Find an equation of the set of all points equidistant from the points $A ( 7 , - 8 , - 9 )$ and $B ( - 6,4 , - 12 )$.
A) $- 13 x - 12 y + 3 z = - 1$
B) $\quad 13 x - 12 y + 3 z = - 1$
C) $- 13 x - 12 y + 3 z = 1$
D) $\quad 13 x - 12 y + 3 z = 1$
E) $13 x - 12 y = 0$

Question

Quizplus · Accepted Answer

The set of points equidistant from two given points is precisely the perpendicular bisector of the line segment between those two points. 
Thus, we first find the midpoint of the line segment joining $A$ and $B$:
$$\left(\frac{7-6}{2},\frac{-8+4}{2},\frac{-9-12}{2}ight) = (0,-2,-10)$$
So the equation of the plane containing this midpoint and perpendicular to the line segment $AB$ will be the set of all points equidistant to $A$ and $B$. The direction vector of the line segment $AB$ is $\overrightarrow{AB} = \begin{pmatrix}-6-7 \ 4-(-8) \ -12-(-9)\end{pmatrix} = \begin{pmatrix}-13 \ 12 \ -3\end{pmatrix}$. 
So, we can find the normal vector to the plane by taking any vector orthogonal to $\overrightarrow{AB}$, such as $\begin{pmatrix} 3 \ 4 \ 13 \end{pmatrix}$. Thus, the equation of the plane is
$$-13x + 12y + 3z = -13(0) + 12(-2) + 3(-10) = -1$$
which simplifies to
$$\boxed{13x - 12y + 3z = -1}.$$

Find an Equation of the Set of All Points Equidistant A(7,−8,−9)A ( 7 , - 8 , - 9 )A(7,−8,−9)

Find an Equation of the Set of All Points Equidistant $A ( 7 , - 8 , - 9 )$