Finding A Relationship For Equidistance

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  • #1
nycmathguy
Homework Statement:
Find a relationship between x and y.
Relevant Equations:
d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Find a relationship between x and y
such that (x, y) is equidistant (the same distance) from the two points.

Our two points: (6, 5), (1, −8)

No sample question is given by Ron Larson for this problem in the textbook. I need a set up. I will do the math, of course.
 

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  • #2
anuttarasammyak
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Distance between (x,y) and (6,5) = Distance between (x,y) and (1,-8)
 
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  • #3
nycmathguy
Distance between (x,y) and (6,5) = Distance between (x,y) and (1,-8)
Are you saying to use the distance formula twice?
 
  • #4
anuttarasammyak
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Yes, three points (x,y) on the figure we explore, (6,5) and (1,-8) have three distances between one of which is the distance between (6,5) and (1,-8) a fixed number and so irrelevant for the problem.
 
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  • #5
nycmathguy
Yes, three points (x,y) on the line, (6,5) and (1,-8) have three distances between one of which is the distance between (6,5) and (1,-8) is a fixed number and irrelevant for the problem.
Is this the set up?

Let d_1 = first distance

Let d_2 = second distance

d_1 = sqrt{(6 - x)^2 + (5 - y)^2}

d_2 = sqrt{(1 - x)^2 + (-8 - y)^2}

If not, can you set up the needed distance formulas? I will then take it from there.

Thanks
 
  • #6
DaveC426913
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I am not sure - it has been a long time since school for me - but is there some reason you can't just find the centre point between the two points - i.e. x=(6+1)/2 , 5+(-8)/2)?

Also there are an infinite number of points that meet the requirements. They form a line that passes through (x,y) and has the negative reciprocal slope.
 
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  • #7
anuttarasammyak
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d_1 = sqrt{(6 - x)^2 + (5 - y)^2}

d_2 = sqrt{(1 - x)^2 + (-8 - y)^2}
Solve d_1=d_2.
 
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  • #8
Delta2
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In case you are wondering what you gonna solve for, since it seems you have two unknowns x and y , solve for either. you should find the one as a function of the other, that is y=f(x) where f some function for example f(x)=Ax+B.
You need to do some algebraic steps starting from ##d_1=d_2## to end up to ##y=f(x)##.
 
  • #9
symbolipoint
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Homework Statement:: Find a relationship between x and y.
Relevant Equations:: d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Find a relationship between x and y
such that (x, y) is equidistant (the same distance) from the two points.

Our two points: (6, 5), (1, −8)

No sample question is given by Ron Larson for this problem in the textbook. I need a set up. I will do the math, of course.
Just substitute your given values into the Distance Formula, and compute the expression. The setup is already understood - it will correspond to the formula and the substitutions of given values made.

Not that. Again, I misunderstood the question because of my not reading carefully enough.
 
  • #10
symbolipoint
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Homework Statement:: Find a relationship between x and y.
Relevant Equations:: d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Find a relationship between x and y
such that (x, y) is equidistant (the same distance) from the two points.

Our two points: (6, 5), (1, −8)

No sample question is given by Ron Larson for this problem in the textbook. I need a set up. I will do the math, of course.
@nycmathguy, I have a question for you. If you have two given, known points (q,r) and (t,v), and you want a set of points (x,y) which are equally distant from (q,r) as from (t,v), you can use the distance formula to represent the relationship.

Distance from (x,y) to (q,r) is the same as the distance from (x,y) to (t,v). Would you apply the distance formula and create and simplify the equation?

Now once you have done that, compare your work to what instruction your book shows for this same situation.
 
  • #11
PeroK
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In the old days before the likes of Ron Larson there was such a thing as the perpendicular bisector!
 
  • #12
PeroK
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I am not sure - it has been a long time since school for me - but is there some reason you can't just find the centre point between the two points - i.e. x=(6+1)/2 , 5+(-8)/2)?

Also there are an infinite number of points that meet the requirements. They form a line that passes through (x,y) and has the negative reciprocal slope.
In other words, we should not just pound away algebraically, but try to understand the geometry of the problem.
 
