Find an equation for the set of pts that are equidistant from the 2 pts; name 2 other pts on this equation?

Question

Find an equation (with integral coefficients and constants) for the set of points that are equidistant from A(8, -3, 2) and B(4,5,-4). Name two points (with integral coordinates) that satisfy this equation.

i got the first part so far. for the first part i did this:
(x-8)^2+(y+3)^2+(z-2)^2=(x-4)^2+(y-5)^2+(z=4)^2
so the equation for the set of pts that are equidistant from A and B is 8x-16y+12z=20 which simplifies to 2x-4y+3z=5, but i've been staring at this for the past hour and i still can't figure out how to get the 2 points that satisfy this equation and are INTEGRAL numbers

Answer 1

EDIT: Check your algebra: should be - 12z instead of +12z, which leads to 2x-4y-3z=5. Building on your work, though, one integer solution is the midpoint of the two points: (6,1,-1). Another is (1,0,-1)

rhsaunde... is incorrect. The set of points equidistant from two points in 3-space is a plane, perpendicularly bisecting the segment between the two points.

Answer 2

The locus of poiints in 3space that are equidistant to two given points is a plane thru the midpoint between the two given points and perpendicular to the line segment between the two given points.

The two given points are:
A(8,-3,2) and B(4,5,-4)

First find the midpoint M.

M = (A + B)/2 = [(8+4)/2,(-3+5)/2,(2-4)/2] = (6,1,-1)

M is contained in the plane. Now find the plane's normal vector. It is the same as the directional vector of the line between A and B.

v = A - B = <8-4,-3-5,2+4> = <4,-8,6>
We can select a mulitple of this. Lets divide by and select:
v = <2,-4,3>

The equation of the plane is

2(x - 6) - 4(y - 1) + 3(z + 1) = 0
2x - 12 - 4y + 4 + 3z + 3 = 0
2x - 4y + 3z - 5 = 0

Find two integral points on the plane.

One point is M(6,1,-1)

Set y = 0. Then we have:

2x + 3z - 5 = 0
x = z = 1
So another point is P(1,0,1).

Two points with integer coefficients are:
M(6,1,-1) and P(1,0,1).

Answer 3

We will first want the distance between the points, which is sqrt(4^2 + 8^2 + 6^2), or sqrt(116), roughly 11.5. Then we can structure a set of ellipses, using (x-8)*2 + (y+3)*2 + (z-2)^2 + (x-4)^2 + (y-5)^2 + (z+4)*2 = k^2, where k is any integer greater than or equal to 12. You can grind through the algebra to find two other points.

Answer 4

this is going to be a plane that is perpendicular to the line joining the 2 points and passing through the middle point of their segment