1. ## Flat

Finding the equation of the plan. Mediator of the segment of extreme A(1,-2,6) and B(3,0,0)

$x+y-3z+8=0$

2. Hi
Originally Posted by Apprentice123
Finding the equation of the plan. Mediator of the segment of extreme A(1,-2,6) and B(3,0,0)

$x+y-3z+8=0$
To find the answer you need to know two things :
• the mediating plane is orthogonal to the line segment $[AB]$ (this boils to saying that the plane is orthogonal to $\vec{AB}$)
• the midpoint of $[AB]$ is a point of the mediating plane.

3. Hello,

Originally Posted by Apprentice123
Finding the equation of the plan. Mediator of the segment of extreme A(1,-2,6) and B(3,0,0)

$x+y-3z+8=0$
A mediator is perpendicular to the segment and passes throught the midpoint.

The coordinates of $\overrightarrow{AB}$ are $(2~,~2~,~-6)$

So the equation of the plan, which has $AB$ as orthogonal vector, will be :

$2x+2y-6z+d=0$

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With d to determine. And for this, you will have to use the fact that it goes throught the midpoint of AB.

Coordinates of the midpoint of [AB] : $\left(\frac{3+1}{2}~,~\frac{-2+0}{2}~,~\frac{6+0}2\right)$, that is to say $(2~,~-1~,~3)$.

This point is on the plane, so substitute and get d.

After that, divide the equation by 2.

4. Thank you