Math Is Fun Forum

runlemmingsrun1

Guest

equidistant point

Please solve for a point on the x-axis that is equidistant from (10,4) and (6,2).

    Thanks

gnitsuk

Member

Registered: 2006-02-09

Posts: 121

Re: equidistant point

Call point on x-axis P. It will have coordinates (x,0). We need distance from P to (6,2) to equal distance from P to (10,4).

Formula for distance between to points (a,b),(c,d) is:

So we require:

So square bot sides and expand giving:

Cancel square of x from both sides and rearrange giving:

therefore

Therefore P is the point:

Last edited by gnitsuk (2006-10-04 03:27:37)

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: equidistant point

Picture two triangles on the graph.   The first triangle will have corner points of (6,2), (6,0) and (z,0).   The second will have (10,4), (10,0), and (z,0).   These are right triangles.   You want to find z so the the hypotenuses (h) of these 2 triangles are equal.  Use the Pythagorean theorem. 

For the first triangle:   

For the second triangle:

Set those two equations to be equal to each other and you have:

So the point is (9.5, 0)

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: equidistant point

Sorry about the duplicate answer.   At least we came up with the same answer!

polylog

Member

Registered: 2006-09-28

Posts: 162

Re: equidistant point

We need a point, call it P, which is on the x-axis, and therefore of the form P = (x, 0)

The distance formula between two points is:

We need the distance between (10, 4) and P to be the same as the distance between (6, 2) and P.

Ie, we need:

to be equal to:

Setting these to be equal:

Square both sides:

Expanding:

Re-arranging:

So the point we need is:

polylog

Member

Registered: 2006-09-28

Posts: 162

Re: equidistant point

Haha sorry it's already done ! oops

runlemmingsrun1

Guest

Re: equidistant point

thanks...i was never good in math..