Find Equation of Plane from 2 Points | Equidistant

[ProPatto16]

Homework Statement

find equation of a plane consisting of all points equidistant from points (2,5,5) and (-6,3,1)

Homework Equations

equation of a plane is
ax + by + cz + d = 0 where d = -(ax₀+by₀+cz₀)

The Attempt at a Solution

can this be solved by finding the midway point of the line made by those two points, then using the vector from the midway to one point as a normal to the plane?

i just don't know how to find coordinates for that midpoint. i only know how to find the length of the line. its not just halfway between 2 and -6 for x and 5 and 3 for y and 5 and 1 for z is it?

 

[Char. Limit]

The midpoint between two points (a,b,c) and (d,e,f) is going to be ((a+d)/2, (b+e)/2, (c+f)/2). And then, yes, use the vector from that point to either of your two points as a normal vector.

 

[ProPatto16]

so the midpoint of those two points, the one on the plane, is (-2,4,3) and so the normal vector is from that point to (2,5,5)
does that make the vector coordinates <4,1,2> ?

and then using equation for plane that passes through point (2,5,5) with normal vector <4,1,2>

a(x-x₀) + b(y-y₀) + c(z-z₀)

becomes

4(x-(-2)) + 1(y-4) + 2(z-3) = 0
4x + 8 + y - 4 + 2z - 6 = 0
4x + y + 2z = 2

and that's the equation of the plane?

 

[Char. Limit]

Seems to check out for me.

 

[ProPatto16]

thank you