of ya, actually, ajb's short explaination already confirmed my thought about the phytagoras "shape" on the graph itself. but the explaination from Appolinaria is wonderful...

 

about other equation/formula, I may derived myself to find the intuitive behind it, though so will visit this post again if had any question...

[Vastor]

heyy guys,

 

there is 1 equation for the point [math](x_p, y_p)[/math] which divides PQ in the ratio m:n, where [math]P(x_1, y_1)[/math] and [math]Q(x_2, y_2)[/math].

 

[math](x_p, y_p) = (\frac{nx_1 + mx_2}{m + n}, \frac{ny_1 + my_2}{m+ n})[/math]

 

 

which if specified for the x-axis only...

 

[math] x_p = \frac{nx_1 + mx_2}{m+n}[/math]

 

[math] x_p(m+n) = nx_1 + mx_2[/math]

 

 

if we put this into variable K

 

[math] K = (x_p)* (m+n) [/math]

 

then

 

K = (horizontal value of the point) * (sum of ratio)

 

so, what is K? why does it must be equal to calculate between the x point?, did I label each variable right?

Edited February 21, 2012 by Vastor

[Vastor]

 

wow, never thought it would be so hard to search for this on the net...

 

but, atleast it would be a good start to explain the whole equation then...

 

EDIT:-

 

Eureka!

 

after tried it with multiple calculation, I found the intuitive behind the formula itself.

 

if [math] P_1M : MP_2 = 2 : 1 [/math]

 

[math] x_M = x_1 + \frac{2}{3}(x_2 - x_1) [/math]

 

which in other words

 

[math] ("target" x-value) = ("start" x-value) + \frac{ratio of P_1M}{ratio sum}(("end" x-value) - ("start" x-value)) [/math]

 

where:-

- when, the "end" x-value minus the "start" x-value, we get the distance between the x-value which is from "start" to "end" of line segment.

 

- the distance times by the ratio related to the line divide by sum of ratio will produce a distant between the "start" point and "target" point.

 

- the distance from "start" to "target" point used to sum with the "start" point to obtain the x-value of the "target" point.

 

thnx everyone

 

ooo wait, I solve this myself, should check out google first next time

Edited February 21, 2012 by Vastor

[Vastor]

hey guys,

 

because Google no help for now, I think I can't help but ask here...

 

 

 

Given that the points A(3, 5), B(6, 2), C(5, -1) and D(p, q) are the vertices of a parallelogram, find the values of p and q.

 

My trial:-

 

I'm assuming AB = CD,

 

so I calculate the distance of line AB (using phytaghoras theorem)

 

[math] AB = \sqrt{8} [/math]

 

I'm assuming BC = AD too

 

[math] BC = \sqrt{10} [/math]

 

 

ok, I'm clarifying that I'm clueless

 

help anyone?

[imatfaal]

Not sure that is how I would go about it.

 

1. Draw a diagram - even a scrappy one to get it straight in your head

2. What do we know about parallelograms that can solve the problem - parallelograms have opposite sides that are --------- by definition and must be ----------

 

hint - you do not need to use pythagoras

[Vastor]

hey guys,

 

there still something that I'm not quite understand on how to calculate locus.

 

this actually help something on intuitive of the locus from a fixed point

 

but not explain much about locus(x, y) from two fixed point, (x1, y1) and (x2, y2), based on ratio of m:n, where the equation is:-

 

[math] \frac{\sqrt{(x - x_1)^2 + (y - y_1)^2}}{\sqrt{(x - x_2)^2 + (y - y_2)^2}} = \frac{m}{n} [/math]

 

can anyone explain more about this?

[imatfaal]

The locus all points P such that the distance AP is m/n times that of distance BP. Now when the ratio is 1/1 then we know that the locus is the perpendicular line at the midpoint of the line linking A and B. Can we assume that the answer is a line offset from the simple answer - No! If you draw a simple diagram it is clear this doesn't work (try it).<br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); "><br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); ">So let's work on equations<br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); ">Let A be (j,k) B (r,s)<br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); "><br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); ">AP = Sqrt[(x-j)^2+(y-k)^2]<br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); "><br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); ">BP = Sqrt[(x-r)^2+(y-s)^2]<br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); "><br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); ">AP = (m/n)BP<br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); "><br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); ">AP/BP = (m/n)<br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); "><br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); ">nAP = mBP<br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); "><br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); ">n Sqrt[(x-j)^2+(y-k)^2]= m Sqrt[(x-r)^2+(y-s)^2]<br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); "><br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); ">Take m and n inside sqrt brackets and square both sides<br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); "><br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); ">n ^2 [(x-j)^2+(y-k)^2]= m ^ 2 [(x-r)^2+(y-s)^2]<br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); "><br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: -webkit-auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; background-color: rgba(255, 255, 255, 0.917969); ">n ^2 [x^2 -2jx + j^2 + y^2 -2ky + k^2]= m ^ 2 [x^2 -2rx + r^2 + y^2 -2sy + s^2]

