Find the point (x,y) on the line y=x that is equidistant from 2 points?

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

equidistant from what points?

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

Find the point (x,y) on the line y=x that is equidistant from the points (9,-1) and (-10,-10)

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

are you familiar with the distance formula between two points in the plane?

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

Yes kind of....

Is it possible that you can answer it and I just see the steps?

I'm a bit confused because of the y=x....

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

Quote:

Originally Posted by

**elifast** Yes kind of....

Is it possible that you can answer it and I just see the steps?

I'm a bit confused because of the y=x....

kind of? you're going to need it to solve this problem.

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

Quote:

Originally Posted by

**elifast** Find the point (x,y) on the line y=x that is equidistant from the points (9,-1) and (-10,-10)

Can you solve $\displaystyle (a-9)^2+(a+1)^2=2(a+10)^2~?$

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

no.. omg i'm so confused T_T

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

Please don't stop helping me though!!!!! I really need this

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

Quote:

Originally Posted by

**elifast** no.. omg i'm so confused T_T

Then you are completely unprepared to do this question.

Go to your notes/textbook. What is the distance between $\displaystyle (a,a)~\&~(-10,-10)~?$

**Report back with what you find.**

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

Y= 1x + 0

and a distance of 14.142135623730951...????

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

Quote:

Originally Posted by

**elifast** Y= 1x + 0 and a distance of 14.142135623730951...????

**You have absolutely no idea about any of this.**

You are wasting your time as well as our time.

We cannot help someone as under prepared as you appear to be.

You need to have a sit-down with a live tutor.

Over and out.

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

I'm just not sure of the processes...

If I can see the process i'll be able to understand it :/

I'm not trying to be bad at math. I really am trying my best! Just it doesn't come that easily to me..

Sorry

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

Hello, I am studying this too, and was very unsure how to solve the same problem. The teacher never gave an example of a similar problem, and the Book does not contain one either.

All it does contain is the distance formula and the procedure to finding the distance between two points.

How you get from knowing how to find the distance of two points to solving this problem, without an example or someone else telling you how, is still a mystery to me. Nevertheless, I feel your frustration, and it is why I am replying.

I will not give you all the calculations and steps, that would be like handing you the answer, the essence here is to understand how to approach the problem.

You could draw a graph. Place the two points on it, lets call these Points A and B. Then draw the line y=x which basically means that a value on y is equal to a value on x. ex: (1,1),(2,2),(-1,-1) etc...

Now, you are seeking to find a point along this line, point C which is the same distance (equidistant) from the other two points on the graph A and B.

If you were to draw lines to join that point from the two original points you would see that an isosceles triangle is formed. What does that mean? It means that the distance between A and C and the distance between B and C are the same.

So using the distance formula, you can conclude that dAC = dBC.

Then, examine what we have as information:

A (9,-1)

B (-10,-10)

C (x,y)

and the line y=x (which implies that the final result of x,y of point C will be the same value, if x=1 then y=1 too)

Now use the values and create an equation using the distance formula between AC = BC and then replace the y by x and then solve for x and you will find the value of x of the point that is equidistance from A and B. And since y=x, the value of y of point C will be the same as the value of x that you find.

Hope this helps.

PS: I am trying to solve a similar problem but my line is y=-2x+2 and trying to figure out the mystery once more :P

Re: Find the point (x,y) on the line y=x that is equidistant from 2 points?

Just a quick update.

By trying to solve my problem in relation to line y=-2x+2, it seems that when we replace y by x, it is actually the value of the coordinate Y that we find when solving for X in the calculation. This is not apparent in the problem with line y=x because both are identical. But with a different line the next step is to plug in that value in the Line's equation and solve for X to find the x Coordinate of point C.

And that is about it. Use Distance Formula to find the Y coordinate first of the equidistant point and the then use that in the equation to find the X coordinate.

You will need to have knowledge of Elementary Algebra notions to be able to do properly the calculations (I think this is what the other posters were trying to determine before discussing the problem with you).

The way Math is taught is procedural, it is assumed that you know how to use one level of notions before solving or learning the next. And it takes lots of practice, if you only are trying to find answers to satisfy some assignment, chances are you will completelly miss the mark in exams, when these throw at you confusing configurations of equations (apparently that is the way to verify if someone trully has learned a notion), I do not like it personally but it is how it is sadly and we just have to follow the flow...(in a more conceptual level maybe this is why we do not have Einstein's anymore in the world, pushing everyone in the same mold, but that question is beyond the scope of this forum and discussion).

Cheers!