Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then

Distance of the point (x1,y1,z1) from the plane Ax + By + Cz = D is


The equation of a plane having intercepts a, b, c on the x-, y-, z- axis respectively is



Comparing with Ax + By + Cz = D, we get -


A = 1/a, B = 1/b, C = 1/c, D = 1


Given, the plane is at a distance of 'p' units from the origin.


So, The point is O(0,0,0)


x1 = 0, y1 = 0, z1 = 0


Now,


Distance


Substituting all values, we get -







squaring both sides, we get -



Hence Proved.


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