 

rearrange

 

(n^2-m^2)x^2 + (n^2-m^2)y^2 + (2rm^2-2n^2j)x + (2m^2s-2kn^2)y + m^2j^2 + m^2k^2 - n^2r^2 - n^2s^2 = 0

 

as m, n , j , k , r ,s are all numbers ie are not variables we can look at the above horrible equations as

 

cx^2 + cy^2 + ex + fy = g

 

and that is a circle!

 

I dont know how much you know about circles and their functions so I won;t go any further. But it is clear from the equation that the locus of points P such that the distance AP is m/n times that of distance BP describes a circle. The next bit of your investigation needs to be what radius is that circle and what is the centre point. Do some numerical examples

 

 

 

 

 

 

 

Re the above garbage

 

sorry about that - don't know what happened to my post

[Vastor]

(n^2-m^2)x^2 + (n^2-m^2)y^2 + (2rm^2-2n^2j)x + (2m^2s-2kn^2)y + m^2j^2 + m^2k^2 - n^2r^2 - n^2s^2 = 0

 

as m, n , j , k , r ,s are all numbers ie are not variables we can look at the above horrible equations as

 

cx^2 + cy^2 + ex + fy = g

 

and that is a circle!

 

I dont know how much you know about circles and their functions so I won;t go any further. But it is clear from the equation that the locus of points P such that the distance AP is m/n times that of distance BP describes a circle. The next bit of your investigation needs to be what radius is that circle and what is the centre point. Do some numerical examples

 

 

 

 

Re the above garbage

 

sorry about that - don't know what happened to my post

 

 

ouch, I'm totally lost on what you are trying to say, why don't you just go on more with "circles and their functions", btw, why not re-post the "bugged text" ??? thnx

[Klaynos]

The locus all points P such that the distance AP is m/n times that of distance BP. Now when the ratio is 1/1 then we know that the locus is the perpendicular line at the midpoint of the line linking A and B. Can we assume that the answer is a line offset from the simple answer - No! If you draw a simple diagram it is clear this doesn't work (try it).<br style="color: rgb(34, 34, 34); font-family: arial, sans-serif; font-size: 13p...

 

 

What were you trying to do when the HTML kicks in, use latex?

[imatfaal]

No the problem was that I wrote the post on my iphone in the pub, the connexion was so poor that I could not log on to SFN from iphone - so emailed it to myself and when I was back at office cut'n'paste it from gmail. Gmail must have introduced some tags that caused the garbage. I will sort it out - when I have given the prob a bit more thought, the explanation was pretty crummy anyway

 

ouch, I'm totally lost on what you are trying to say, why don't you just go on more with "circles and their functions", btw, why not re-post the "bugged text" ??? thnx

 

 

 

Vastor - will come back to it when I have a second. Basically - what I was saying was that if you multiply out that equation you get an equation for a circle. I have never found the locus of all points which share a common ratio of distances from two points - so it was a bit of a surprise for me.

 

I think the first thing you should do - is multiply out that equation and gather it together in a more usual form. If you are struggling on algebra let me know, you can work from this start

 

n^2(distance_AP) = m^2(distance_BP)

 

use two actual points for A and B and a nice simple ratio of m/n. The more I think about this the nicer it is a problem

[Vastor]

n^2(distance_AP) = m^2(distance_BP)

 

do you mean n^2(distance_AP)^2 = m^2(distance_BP)^2 ?

 

I think I understand now... tyvm

[imatfaal]

do you mean n^2(distance_AP)^2 = m^2(distance_BP)^2 ?

 

I think I understand now... tyvm

 

Nope - cos the distances were roots - but the ratios weren't. So when we get rid of the roots the ratio becomes squared.

