# Find an equation for the plane consisting of all points that are equidistant from the points (1,0,-2) and (3,4,0).

This problem aims to familiarize us with geometrical calculations. The concept required to solve this problem is the distance formula in 3-dimensional space, and some square and cubic algebraic formulas.

The formula for distance states that the distance between two points in xyz-space is the sum of the squares of the differences between similar xyz coordinates under a square root. Let’s say we have points:

$P_1 = (x_1,y_1,z_1)\space and\space P_2 = (x_2,y_2,z_2)$

The total distance between $P_1$ and $P_2$ is yielded as:

$d(P_1,P_2) = \sqrt{(x_2 x_1)^2 + (y_2 y_1)^2 + (z_2 z_1)^2}$

Given points are $(1,0,-2)$ and $(3,4,0)$.

We have to generate an equation for the plane consisting of all points that are equidistant from the points $(1,0,-2)$ and $(3,4,0)$.

Let’s assume the point $(x,y,z)$ on the plane that is equidistant from the given points. To calculate the distance of the given points with the $(x,y,z)$, we will use the distance formula.

Distance Formula is given as:

$\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 +(z_2 – z_1)^2 }$

Applying this formula on points $(x,y,z)$ and $(1,0,-2)$ to calculate the distance:

$\sqrt{(x – 1)^2 + (y – 0)^2 +(z + 2)^2 }$

Expanding the expression using the algebraic formulas:

$(a-b)^2=a^2+b^2-2ab$

$(a+b)^2=a^2+b^2+2ab$

$\sqrt{(x^2 -2x +1) + y^2 +(z^2 +4z+4)}$

$\sqrt{(x^2 + y^2 + z^2 -2x +4z +5)}$

Now calculating the distance of the point $(3,4,0)$ with the $(x,y,z)$.

$\sqrt{(x – 3)^2 + (y – 4)^2 + z^2 }$

Expanding the expression using the algebraic formulas:

$\sqrt{(x^2 -6x +9) + (y^2 -8y+16) + z^2 }$

$\sqrt{(x^2 + y^2 + z^2 -6x – 8y + 25)}$

As both distances are equidistant, equating them and then simplifying:

$\sqrt{(x^2 + y^2 + z^2 -2x +4z +5)} = \sqrt{(x^2 + y^2 + z^2 -6x – 8y + 25)}$

The expression is re-written as:

$x^2 + y^2 + z^2 -2x +4z +5 = x^2 + y^2 + z^2 -6x -8y + 25$

$\cancel{x^2}+\cancel{y^2}+\cancel{z^2}-2x+4z+5 = \cancel{x^2}+\cancel{y^2}+\cancel{z^2}-6x-8y+25$

$-2x+4z+5=-6x-8y+25$

$-2x+6x +8y+4z +5-25 = 0$

$4x +8y+4z -20=0$

Dividing the equation with $4$:

$x+2y+z=5$

So the equation of the plane that consists of all the points that are equidistant from the given points is calculated to be:

$(1,0,-2)$ and $(3,4,0)$ is $x +2y+z = 5$.

## Example

What is the equation of the plane consisting of all points that are equidistant from $(-5, 5, -3)$ and $(4,5,3)$?

Calculating the distance between $(x,y,z)$ and $(-5,5,-3)$:

$\sqrt{(x + 5)^2 + (y – 5)^2 +(z + 3)^2 }$

$\sqrt{(x^2 + y^2 + z^2 +10x -10y +6z + 59)}$

Now calculating the distance between $(4,5,3)$ with $(x,y,z)$.

$\sqrt{(x – 4)^2 + (y – 5)^2 + (z-3)^2 }$

$\sqrt{(x^2 + y^2 + z^2 -8x – 10y -6z+ 50)}$

As both distances are equidistant, putting them equal to each other and simplifying:

$\sqrt{(x^2 + y^2 + z^2 +10x -10y +6z + 59)} = \sqrt{(x^2 + y^2 + z^2 -8x – 10y -6z+ 50)}$

Re-writing:

$10x + 8x -10y + 10y +6z +6z +59 -50 = 0$

$6x + 4z = -3$