Chapter 9.3, Problem 23E

To determine

To find:The equation for the surface using distance formula.

Answer to Problem 23E

The equation for the surface using distance formula is $2y-3=0$ and it is an equation of the plane as it is a linear equation.

Explanation of Solution

Given information:

The givensets of all points $\left(x,y,z\right)$ in a coordinate space that are equidistant from the points $P\left(0,0,0,\right)$ and $Q\left(0,3,0,\right)$ .

Calculation:

The givensets of all points $\left(x,y,z\right)$ in a coordinate space that are equidistant from the points $P\left(0,0,0,\right)$ and $Q\left(0,3,0,\right)$ .

The distance between the points $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ .

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Calculate the distance between the points $R\left(x,y,z\right)$ and $P\left(0,0,0,\right)$ .

$\begin{array}{l}{d}_{PR}=\sqrt{{\left(0-x\right)}^2+{\left(0-y\right)}^2+{\left(0-z\right)}^2}\\ d_{PR}=\sqrt{x^2+y^2+z^2}\end{array}$

Calculate the distance between the points $R\left(x,y,z\right)$ and $Q\left(0,3,0,\right)$ .

$\begin{array}{l}{d}_{QR}=\sqrt{{\left(0-x\right)}^2+{\left(3-y\right)}^2+{\left(0-z\right)}^2}\\ d_{QR}=\sqrt{x^2+{\left(3-y\right)}^2+z^2}\end{array}$

The points $\left(x,y,z\right)$ in a coordinate space are equidistant from the points $P\left(0,0,0,\right)$ and $Q\left(0,3,0,\right)$ .

$\begin{array}{c}{d}_{PR}=d_{QR}\\ \sqrt{x^2+y^2+z^2}=\sqrt{x^2+{\left(3-y\right)}^2+z^2}\end{array}$

Squaring on both sides of the above equation.

  $\begin{array}{c}{x}^2+y^2+z^2=x^2+{\left(3-y\right)}^2+z^2\\ x^2+y^2+z^2=x^2+9+y^2-6y+z^2\\ 0=9-6y\\ 2y-3=0\end{array}$

The equation of the surface is $2y-3=0$ and it is an equation of the plane as it is a linear equation.

Therefore, theequation for the surface using distance formulais $2y-3=0$ and it is an equation of the plane as it is a linear equation.