Find an equation of the plane consisting of all points that are equidistant from (3, 0, 0) and (-2, -4, 0)

Solution:

The plane will have a direction vector which is equal to the direction between the points (3, 0, 0) and (-2, -4, 0)

It will pass through the midpoint of segment between these two points

[(3, 0, 0) + (-2, -4, 0)]/2 = (1/2, -2, 0)

The equation of plane is

5(x - 1/2) + 4(y + 2) + 0(z) = 0

Which is the plane through the given direction and point

Using the multiplicative distributive property

5x - 5/2 + 4y + 8 = 0

5x + 4y + 11/2 = 0

Taking LCM

10x + 8y + 11 = 0

Therefore, the equation of the plane is 10x + 8y + 11 = 0.

Find an equation of the plane consisting of all points that are equidistant from (3, 0, 0) and (-2, -4, 0)

Summary:

An equation of the plane consisting of all points that are equidistant from (3, 0, 0) and (-2, -4, 0) is 10x + 8y + 11 = 0.