Show that the points and are equidistant from the plane and lies on opposite side of it.
We are given with points,
Also, .
Where,
So,
⇒ 5x + 2y – 7z + 9 = 0
We need to show that the points A and B are equidistant from the plane .
Also that, the points lie on the opposite side of the plane.
Normal of the plane 5x + 2y – 7z + 9 = 0 is,
We know that,
Perpendicular distance of the position vector of a point, to the the plane, P: ax + by + cz + d = 0 is given as
Where,
So, perpendicular distance of the point to the plane 5x + 2y – 7z + 9 = 0 having normal is,
So, perpendicular distance of the point to the plane 5x + 2y – 7z + 9 = 0 having normal is,
∴, |D1| = |D2|
But, since D1 and D2 have different signs.
⇒ The points A and B lie on the opposite sides of the plane.
Thus, we have shown that the given points are equidistant from the plane and lies on the opposite side of the plane.