Question

Find the poin on x axis which is equidistant from (6,3) and( 3,0)

Answer 1

so the eq become
(x-6)²+(0-3)²=(3-x)²+(0-0)²
x²+36-12x+9=9+x²-6x
36-6x=0
x=6
so the coordinates become
(6,0)
hope it help u

Answer 2

Answer:

Let the two points be A ( 6 , 3 ) and B ( 3 , 0 )

and the point be C ( x' , 0 )

AC = BC

( x' - 6 )² + ( - 3 ) ² = ( x' - 3 ) ² + ( 0 )

x'² - 12x' + 36 + 9 = x'² - 6x’ + 9

6x' = 36

x' = 6

So the point is C ( 6 , 0 )