  • #13
caz
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The above discussions give two different ways of solving the problem
1) a) pick an arbitrary point (x,y)
b) calculate the distances from (x,y) to (6, 5) and (1, −8)
c) set the distances equal and solve for y=f(x)
2) a) Think about the geometry of the situation
b) Recognize that a line is a solution to this problem and that it is at a right angle to the line connecting (6, 5) and (1, −8)
c) Find the midpoint on line connecting (6, 5) and (1, −8)
d) Calculate the slope on line connecting (6, 5) and (1, −8)
e) Calculate the negative reciprocal slope
f) Using the negative reciprocal slope and the midpoint, determine the equation for the line

I've attached a sketch

picture.jpeg
 
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  • #14
PeroK
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... in my opinion, this is definitely a problem that should be done both ways, with the aim of getting the same answer by two contrasting methods. That's what's needed to develop confidence in both methods.
 
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  • #15
DaveC426913
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I still don't get it. Why not simply assume (x,y) is the midpoint of the two known points?

x=(6+1)/2 = 3.5
y=(5+(-8))/2 = -1.5
Boom.
 
  • #16
PeroK
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I still don't get it. Why not simply assume (x,y) is the midpoint of the two known points?

x=(6+1)/2 = 3.5
y=(5+(-8))/2 = -1.5
Boom.
You need all points on a line, not only that one.
 
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  • #17
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I still don't get it. Why not simply assume (x,y) is the midpoint of the two known points?

You need all points on a line, not only that one.
I agree. Getting all the points on that bisector gets you the x-y relationship that is asked for in the problem statement.
 
  • #18
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Are you saying to use the distance formula twice?
How else are you going to find the distance between (x, y) and (6,5) and between (x, y) and (1, -8)?
 
  • #19
DaveC426913
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You need all points on a line, not only that one.
Yes. As I clarify*, way back in post 6.

* the first to do so

But that's not the point. You only need one point that's equidistant and the slope. And this is the easiest point to find - which is the point.

I agree. Getting all the points on that bisector gets you the x-y relationship that is asked for in the problem statement.
OMG, we're going around in circles here. All this was clarified in post 6.

Yes, you need the line. No one argues this.


My question is: why bother using the distance formula twice and equating them to get an unknown point?

The easiest point to find on the line is simply the one that is halfway between the two known points. And you know its coords simply by averaging the x and y of the known points.

x=(6+1)/2 = 3.5
y=(5+(-8))/2 = -1.5

And then the slope of the line is simply the inverse reciprocal of the line between the two known points.
slope = rise over run
Between to known points: (6-1) / (5+8) = 5/13
New slope of all equidistant points = -13/5
 
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  • #20
nycmathguy
How else are you going to find the distance between (x, y) and (6,5) and between (x, y) and (1, -8)?
Hey, it's just a question. No need to be
Yes. As I clarify*, way back in post 6.

* the first to do so

But that's not the point. You only need one point that's equidistant and the slope. And this is the easiest point to find - which is the point.


OMG, we're going around in circles here. All this was clarified in post 6.

Yes, you need the line. No one argues this.


My question is: why bother using the distance formula twice and equating them to get an unknown point?

The easiest point to find on the line is simply the one that is halfway between the two known points. And you know its coords simply by averaging the x and y of the known points.

(And then the slope of the line is simply the inverse reciprocal of the line between the two known points.)
I will work on it.
 
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  • #21
nycmathguy
The above discussions give two different ways of solving the problem
1) a) pick an arbitrary point (x,y)
b) calculate the distances from (x,y) to (6, 5) and (1, −8)
c) set the distances equal and solve for y=f(x)
2) a) Think about the geometry of the situation
b) Recognize that a line is a solution to this problem and that it is at a right angle to the line connecting (6, 5) and (1, −8)
c) Find the midpoint on line connecting (6, 5) and (1, −8)
d) Calculate the slope on line connecting (6, 5) and (1, −8)
e) Calculate the negative reciprocal slope
f) Using the negative reciprocal slope and the midpoint, determine the equation for the line

I've attached a sketch

View attachment 284756
Beautifully done. Thank you.
 

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