Question

prove that the points (1,2,3) (-1,1,4) and (0,3,3) (1,3,2) are equidistant from the point (-1,1,1)​

Answer 1

Step-by-step explanation:

We know that the distance of the point (x

1

,y

1

,z

1

) from the plane ax+by+cz+d=0 is given by

a

2

+b

2

+c

2

∣ax

1

+by

1

+cz

1

+d∣

Distance of the point (1,1,1) from the plane 3x+4y−12z+13=0

The required distance =

(3)

2

+(4)

2

+(−12)

2

∣3(1)+4(1)−12(1)+13∣

=

9+16+144

∣3+4−12+13∣

=

13

8

units ---- ( 1 )

Distance of the point (−3,0,1) from the plane 3x+4y−12z+13=0

The required distance =

(3)

2

+(4)

2

+(−12)

2

∣3(−3)+4(0)−12(1)+13∣

=

9+16+144

∣−9+0−12+13∣

=

13

8

units ---- ( 2 )

So, from ( 1 ) and ( 2 ) we can say that the given points are equidistant from the given plane.