The set of all points equidistant from (0, 1, 5) and 5, -1, 3) is a plane. Find the equation.

1 answer

Let P(x, y, z) be a point equidistant from (0, 1, 5) and (5, -1, 3). The distance between P and (0, 1, 5) is equal to the distance between P and (5, -1, 3).

Using the distance formula, we can write the following equation:

√((x - 0)^2 + (y - 1)^2 + (z - 5)^2) = √((x - 5)^2 + (y + 1)^2 + (z - 3)^2)

Simplifying this equation, we get:

(x^2 + y^2 + z^2 - 10z + 26) = (x^2 - 10x + 25 + y^2 + 2y + 1 + z^2 - 6z + 9)

Canceling out the common terms, we get:

-10z + 26 = -10x + 25 + 2y + 1 - 6z + 9

Rearranging this equation, we get:

-10z + 10x + 6z - 2y = 9 - 25 - 1 - 9 + 26

Combining like terms, we have:

10x - 4z - 2y = 1

Dividing both sides by 2, we get:

5x - 2z - y = 1

Therefore, the equation of the plane is 5x - 2z - y = 1.