Show that the points and
are equidistant from the plane
and lies on the opposite of it.
Given two points,
Also,
Where,
Therefore,
We must show that the points A and B are equidistant from the plane
5x + 2y - 7z + 9 = 0
We also need to show that the points lie on the opposite side of the plane.
Normal of the plane is,
We know, the perpendicular distance of the position vector of a point
to the plane, p: ax + by + cz + d = 0 is given as:
Where
Thus, the perpendicular distance of the point to the plane 5x + 2y - 7z + 9 = 0 having normal
is given by,
Hence, the perpendicular distance of the point to the plane 5x + 2y - 7z + 9 = 0 having normal
Therefore, |D1| = |D2|
However, D1 and D2 have different signs.
Therefore, the points A and B will lie on opposite sides of the plane.
Hence, we have successfully shown that the points are equidistant from the plane and lie on opposite sides of the plane.