Question

find a point on x axis which is equidistant from points (7,6)and(3,4)

Answer 1

so we basically need a pt which is equidistant from (7,6) and (3,4)
let it be point P(x,y) and because it lies on x-axis the coordinates of P is
(x,o)
as the distance from P is equal so, let A(7,6) and B(3,4),
PA=PB
PA^2= PB^2 [ PA.PA=PB.PB]
apply distance formula, we get
(x-7)^2 + (0-6)^2= (x-3)^2 + (4-0)^2
 x^2 + 49 – 14x + 36 = x^2 + 9 – 6x + 16
  85 – 25 = 8x
 8x= 60 X = 60/8 = 7.5
P(7.5,0)