Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2021-05-06 17:58:33

mathland
Member
Registered: 2021-03-25
Posts: 444

Equidistance

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

Given two points: A(6, 5), B(1, −8)

Is this question asking for the midpoint of AB?

Offline

#2 2021-05-07 05:53:19

Bob
Administrator
Registered: 2010-06-20
Posts: 10,140

Re: Equidistance

That's the correct place to start.

But we need all the points that are equidistant.  So you need the perpendicular bisector of AB

https://www.mathsisfun.com/geometry/con … isect.html

Steps: 1.  work out the gradient of AB

2. Hence the gradient of a line that is perpendicular to AB.

3. Using that and the midpoint, work out the equation of that line.

B.


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#3 2021-05-07 11:13:19

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Equidistance

Bob wrote:

That's the correct place to start.

But we need all the points that are equidistant.  So you need the perpendicular bisector of AB

https://www.mathsisfun.com/geometry/con … isect.html

Steps: 1.  work out the gradient of AB

2. Hence the gradient of a line that is perpendicular to AB.

3. Using that and the midpoint, work out the equation of that line.

B.

Good steps and should be easy to do.

Offline

Board footer

Powered by FluxBB