Question

Find an equation of the plane consisting of all points that are equidistant from (0, -2, 0) and (4, -3, 0), and having 4 as the coefficient of x.

Answer 1

SOLUTION -  GIVEN = Two  Points  (0,-2,0) & (4, -3,0 ) which are inequidistant  

                                     from the all the points of plane.

                    Find out = Equation of plane

TO PROOF - The general equation of plane .

                      a ( x-x₀) + b ( y- y₀) + c(z-z₀) =0

                     let  A = (0,-2,0) & B = (4, -3, 0 )

$\vec{AB}\vec{AB}= (4,-3,0)- ( 0,-2, 0)

= (4-0,-3+2,0-0)

= (4,-1,0)

                 ( All the points onthe plane are equidistant from A & B Then                 [tex]\vec{AB}$  is orthogonal to plane and become the normal vector.)

               let O be midpoint $\vec{AB}$

               O = $\frac{1}{2}$( 0+4, -2-3, 0-0 )

                  =\frac{1}{2} ( 4, -5, 0)

                 =$(2, \frac{-5}{2},0)$

        Equation of the plane by using the point & normal vector

         by using general equation of a plane

           4 ( x- 2) -1 (y +$\frac{5}{2}$ = 0

$8x - 16 - 2y -5 =0

8x- 2y = 21$

        Hence Proved