 

I had forgotten about this one - let me know if you cannot get to the bottom of it

[Vastor]

Nope - cos the distances were roots - but the ratios weren't. So when we get rid of the roots the ratio becomes squared.

 

I had forgotten about this one - let me know if you cannot get to the bottom of it

 

that's why!

 

distance = [math]\sqrt{(x_1- x_2)^2 + (y_1 - y_2)^2} [/math]

 

where n:m is ratio,

 

[math]m(distance_{AP}) = n(distance_{BP}) [/math]

 

[math]m(\sqrt{(x - x_1)^2 + (y - y_1)^2}) = n(\sqrt{(x - x_2)^2 + (y - y_2)^2}) [/math]

 

right?!

[imatfaal]

Yep - that's the ticket. Square both sides and you can simplify. I really would use actual values for (x1y1) and (x2y2) the first time around - the equation is pretty monstrous, I haven't got my head around exactly what it means apart from the form of the locus.

[Schrödinger's hat]

Any luck with that ratio of distances one Vastor?

Imatfaal and I nutted it out in another thread. It gets pretty gruesome without a few tricks to simplify it.

Some things that may help:

Pick new coordinates so that the origin is at the centre of one of the circles. This is equivalent to shifting everything over.

[math]x' = x-x_1,\quad y' = y - y_1[/math]

Put frequently appearing variables in another variable:

[math]a = x_1 - x_2,\quad b = y_1 - y_2,\quad c = \frac{m}{n}[/math]

 

A re-scaling may also be useful later on. You should see where to do it if you need it.

[math]x'' = (c^2 - 1)x'[/math]

 

Remember the method of completing the square. If you see this:

[math] x^2 + 2kx [/math]

You can look at this:

[math] (x + k)^2 [/math]

And see that it expands to:

[math] x^2 + 2kx + k^2[/math]

So then you know:

[math] x^2 + 2kx = x^2 + 2kx + k^2 - k^2 = (x + k)^2 - k^2[/math]

 

Combine these methods carefully and you should be able to prevent it turning into a giant mess.

Edited March 10, 2012 by Schrödinger's hat

[Vastor]

huh?! you make me lost with that high level of math.... xD

[Schrödinger's hat]

huh?! you make me lost with that high level of math.... xD

 

 

Well our goal is to make this:

[math]m(\sqrt{(x - x_1)^2 + (y - y_1)^2}) = n(\sqrt{(x - x_2)^2 + (y - y_2)^2}) [/math]

Look like this:

[math] (x-x_3)^2 + (y-y_3)^2 = r[/math]

Where

[math]x_3, y_3 \mbox{and} r[/math]

are some combinations of our original variables.

 

We can start by shifting everything so that one of our circles is at the origin. This doesn't change the problem because it's just the same as re-labelling our axes (everywhere you see an x, you replace it with that number - x_1 and same w/ y).

I put a little dash next to them so we know they're not the original x and y, but any results with these new coordinates hold for the old ones. It's a trick to get rid of some of the mess.

This is my new set of variables:

[math]x' = x-x_1,\quad y' = y - y_1[/math]

So everywhere I see x,y I can replace them with

[math]x = x'+x_1,\quad y = y' + y_1[/math]

Also divide everything by n at the same time.

[math]\frac{m}{n}\left(\sqrt{(x' + x_1 - x_1)^2 + (y' + y_1 - y_1)^2}\right) = \sqrt{(x' + x_1 - x_2)^2 + (y' + y_1 - y_2)^2} [/math]

Next step, I cancel the numbers with opposite signs and replace some numbers with shorter names for them

[math]a = x_2 - x_1,\quad b = y_2 - y_1,\quad c = \frac{m}{n}[/math]

[math]c\left(\sqrt{(x')^2 + (y')^2}\right) = \sqrt{(x' - a)^2 + (y - b)^2} [/math]

This following isn't really necessary, but if you square both sides and start rearranging you might see that it makes things a bit tidier.

[math]x'' = (c^2 - 1)x'[/math]

 

 

Then the last part is completing the square. This is a common method of dealing with quadratics. Read over what I said carefully, look up the derivation for the quadratic formula and/or ask for further help if you need it.

 

None of these tricks are necessary to solve the problem, you can just have at it with algebra. It gets very big and messy doing it that way though and might be a little intimidating.

Edited March 10, 2012 by Schrödinger's hat