Show that the points ( - 3, 0, 1) and (1, 1, 1) are equidistant from the plane 3x + 4y – 12z + 13 = 0.

Question

Philoid · Accepted Answer

Formula :

where () is point from which distance is to be calculated

Therefore ,

Plane = 3x + 4y - 12z + 13 = 0

First Point = ( - 3 , 0 , 1 )

Distance for first point

Plane = 3x + 4y - 12z + 13 = 0

Second Point = ( 1 , 1 , 1 )

Distance for first point

Hence proved.