Finding The Equidistant of 2 Points

In summary, the formula for finding the coordinates of the point on the y-axis that are equidistant from two other pairs of points (3,0) and (3,6) involves finding the midpoint of the line joining these points and writing the equation of the perpendicular bisector. The y-component of the midpoint can be found by determining the point that is halfway between 0 and 6 on the number line. This problem can be simplified since the line is vertical, making any perpendicular line a horizontal line with the equation "y=constant".
  • #1
Ecom
3
0

Homework Statement



I need the formula for finding the coordinates of the point on the y-axis that are equildistant from two other pair of points (3,0) and
(3,6).

Homework Equations


i don't know, but these might have something to do with it.

[tex]\sqrt{}[/tex] (y2-y1)2+(x2-x1)2

(x2+x1[tex]/[/tex]2, y2+y1[tex]/2[/tex])

The Attempt at a Solution


I found the midpoint of (3,0)-(3,6), and found the distance between the midpoint to the endpoints, but i don't know were to got from there. I also tried writing random scribbles on my graph papers, crying, begging God to take me out of IB extended math, and finally, posting on Physics Forums for help.
 
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  • #2
Hint: Points equidistant from your two points would be on the perpendicular bisector of the line joining them. Draw a sketch.
 
  • #3
LCKurtz said:
Hint: Points equidistant from your two points would be on the perpendicular bisector of the line joining them. Draw a sketch.

Ah Yes, but that would be too easy!:wink: I need to find the anwser using an equation.
 
  • #4
Sure. Solve for the mid-point between your points. Write the equation of the perpendicular bisector. Solve for where it hits the y axis. The picture was just to lead the way.
 
  • #5
ohhhhhhhhhhh...

Probably should have looked up "perpendicular bisector". I knew what perpendicular meant, but you threw me off at bisector.

Anyway, thanks for the information. Saved me hours of going over "attempt at solution" again
 
  • #6
This is particularly simple because it is a vertical line. Any line perpendicular to it is a horizontal line and has the equation "y= constant". But the problem is finding that constant. To do that you need to know the y-component of the midpoint, which was what the original question asked!

It should be obvious what the x-component is. What point is half way between 0 and 6 on the number line?
 

Related to Finding The Equidistant of 2 Points

1. What is the equidistant of two points?

The equidistant of two points refers to the point or line that is equidistant (equal distance) from both given points. This means that the distance between the equidistant point and each of the two given points is the same.

2. How do you find the equidistant of two points?

To find the equidistant of two points, you first need to determine the coordinates of the two given points. Then, you can use the midpoint formula to find the coordinates of the equidistant point, which is the average of the x-coordinates and the average of the y-coordinates of the two given points.

3. What is the midpoint formula?

The midpoint formula is a mathematical formula used to find the coordinates of the midpoint between two given points. It is (x1 + x2)/2 for the x-coordinate and (y1 + y2)/2 for the y-coordinate, where (x1, y1) and (x2, y2) are the coordinates of the two given points.

4. Can there be more than one equidistant point between two given points?

Yes, there can be more than one equidistant point between two given points. This is because the equidistant point can be a line segment, and there can be multiple points on that line segment that are equidistant from the given points.

5. What is the significance of finding the equidistant of two points?

Finding the equidistant of two points can be useful in various applications, such as mapping and navigation. It can also help in geometric constructions and determining symmetrical points. Additionally, it is a fundamental concept in mathematics and can help in understanding and solving more complex problems.

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