Question

Find a point on x axis which is equidistant from( 6,3) and (3,0)

Answer 1

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

and the point on x axis is P ( x,0 )

They are equidistant from each other

AP = BP

by applying the distance formula

Answer 2

A(6,3) (x1 y1)
B(3,0) (x2 y2)
P(x,0) (x y)
AP = BP

by distance formula,
AP = √[(x1-x)²+(y1-y)²]
= √[(6-x)²+(3-0)²]
= √[36+x²-12x+9]
= √[x²-12x+45]

BP = √[(x2-x)²+(y2-y)²]
= √[(3-x)²+(0-0)²]
= √[x²-6x+9]

as AP = BP
therefore
√[x²-12x+45] = √[x²-6x+9]
x²-12x+45 = x²-6x+9
12x-45 = 6x-9
4x-15 = 2x-3
2x = 12
x = 6

therefore the point is P(6,0)