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.


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